Algebra · Qual Corpus

Four sessions, in order: a warm-up on groups, finite group theory, Sylow theory, and rings.

Cosets of a subgroup all have the same cardinality (Tag P-SERVA)

Let $H\leq G$ be a subgroup (not necessarily normal). Prove that any two cosets $xH, yH\in G/H$ have the same cardinality.

Define a map $m_g: G\to G$ where $x\mapsto gx$, restrict to $m_h:H\surjects gH$, inverse $\inverseof{(m_g)} = m_{\inverseof{g}}$

Solution

Theorem.
Any two left cosets of a subgroup have equal cardinality.

Proof.

Fix $x,y\in G$. Define $\tau:xH\to yH$ by $\tau(xh)=yh$ for $h\in H$.
$\tau$ is well defined since multiplying $h$ by an element of $H$ gives another element of $H$.
For $u\in H$, define $\tau^{-1}(yu)=xu$; then $\tau^{-1}(yu)\in xH$ and \[ \tau(\tau^{-1}(yu))=y u,\qquad \tau^{-1}(\tau(xh))=x h. \] So $\tau^{-1}$ is an inverse.
Hence $\tau$ is a bijection and $|xH|=|yH|$.

Fundamental theorem of cosets: $xH=yH\iff x^{-1}y\in H\iff y^{-1}x\in H$ (Tag P-SNEIS)

Let $H\le G$. Prove that \[ xH=yH\iff x^{-1}y\in H\iff y^{-1}x\in H. \]

Solution

Suppose first that $xH=yH$. Since $y\in yH=xH$, there is some $h\in H$ with \[ y=xh. \] Hence \[ x^{-1}y=h\in H. \] Conversely, if $x^{-1}y=h\in H$, then $y=xh$, and therefore \[ yH=xhH=xH. \] Thus \[ xH=yH\iff x^{-1}y\in H. \]

Finally, \[ x^{-1}y\in H\iff (x^{-1}y)^{-1}=y^{-1}x\in H, \] because subgroups are closed under inverses. Hence all three conditions are equivalent.

Proper subgroups of a group of order $pq$ are cyclic (Tag P-YNF6T)

Suppose $|G| = pq$ with $p, q \geq 2$ prime, and let $H \le G$ be a proper subgroup. Prove that $H$ must be cyclic.

Solution

By Lagrange’s theorem, \[ |H|\mid pq. \] Since $H$ is proper, \[ |H|\in\{1,p,q\}. \] The trivial group is cyclic. If $|H|$ is prime, choose $h\in H$ with $h\neq e$. Then the order of $h$ divides $|H|$ and is greater than $1$, hence equals $|H|$. Therefore \[ H=\langle h\rangle. \] Thus every proper subgroup of $G$ is cyclic.

Lagrange's theorem (Tag P-WY2K6)

State and prove Lagrange’s Theorem for finite groups: if $G$ is a finite group and $H \le G$ is a subgroup, then $|H|$ divides $|G|$ and $|G| = [G : H] \cdot |H|$.

Solution

The left cosets of $H$ in $G$ partition $G$. Indeed, define \[ x\sim y\iff x^{-1}y\in H. \] This is an equivalence relation, and the equivalence class of $g\in G$ is exactly \[ gH=\{gh:h\in H\}. \] Hence \[ G=\bigsqcup_{i=1}^{[G:H]} g_iH \] for any choice of left-coset representatives $g_i$.

For each $g\in G$, left multiplication \[ H\to gH,\qquad h\mapsto gh \] is a bijection, so every left coset has cardinality $|H|$. Therefore \[ |G| =\sum_{i=1}^{[G:H]}|g_iH| =[G:H]|H|. \] In particular, $|H|$ divides $|G|$.

Cauchy's theorem (Tag P-HPTFQ)

Prove Cauchy’s theorem.

Induce on $\size G$. Assume $\size G > p$ and pick $g\neq 1$. If $p\divides \size g$, use cyclic group theory, so assume otherwise. Use that $\size G = \size G/N \size N$ so $p$ divides $\size G/N$, apply IH to get an element of order $p$ in the quotient. Then $y\not\in N$ but $y^p\in N$, so $\gens{y}\neq \gens{y^p}$ since $y^p\in N \implies \gens{y^p} \subseteq N$. Get $p\divides \size \gens{y}$, apply IH.

Solution

Cauchy’s theorem: if $G$ is a finite group and $p$ is a prime dividing $|G|$, then $G$ has an element of order $p$.

Let $X=\{(g_1,\dots,g_p)\in G^p: g_1g_2\cdots g_p=1\}$, and let $C_p=\langle\sigma\rangle$ act on $X$ by $\sigma(g_1,\dots,g_p)=(g_2,\dots,g_p,g_1)$.

<1>1. $|X|=|G|^{p-1}$, so $p\mid|X|$.

$g_1,\dots,g_{p-1}$ are arbitrary and determine $g_p=(g_1\cdots g_{p-1})^{-1}$. The cyclic shift preserves $X$, since $g_2\cdots g_pg_1=g_1^{-1}(g_1\cdots g_p)g_1=1$.

<1>2. Every $C_p$-orbit in $X$ has size $1$ or $p$, and the orbits of size $1$ are the tuples $(g,\dots,g)$ with $g^p=1$.

Orbit sizes divide $|C_p|=p$. A tuple is fixed by $\sigma$ exactly when all its entries are equal, and $(g,\dots,g)\in X$ exactly when $g^p=1$.

<1>3. Q.E.D.

Let $F$ be the set of fixed tuples. By steps <1>1 and <1>2, $|F|\equiv|X|\equiv0\pmod p$, and $(1,\dots,1)\in F$, so $|F|\ge p\ge2$. Hence some $(g,\dots,g)\in F$ has $g\neq1$ and $g^p=1$, and $g$ has order $p$.

Groups of prime order are cyclic (Tag P-V7BGS)

Show that every group of prime order is cyclic.

Solution

Let $|G|=p$ with $p$ prime, and choose any nonidentity element $g\in G$. By Lagrange’s theorem, \[ |g|\mid p. \] Since $g\ne e$, its order is not $1$, so \[ |g|=p. \] Therefore \[ |\langle g\rangle|=p=|G|, \] and hence \[ G=\langle g\rangle. \] Thus $G$ is cyclic.

The order of an element divides the order of the group (Tag P-SVM3O)

Let $G$ be a finite group and let $g \in G$. Prove that the order of $g$, denoted $\operatorname{ord}(g)$, divides the order of the group $|G|$.

Solution

The cyclic subgroup generated by $g$ has order \[ |\langle g\rangle|=\operatorname{ord}(g). \] Since $\langle g\rangle\le G$, Lagrange’s theorem gives \[ |G|=[G:\langle g\rangle]\,|\langle g\rangle|. \] Therefore \[ \boxed{\operatorname{ord}(g)\mid |G|}. \] In particular, writing $|G|=m\operatorname{ord}(g)$ gives \[ g^{|G|}=(g^{\operatorname{ord}(g)})^m=e. \]

$g^{\lvert G\rvert}=e$ for every $g\in G$ (Tag P-OBO4F)

Let $G$ be a finite group of order $n$. Prove that \[ g^n=e \] for every $g\in G$.

Solution

For any $g\in G$, the cyclic subgroup $\langle g\rangle$ has order $|g|$. By Lagrange’s theorem, \[ |g|\mid |G|=n. \] Write $n=k|g|$. Then \[ g^n=g^{k|g|}=(g^{|g|})^k=e. \] Thus every element satisfies $g^{|G|}=e$.

The centralizer $C_G(s)$ is a subgroup in which $\langle s\rangle$ is normal (Tag P-SXM4Q)

Let $s\in G$. Define $C_G(s)$, prove that it is a subgroup of $G$, and show that $\langle s\rangle\normal C_G(s)$.

Solution

The centralizer of $s$ in $G$ is \[ C_G(s)=\{g\in G:gs=sg\}. \]

It is a subgroup. The identity commutes with $s$. If $g,h\in C_G(s)$, then \[ (gh)s=g(hs)=g(sh)=(gs)h=(sg)h=s(gh), \] so $gh\in C_G(s)$. If $g\in C_G(s)$, then $gs=sg$ implies \[ g^{-1}s=sg^{-1}, \] so $g^{-1}\in C_G(s)$.

Also $s\in C_G(s)$, hence $\langle s\rangle\le C_G(s)$. For $g\in C_G(s)$ and any integer $k$, \[ gs^kg^{-1}=s^kgg^{-1}=s^k. \] Thus every element of $C_G(s)$ normalizes $\langle s\rangle$, and in fact conjugation acts trivially on it. Therefore \[ \langle s\rangle\normal C_G(s). \]

$NH$ is a subgroup when $N$ is normal (Tag P-DUUFY)

Let $H\leq G$ be a subgroup and $N\normal G$ be a normal subgroup. Show that $NH \leq G$ is a subgroup.

Solution

We use the one-step subgroup criterion.

<1>1. The set $NH$ is nonempty.

Since $e\in N\cap H$, one has $e=ee\in NH$.

<1>2. If $n_1h_1,n_2h_2\in NH$, then \[ (n_1h_1)(n_2h_2)^{-1}\in NH. \]

Compute \[ (n_1h_1)(n_2h_2)^{-1} =n_1h_1h_2^{-1}n_2^{-1}. \] Put $h=h_1h_2^{-1}\in H$. Since $N\trianglelefteq G$, \[ hn_2^{-1}h^{-1}\in N. \] Hence \[ n_1h n_2^{-1} =n_1(hn_2^{-1}h^{-1})h, \] which is a product of an element of $N$ and an element of $H$, so it lies in $NH$.

<1>3. Therefore $NH\le G$.

Apply the subgroup criterion using <1>1 and <1>2.

Image, kernel, cosets, and the first isomorphism theorem (Tag P-L7SEG)

Let $\phi:G_1\to G_2$ be a group homomorphism and let $H_2\le G_2$. Put \[ H_1=\phi^{-1}(H_2). \]

Show that $\phi(G_1)\le G_2$ and $H_1\le G_1$.
Show that $\ker\phi\trianglelefteq G_1$.
Prove \[ xH_1=yH_1 \iff \phi(x)H_2=\phi(y)H_2. \]
Show that $\phi$ is injective iff $\ker\phi=\{e\}$.
Prove the first isomorphism theorem \[ G_1/\ker\phi\cong\operatorname{im}\phi. \]

Solution

<1>1. Images and preimages are subgroups.

If $a=\phi(x)$ and $b=\phi(y)$ lie in $\phi(G_1)$, then \[ ab^{-1}=\phi(xy^{-1})\in\phi(G_1), \] so the image is a subgroup.

If $x,y\in H_1$, then $\phi(x),\phi(y)\in H_2$, hence \[ \phi(xy^{-1})=\phi(x)\phi(y)^{-1}\in H_2. \] Thus $xy^{-1}\in H_1$.

<1>2. The kernel is normal.

If $k\in\ker\phi$ and $g\in G_1$, then \[ \phi(gkg^{-1})=\phi(g)e\phi(g)^{-1}=e, \] so $gkg^{-1}\in\ker\phi$.

<1>3. The coset criterion holds.

We have \[ xH_1=yH_1 \iff y^{-1}x\in H_1 \iff \phi(y)^{-1}\phi(x)\in H_2 \iff \phi(x)H_2=\phi(y)H_2. \]

<1>4. Injectivity is equivalent to trivial kernel.

If $\phi$ is injective, then $\phi(g)=e$ implies $g=e$. Conversely, if $\ker\phi=\{e\}$ and $\phi(x)=\phi(y)$, then \[ \phi(y^{-1}x)=e, \] so $y^{-1}x=e$ and $x=y$.

<1>5. First isomorphism theorem.

Define \[ \bar\phi:G_1/\ker\phi\to\operatorname{im}\phi, \qquad \bar\phi(g\ker\phi)=\phi(g). \] This is well defined because two cosets are equal exactly when their quotient lies in the kernel. It is a surjective homomorphism by construction, and its kernel is trivial by <1>4. Hence it is an isomorphism.

Cycle type, order, and sign of $(4\,2\,1)(6\,1\,3\,2)$ (Tag P-DUVZD)

Let $\sigma = (4\, 2\, 1)(6\, 1\, 3\, 2) \in S_6$ in cycle notation.
- Write $\sigma$ as a product of disjoint cycles.
- Compute the order of $\sigma$. What is the general theorem about the order of cycles?
- Determine if $\sigma$ is even or odd. What is the general theorem?

Solution

Permutations are composed from right to left.

<1>1. $\sigma=(1\,3)(2\,6\,4)$.

Applying $(6\,1\,3\,2)$ and then $(4\,2\,1)$: $1\mapsto3\mapsto3$, $3\mapsto2\mapsto1$, $2\mapsto6\mapsto6$, $6\mapsto1\mapsto4$, $4\mapsto4\mapsto2$, $5\mapsto5$. So $\sigma$ swaps $1$ and $3$, cycles $2\mapsto6\mapsto4\mapsto2$, and fixes $5$.

<1>2. If $\pi=c_1\cdots c_k$ is a product of disjoint cycles of lengths $\ell_1,\dots,\ell_k$, then $\operatorname{ord}(\pi)=\operatorname{lcm}(\ell_1,\dots,\ell_k)$; hence $\operatorname{ord}(\sigma)=\operatorname{lcm}(2,3)=\boxed{6}$.

Disjoint cycles commute, so $\pi^N=c_1^N\cdots c_k^N$, and the $c_i^N$ have disjoint supports. Thus $\pi^N=1$ exactly when every $c_i^N=1$, that is, when $\ell_i\mid N$ for every $i$.

<1>3. With $\pi$ as in step <1>2, $\operatorname{sgn}(\pi)=(-1)^{\sum_i(\ell_i-1)}$; hence $\operatorname{sgn}(\sigma)=(-1)^{1+2}=-1$ and $\sigma$ is odd.

A $k$-cycle is the product $(a_1\,a_k)(a_1\,a_{k-1})\cdots(a_1\,a_2)$ of $k-1$ transpositions, so it has sign $(-1)^{k-1}$, and $\operatorname{sgn}$ is a homomorphism.

Homomorphisms from $S_n$ to a group of odd order (Tag P-VGFA7)

Let $n \ge 2$, and let $G$ be a finite group of odd order $|G| = m$. Let $\phi: S_n \to G$ be a group homomorphism. (1) Prove that if $\tau \in S_n$ is a transposition, then $\tau \in \ker\phi$. (2) Prove that $\phi$ is the trivial homomorphism ($\ker\phi = S_n$). (3) Does this result depend on whether $n$ is even or odd?

Solution

Let $\tau\in S_n$ be a transposition. Since $\tau^2=1$, \[ \phi(\tau)^2=1. \] Thus the order of $\phi(\tau)$ divides $2$. It also divides $|G|$, which is odd. Hence \[ \operatorname{ord}(\phi(\tau))=1, \] so $\phi(\tau)=1$ and every transposition lies in $\ker\phi$.

Transpositions generate $S_n$. Therefore \[ \ker\phi=S_n, \] and $\phi$ is trivial.

The parity of $n$ plays no role; the argument works for every $n\ge2$.

Centers of $\GL_n(\FF_p)$ and $\SL_n(\FF_p)$ are scalar matrices (Tag P-4VJTF)

Let $\FF_p$ be the finite field with $p$ elements, where $p$ is a prime. Show that the centers of $\GL_n(\FF_p)$ and $\SL_n(\FF_p)$ consist only of scalar matrices.
- Show that the scalars $\zeta$ that appear in scalar matrices $Z(\SL_n(\FF_p))$ are roots of unity in $\FF_p$, i.e. $\zeta^p = 1$.

Solution

For $i\neq j$ let $e_{ij}$ be the matrix unit and $T_{ij}=I_n+e_{ij}$, a transvection with $\det T_{ij}=1$, so $T_{ij}\in\SL_n(\FF_p)\subseteq\GL_n(\FF_p)$.

<1>1. A matrix $A=(a_{k\ell})$ commuting with every $T_{ij}$ is scalar.

$AT_{ij}=T_{ij}A$ gives $Ae_{ij}=e_{ij}A$. The $(k,\ell)$ entries are $a_{ki}\delta_{j\ell}$ and $\delta_{ki}a_{j\ell}$. Taking $\ell=j$ and $k\neq i$ gives $a_{ki}=0$; taking $k=i$, $\ell=j$ gives $a_{ii}=a_{jj}$. As $i\neq j$ vary, $A$ is diagonal with equal diagonal entries.

<1>2. $Z(\GL_n(\FF_p))=\{\lambda I_n:\lambda\in\FF_p^\times\}$.

A central element commutes with every $T_{ij}$, so it is scalar by step <1>1, and every invertible scalar matrix is central.

<1>3. $Z(\SL_n(\FF_p))=\{\zeta I_n:\zeta\in\FF_p,\ \zeta^n=1\}\cong\ZZ/\gcd(n,p-1)\ZZ$.

A central element of $\SL_n(\FF_p)$ commutes with every $T_{ij}\in\SL_n(\FF_p)$, so it is scalar by step <1>1; the scalar matrix $\zeta I_n$ lies in $\SL_n(\FF_p)$ exactly when $\det(\zeta I_n)=\zeta^n=1$, and it is then central. The solutions of $\zeta^n=1$ in the cyclic group $\FF_p^\times$ of order $p-1$ form its subgroup of order $\gcd(n,p-1)$.

The scalars in $Z(\SL_n(\FF_p))$ are the $n$th roots of unity in $\FF_p$, $\zeta^n=1$. The condition $\zeta^p=1$ in the statement holds only for $\zeta=1$, since $\zeta^p=\zeta$ for every $\zeta\in\FF_p$.

Orders of $\GL_n(\FF_p)$ and $\SL_n(\FF_p)$ (Tag P-PSAET)

Determine the orders $|\operatorname{GL}_n(\mathbb{F}_p)|$ and $|\operatorname{SL}_n(\mathbb{F}_p)|$ for any prime $p$ and integer $n \ge 1$.

Solution

An invertible $n\times n$ matrix over $\mathbb F_p$ is exactly an ordered basis of $\mathbb F_p^n$ written as columns. The first column has $p^n-1$ choices, the second has $p^n-p$ choices, and in general the $(k+1)$-st column has $p^n-p^k$ choices. Hence \[ \boxed{|\operatorname{GL}_n(\mathbb F_p)|=\prod_{k=0}^{n-1}(p^n-p^k).} \] Equivalently, \[ |\operatorname{GL}_n(\mathbb F_p)| =p^{n(n-1)/2}\prod_{j=1}^n(p^j-1). \]

The determinant map \[ \det:\operatorname{GL}_n(\mathbb F_p)\to\mathbb F_p^\times \] is surjective, because \[ \det\operatorname{diag}(a,1,\dots,1)=a \] for every $a\in\mathbb F_p^\times$. Its kernel is $\operatorname{SL}_n(\mathbb F_p)$. Therefore \[ \boxed{|\operatorname{SL}_n(\mathbb F_p)|= \frac1{p-1}\prod_{k=0}^{n-1}(p^n-p^k).} \]

$G/Z(G)$ cyclic implies $G$ abelian (Tag P-6HPKO)

Prove that if $G/Z(G)$ is cyclic, then $G$ is abelian.

Solution

Write $Z=Z(G)$ and suppose \[ G/Z=\langle gZ\rangle. \]

<1>1. Every element of $G$ can be written as $g^m z$ for some $m\in\ZZ$ and $z\in Z$.

If $x\in G$, then its coset lies in the cyclic quotient, so \[ xZ=(gZ)^m=g^mZ \] for some $m$. Hence \[ g^{-m}x\in Z, \] so $x=g^m z$ for some $z\in Z$.

<1>2. Any two elements of $G$ commute.

Take \[ x=g^m z_1, \qquad y=g^n z_2 \] with $z_1,z_2\in Z$ by <1>1. Since the $z_i$ are central, \[ xy=g^m z_1g^n z_2=g^{m+n}z_1z_2 =g^{m+n}z_2z_1=g^n z_2g^m z_1=yx. \]

<1>3. Therefore $G$ is abelian.

By <1>2, every pair of elements of $G$ commutes.

Groups of order $p^2$ (Tag P-YCLOT)

(Important) Classify all groups of order $p^2$.

Must be abelian since quotient is cyclic. If there’s an element of order $p^2$, cyclic, done. Else every element $a\neq 1$ must have order $p$. Then $\gens{a}\neq G$, so pick $b$ in its complement, it has order $p$. Call these two subgroups $H, K$ Recognize direct products: abelian implies both are normal, $H \intersect K = \ts{1}$. and $\size HK = \size H \size K / \size(H \intersect K) = p\cdot p/1 = p^2$

Solution

Goal: Prove that every group of order $p^2$ (for $p$ prime) is abelian and isomorphic to either $C_{p^2}$ or $C_p \times C_p$.

<1>1. Every group $G$ of order $p^2$ is abelian.

<2>1. By the class equation for the action of $G$ on itself by conjugation, \[|G| = |Z(G)| + \sum_{i=1}^k [G : C_G(x_i)],\] where the sum runs over representatives of non-central conjugacy classes. <2>2. For each non-central $x_i$, the centralizer $C_G(x_i)$ is a proper subgroup containing $Z(G)$ and $x_i$, so $[G : C_G(x_i)] > 1$. <2>3. Since $|G| = p^2$, each index $[G : C_G(x_i)]$ is a power of $p$ dividing $|G|$, hence $p \mid [G : C_G(x_i)]$. <2>4. Since $p \mid |G|$ and $p$ divides each index term, $p$ divides $|Z(G)|$. Since $e \in Z(G)$, $|Z(G)| \ge p > 0$, so $Z(G)$ is nontrivial. <2>5. By Lagrange’s theorem, $|Z(G)| \in \{p, p^2\}$. <2>6. If $|Z(G)| = p$, then the quotient group $G/Z(G)$ has order $p^2/p = p$, so $G/Z(G)$ is cyclic. <2>7. Lemma: If $G/Z(G)$ is cyclic, then $G$ is abelian. Proof of Lemma: Let $G/Z(G) = \langle g Z(G) \rangle$. Any elements $x, y \in G$ can be written as $x = g^m z_1$ and $y = g^n z_2$ for some $m, n \in \mathbb{Z}$ and $z_1, z_2 \in Z(G)$. Then \[xy = (g^m z_1)(g^n z_2) = g^m g^n z_1 z_2 = g^{m+n} z_1 z_2 = g^n g^m z_2 z_1 = (g^n z_2)(g^m z_1) = yx.\] Thus $G$ is abelian, so $Z(G) = G$, contradicting $|Z(G)| = p$. <2>8. Therefore $|Z(G)| = p^2$, which means $Z(G) = G$, so $G$ is abelian.

<1>2. Classification of abelian groups of order $p^2$.

<2>1. By Lagrange’s theorem, the order of every element of $G$ divides $p^2$, so $o(g) \in \{1, p, p^2\}$ for all $g \in G$. <2>2. Case 1: $G$ contains an element $a$ of order $p^2$. Then the cyclic subgroup $\langle a \rangle$ has order $p^2 = |G|$, so $G = \langle a \rangle \cong C_{p^2}$. <2>3. Case 2: Every non-identity element of $G$ has order $p$. Pick any $a \in G$ with $a \neq e$, and set $H = \langle a \rangle \cong C_p$. Since $|H| = p < p^2$, choose an element $b \in G \setminus H$. The subgroup $K = \langle b \rangle$ has order $p$. <2>4. The intersection $H \cap K$ is a subgroup of $H$. Since $|H| = p$ is prime, $|H \cap K| \in \{1, p\}$. Because $b \in K \setminus H$, $H \cap K \neq K$, so $H \cap K = \{e\}$. <2>5. Since $G$ is abelian, every subgroup is normal, so $H \trianglelefteq G$ and $K \trianglelefteq G$. <2>6. By the product formula, \[|HK| = \frac{|H| \cdot |K|}{|H \cap K|} = \frac{p \cdot p}{1} = p^2 = |G|,\] so $G = HK$. <2>7. The map $\psi: H \times K \to G$ defined by $\psi(h, k) = hk$ is a homomorphism (since $G$ is abelian) with trivial kernel (since $H \cap K = \{e\}$) and full image $HK = G$. <2>8. Thus $\psi$ is an isomorphism, so $G \cong H \times K \cong C_p \times C_p$.

<1>3. Conclusion:

Up to isomorphism, the only groups of order $p^2$ are $C_{p^2}$ and $C_p \times C_p$. Both are abelian.

A subgroup of index $2$ is normal; the same for index the smallest prime dividing $|G|$ (Tag P-QJ7MD)

Prove that every subgroup of index $2$ is normal.
More generally, let $G$ be finite and let $p$ be the smallest prime dividing $|G|$. Prove that every subgroup of index $p$ is normal.

Solution

<1>1. Index $2$.

If $H\le G$ has index $2$, then there are exactly two left cosets and two right cosets. For $g\notin H$, both $gH$ and $Hg$ are the complement $G\setminus H$. Hence \[ gH=Hg \] for every $g\in G$, so $H\trianglelefteq G$.

<1>2. Index equal to the smallest prime divisor.

Every prime divisor of $|G|$ is at least $p$, while every prime divisor of $p!$ is at most $p$. Therefore the only possible prime divisor of $|\rho(G)|$ is $p$. Since $p^2\nmid p!$, we get \[ |\rho(G)|=p. \] A transitive group of prime order acts regularly, so the stabilizer of the coset $H$ in $\rho(G)$ is trivial. But the stabilizer upstairs is $H$, hence \[ H=\ker\rho. \] Therefore $H$ is normal in $G$.

A finite group is not the union of conjugates of a proper subgroup; $S_n$ has a derangement (Tag P-6XW74)

Prove that if $G$ is a finite group and $H<G$ is a proper subgroup, then $G$ cannot be written as a union of conjugates of $H$.
Use this to prove that if $G=\Sym(X)$ is the group of permutations of a finite set $X$ with $|X|=n>1$, then there exists $g\in G$ with no fixed point in $X$.

Solution

<1>1. Let $G$ act transitively on the coset space $X=G/H$ by left multiplication. Then an element $g\in G$ fixes some coset if and only if it lies in a conjugate of $H$.

A coset $aH$ is fixed by $g$ exactly when \[ gaH=aH, \] which is equivalent to \[ a^{-1}ga\in H. \] Equivalently, \[ g\in aHa^{-1}. \] Thus the elements lying in at least one conjugate of $H$ are exactly the elements having at least one fixed point in the coset action.

<1>2. Every finite transitive action on a set with more than one point has a fixed-point-free element.

Let a finite group $G$ act transitively on a finite set $X$ with $|X|>1$. For $g\in G$, write $X^g$ for the fixed-point set. Burnside’s orbit-counting lemma gives \[ 1=|X/G|=\frac1{|G|}\sum_{g\in G}|X^g|, \] so \[ \sum_{g\in G}|X^g|=|G|. \] The identity contributes $|X|>1$. If every nonidentity element fixed at least one point, then \[ \sum_{g\in G}|X^g| \ge |X|+(|G|-1)>|G|, \] a contradiction. Hence some element has no fixed points.

<1>3. A finite group is not the union of the conjugates of a proper subgroup.

If $H<G$, then the coset space $G/H$ has more than one point and the action is transitive. By <1>2, some $g\in G$ fixes no coset. By <1>1, this $g$ belongs to no conjugate of $H$. Therefore \[ \bigcup_{a\in G}aHa^{-1}\ne G. \]

<1>4. For every finite set $X$ with $|X|=n>1$, the symmetric group $\Sym(X)$ contains a derangement.

Fix $x\in X$ and let \[ H=\operatorname{Stab}_{\Sym(X)}(x). \] This is a proper subgroup. Since $\Sym(X)$ acts transitively on $X$, the conjugates of $H$ are exactly the point stabilizers \[ \operatorname{Stab}_{\Sym(X)}(y),\qquad y\in X. \] By <1>3, choose \[ g\notin\bigcup_{y\in X}\operatorname{Stab}(y). \] Then $g$ fixes no point of $X$, so $g$ is a derangement.

A subgroup of a finite $p$-group generating $H/[H,H]$ equals $H$ (Tag P-KXSW3)

Let $H$ be a finite $p$-group and let $G\le H$. Suppose the composite \[ G\hookrightarrow H\twoheadrightarrow H/[H,H] \] is surjective. Prove that $G=H$.

Solution

Surjectivity is equivalent to \[ H=G[H,H]. \] Suppose for contradiction that $G<H$. Since $H$ is finite, $G$ is contained in a maximal subgroup $M<H$.

<1>1. Every maximal subgroup of a finite $p$-group has index $p$ and is normal.

The quotient action of $H$ on the left cosets $H/M$ shows that the index is a power of $p$. Maximality forces the quotient to have no nontrivial proper subgroup, hence \[ [H:M]=p. \] A subgroup of index $p$ in a finite $p$-group is normal.

<1>2. We have $[H,H]\subseteq M$.

Because $H/M$ has order $p$, it is cyclic and therefore abelian. The derived subgroup is contained in the kernel of every homomorphism to an abelian group, so \[ [H,H]\subseteq M. \]

Now $G\subseteq M$ and $[H,H]\subseteq M$, so \[ G[H,H]\subseteq M<H, \] contradicting the surjectivity identity $G[H,H]=H$. Therefore $G=H$.

Pairs $(n,p)$ for which $\SL_n(\FF_p)$ is solvable (Tag P-KPEPM)

Let $p$ be prime and $n\ge1$. Determine all pairs $(n,p)$ for which \[ \SL_n(\FF_p) \] is solvable.

Solution

The solvable cases are \[ n=1\quad\text{for every prime }p, \] and \[ (n,p)=(2,2),(2,3). \]

<1>1. If $n=1$, then $\SL_1(\FF_p)=1$, so it is solvable.

<1>2. The groups $\SL_2(\FF_2)$ and $\SL_2(\FF_3)$ are solvable.

We have \[ \SL_2(\FF_2)\cong S_3, \] which is solvable.

For $p=3$, the center of $\SL_2(\FF_3)$ is $\{\pm I\}$ and \[ \PSL_2(\FF_3)\cong A_4. \] Thus $\SL_2(\FF_3)$ is an extension of the solvable group $A_4$ by the abelian group $C_2$, hence is solvable.

<1>3. All other cases with $n\ge2$ are nonsolvable.

The classical simplicity theorem for projective special linear groups says that \[ \PSL_n(\FF_p) \] is nonabelian simple for $n\ge2$, except for \[ (n,p)=(2,2),(2,3). \] Since $\PSL_n(\FF_p)$ is a quotient of $\SL_n(\FF_p)$, solvability of $\SL_n(\FF_p)$ would force solvability of $\PSL_n(\FF_p)$. A nonabelian simple group is not solvable, so no other pair occurs.

$HK$ is a subgroup when $H\le N_G(K)$; a counterexample when $K$ is not a subgroup (Tag P-O2H7G)

Let $H,K\le G$ and suppose $H\le N_G(K)$. Prove that $HK$ is a subgroup of $G$. Give a counterexample if $K$ is only a subset, not a subgroup.

Solution

<1>1. If $H\le N_G(K)$, then $HK\le G$.

The identity belongs to $HK$. Let $h_1k_1,h_2k_2\in HK$. Since $h_2\in N_G(K)$, \[ k_1h_2=h_2(h_2^{-1}k_1h_2) \] with $h_2^{-1}k_1h_2\in K$. Hence \[ (h_1k_1)(h_2k_2)=h_1h_2\,(h_2^{-1}k_1h_2)k_2\in HK. \] Also, for $hk\in HK$, \[ (hk)^{-1}=k^{-1}h^{-1}=h^{-1}(hk^{-1}h^{-1})\in HK \] because $h^{-1}\in N_G(K)$. Therefore $HK$ is a subgroup.

<1>2. The subgroup hypothesis on $K$ is necessary.

Take the additive group $G=\ZZ$, let $H=2\ZZ$, and let $K=\{0,1\}$. Since $G$ is abelian, every element normalizes every subset in the setwise sense, but \[ H+K=2\ZZ\cup(2\ZZ+1)=\ZZ. \] This particular choice still gives a subgroup, so choose instead $H=\{0\}$ and $K=\{0,1\}$. Then $H+K=\{0,1\}$ is not a subgroup of $\ZZ$ because $1+1=2\notin\{0,1\}$.

Recognizing direct products (Tag P-P4KA6)

Let $H,K\trianglelefteq G$ satisfy \[ H\cap K=\{e\},\qquad HK=G. \] Prove that $G\cong H\times K$. Generalize the criterion to finitely many normal subgroups.

Solution

<1>1. The subgroups $H$ and $K$ commute elementwise.

For $h\in H$ and $k\in K$, the commutator \[ [h,k]=hkh^{-1}k^{-1} \] lies in $H$ because $H$ is normal, and lies in $K$ because $K$ is normal. Hence \[ [h,k]\in H\cap K=\{e\}. \] Thus $hk=kh$.

<1>2. Multiplication gives an isomorphism.

Define \[ \mu:H\times K\to G,\qquad (h,k)\mapsto hk. \] By <1>1 it is a homomorphism. It is surjective because $HK=G$. If $hk=e$, then $h=k^{-1}\in H\cap K$, so $h=k=e$. Thus $\mu$ is injective.

<1>3. Finite-family version.

Let $H_1,\dots,H_r\trianglelefteq G$. If \[ G=H_1\cdots H_r \] and for every $i$, \[ H_i\cap \prod_{j\ne i}H_j=\{e\}, \] then the subgroups commute pairwise and the multiplication map \[ H_1\times\cdots\times H_r\longrightarrow G \] is an isomorphism.

Equivalently, the internal direct-product condition is precisely that the multiplication map be surjective with trivial kernel. Pairwise trivial intersections alone are not sufficient for three or more factors.

Group action, orbit, stabilizer, and fixed points (Tag P-WC3HY)

State definitions of the following:
- Group action
- Orbit
- Stabilizer
- Fixed points

Solution

Let $G$ be a group acting on a set $X$.

A group action is a map \[ G\times X\to X,\qquad (g,x)\mapsto g\cdot x, \] such that \[ e\cdot x=x, \qquad (gh)\cdot x=g\cdot(h\cdot x). \]

The orbit of $x\in X$ is \[ Gx=\{g\cdot x:g\in G\}. \]

The stabilizer of $x$ is \[ G_x=\{g\in G:g\cdot x=x\}. \]

For $g\in G$, its fixed-point set is \[ X^g=\{x\in X:g\cdot x=x\}. \] The fixed points of the whole action are \[ X^G=\{x\in X:g\cdot x=x\text{ for every }g\in G\}. \]

Orbit-stabilizer theorem (Tag P-5YUW6)

State the orbit-stabilizer theorem

Solution

Let a group $G$ act on a set $X$, and let $x\in X$. Write \[ G_x=\{g\in G:g\cdot x=x\} \] for the stabilizer and $G\cdot x$ for the orbit.

<1>1. The map \[ \Phi:G/G_x\longrightarrow G\cdot x, \qquad gG_x\longmapsto g\cdot x \] is a bijection.

It is well-defined: if $gG_x=hG_x$, then $h^{-1}g\in G_x$, so \[ g\cdot x=h(h^{-1}g)\cdot x=h\cdot x. \] It is surjective by the definition of the orbit. If $g\cdot x=h\cdot x$, then \[ h^{-1}g\cdot x=x, \] so $h^{-1}g\in G_x$ and therefore $gG_x=hG_x$. Hence it is injective.

<1>2. Therefore \[ |G\cdot x|=[G:G_x] \] whenever these cardinalities are finite.

By <1>1, the orbit is in bijection with the left coset space $G/G_x$.

<1>3. If $G$ is finite, then \[ |G|=|G\cdot x|\,|G_x|. \]

By Lagrange’s theorem, \[ |G|=[G:G_x]|G_x|. \] Substitute <1>2. This is the orbit–stabilizer theorem.

The class equation (Tag P-JZM4U)

State the class equation and derive it from orbit-stabilizer.

Solution

Let a finite group $G$ act on itself by conjugation, \[ g\cdot x=gxg^{-1}. \] The orbit of $x$ is its conjugacy class, and its stabilizer is its centralizer \[ C_G(x)=\{g\in G:gx=xg\}. \] By orbit-stabilizer, \[ |\operatorname{Cl}_G(x)|=[G:C_G(x)]. \]

The conjugacy classes partition $G$. The classes of size $1$ are exactly the central elements, because \[ |\operatorname{Cl}_G(x)|=1 \iff C_G(x)=G \iff x\in Z(G). \] Choosing one representative $x_i$ from each noncentral conjugacy class gives the class equation \[ |G|=|Z(G)|+\sum_i [G:C_G(x_i)]. \]

The center of a $p$-group is nontrivial (Tag P-3PD4W)

Prove that the centre of a group of order $p^r$ ($p$ prime) is not trivial.

Solution

Let $|G|=p^r$. Under conjugation, the class equation is \[ |G|=|Z(G)|+\sum_i [G:C_G(x_i)], \] where the $x_i$ represent the noncentral conjugacy classes.

<1>1. Every noncentral conjugacy-class size is divisible by $p$.

For noncentral $x_i$, the centralizer $C_G(x_i)$ is a proper subgroup of the $p$-group $G$. Hence \[ [G:C_G(x_i)]>1 \] and, being a divisor of $|G|=p^r$, it is a positive power of $p$. Therefore it is divisible by $p$.

<1>2. The order $|Z(G)|$ is divisible by $p$.

Reducing the class equation modulo $p$ and using <1>1 gives \[ 0\equiv |G|\equiv |Z(G)|\pmod p. \] Thus $p\mid |Z(G)|$.

<1>3. Therefore $Z(G)$ is nontrivial.

Since $p\mid |Z(G)|$, one has $|Z(G)|\ge p>1$. Hence $Z(G)\ne\{e\}$.

Classification of abelian groups of a given composite order (Tag P-2TASE)

Important: Pick your favorite composite number $m = \prod p_i^{e_i}$ and classify all abelian groups of that order.
- Write their invariant factor decompositions and their elementary divisor decompositions. Come up with an algorithm for converting back and forth between these.

Solution

Take \[ m=72=2^3\cdot3^2. \] By the classification theorem for finite abelian groups, the $2$-primary part is determined by a partition of $3$, and the $3$-primary part by a partition of $2$.

<1>1. The possible primary components are \[ \begin{aligned} G_{(2)}&\cong C_8,\quad C_4\oplus C_2,\quad C_2\oplus C_2\oplus C_2,\\ G_{(3)}&\cong C_9,\quad C_3\oplus C_3. \end{aligned} \]

The partitions of $3$ are \[ 3,\qquad2+1,\qquad1+1+1, \] which give the three $2$-primary groups. The partitions of $2$ are \[ 2,\qquad1+1, \] which give the two $3$-primary groups.

<1>2. Hence there are exactly six abelian groups of order $72$.

The primary decomposition is unique, so each choice of a $2$-primary type and a $3$-primary type gives one isomorphism class, and every class occurs this way. Thus there are \[ 3\cdot2=6 \] classes.

<1>3. Their elementary-divisor and invariant-factor forms are as follows: \[ \begin{array}{c|c} \text{elementary divisors}&\text{invariant factors}\\ \hline C_8\oplus C_9&C_{72}\\ C_8\oplus C_3\oplus C_3&C_3\oplus C_{24}\\ C_4\oplus C_2\oplus C_9&C_2\oplus C_{36}\\ C_4\oplus C_2\oplus C_3\oplus C_3&C_6\oplus C_{12}\\ C_2\oplus C_2\oplus C_2\oplus C_9&C_2\oplus C_2\oplus C_{18}\\ C_2\oplus C_2\oplus C_2\oplus C_3\oplus C_3&C_2\oplus C_6\oplus C_6 \end{array} \]

For relatively prime integers $a,b$, one has \[ C_a\oplus C_b\cong C_{ab}. \] To obtain invariant factors, align the prime-power cyclic factors in nondecreasing order, pad the shorter lists on the left by $1$’s, and multiply columnwise. For example, \[ (C_4,C_2)=(C_2,C_4), \qquad (C_9)=(C_1,C_9), \] so \[ (C_2\oplus C_4)\oplus C_9 \cong C_{2\cdot1}\oplus C_{4\cdot9} =C_2\oplus C_{36}. \] The other rows follow identically. In every invariant-factor row, each factor divides the next.

<1>4. The conversion algorithm from elementary divisors to invariant factors is columnwise multiplication of aligned prime powers.

For each prime $p$, write the $p$-primary cyclic factors as \[ p^{a_{p,1}}\mid p^{a_{p,2}}\mid\cdots\mid p^{a_{p,r_p}}. \] Let $r=\max_p r_p$ and pad each list on the left by exponent $0$. For each column $j$, define \[ d_j=\prod_p p^{a_{p,j}}. \] Then \[ d_1\mid d_2\mid\cdots\mid d_r, \] and the direct sum of the $C_{d_j}$ is the invariant-factor decomposition.

<1>5. The reverse algorithm factors each invariant factor into its prime-power parts.

Given \[ C_{d_1}\oplus\cdots\oplus C_{d_r}, \qquad d_1\mid\cdots\mid d_r, \] factor each \[ d_j=\prod_p p^{v_p(d_j)}. \] Using the Chinese remainder theorem, \[ C_{d_j}\cong\bigoplus_{p\mid d_j}C_{p^{v_p(d_j)}}. \] Collecting these prime-power cyclic summands gives the elementary-divisor decomposition. The two algorithms are inverse because they merely regroup the same primary cyclic factors.

A finite group is not the union of conjugates of a proper subgroup; $S_n$ has a derangement (Tag P-6XW74)

Prove that if $G$ is a finite group and $H<G$ is a proper subgroup, then $G$ cannot be written as a union of conjugates of $H$.
Use this to prove that if $G=\Sym(X)$ is the group of permutations of a finite set $X$ with $|X|=n>1$, then there exists $g\in G$ with no fixed point in $X$.

Solution

<1>1. Let $G$ act transitively on the coset space $X=G/H$ by left multiplication. Then an element $g\in G$ fixes some coset if and only if it lies in a conjugate of $H$.

<1>2. Every finite transitive action on a set with more than one point has a fixed-point-free element.

<1>3. A finite group is not the union of the conjugates of a proper subgroup.

<1>4. For every finite set $X$ with $|X|=n>1$, the symmetric group $\Sym(X)$ contains a derangement.

Composition series, solvability, and nilpotence (Tag P-HYQCG)

Define what a composition series is, and state what it means for a group to be simple, solvable, or nilpotent.
- How are the derived and lower/upper central series defined? What type(s) of the groups above does each series correspond to?

Solution

A nontrivial group is simple if its only normal subgroups are $1$ and itself.

A composition series of a finite group $G$ is a subnormal series \[G=G_0\triangleright G_1\triangleright\cdots\triangleright G_n=1\] in which each factor $G_i/G_{i+1}$ is simple; equivalently, each $G_{i+1}$ is a maximal proper normal subgroup of $G_i$. By the Jordan–Hölder theorem, the multiset of composition factors, up to isomorphism, does not depend on the composition series.

The derived series is $G^{(0)}=G$, $G^{(i+1)}=[G^{(i)},G^{(i)}]$, and $G$ is solvable if $G^{(n)}=1$ for some $n$. Each $G^{(i)}/G^{(i+1)}$ is abelian; conversely, if $G=H_0\trianglerighteq H_1\trianglerighteq\cdots\trianglerighteq H_r=1$ has abelian factors, induction gives $G^{(i)}\le H_i$, so $G^{(r)}=1$. Thus $G$ is solvable exactly when it has a finite subnormal series with abelian factors.

The lower central series is $\gamma_1(G)=G$, $\gamma_{i+1}(G)=[\gamma_i(G),G]$, and the upper central series is $Z_0(G)=1$, $Z_{i+1}(G)/Z_i(G)=Z(G/Z_i(G))$. The group $G$ is nilpotent if $\gamma_{c+1}(G)=1$ for some $c$; equivalently, $Z_c(G)=G$ for some $c$, and the least such $c$ is the same for both series, the nilpotency class.

Thus composition series record the simple factors, the derived series detects solvability, and the lower and upper central series detect nilpotence. Every nilpotent group is solvable, since $G^{(i)}\le\gamma_{i+1}(G)$; the converse fails for $S_3$, which is solvable with trivial center.

Orbits, stabilizers, and kernels of the standard $G$-actions (Tag P-N2TM2)

For each standard action below, identify the orbits, stabilizers, global fixed points, kernel, and image in the relevant symmetric group:

left translation of $G$ on itself;
conjugation of $G$ on itself;
conjugation of $G$ on its set of subgroups;
left translation of $G$ on $G/H$ for a fixed subgroup $H\le G$.

Solution

<1>1. Left translation on $G$.

The action is transitive, so there is one orbit, namely $G$. The stabilizer of every $x\in G$ is trivial because \[ gx=x\iff g=e. \] Hence the kernel is trivial and the image is the left regular copy of $G$ in $\Sym(G)$. If $G\ne1$, there are no global fixed points.

<1>2. Conjugation on $G$.

The orbit of $x$ is its conjugacy class \[ \operatorname{Cl}_G(x). \] Its stabilizer is the centralizer \[ C_G(x). \] The global fixed points are exactly $Z(G)$. The kernel consists of elements conjugating every $x$ trivially, hence is also $Z(G)$. Therefore the image is \[ \Inn(G)\cong G/Z(G). \]

<1>3. Conjugation on the set of subgroups.

The orbit of $H$ is the conjugacy class of the subgroup, \[ \{gHg^{-1}:g\in G\}. \] The stabilizer is its normalizer \[ N_G(H). \] The global fixed points are precisely the normal subgroups of $G$. The kernel is \[ \bigcap_{H\le G}N_G(H), \] the subgroup of elements normalizing every subgroup of $G$. Thus the image is the quotient of $G$ by this kernel.

<1>4. Left translation on $G/H$.

The action is transitive. The stabilizer of the coset $xH$ is \[ xHx^{-1}. \] The kernel is the core of $H$, \[ \operatorname{core}_G(H) =\bigcap_{x\in G}xHx^{-1}, \] the largest normal subgroup of $G$ contained in $H$. Hence the image is \[ G/\operatorname{core}_G(H). \] If $H<G$, there is no global fixed coset; if $H=G$, the unique coset is fixed.

Stabilizer subgroups of two points in a $G$-set are conjugate (Tag P-YMVNK)

Let $X$ be a $G$-set with group action $G \times X \to X$. (1) Prove that if $x_1, x_2 \in X$ belong to the same $G$-orbit ($x_2 = g \cdot x_1$ for some $g \in G$), then their stabilizer subgroups $\operatorname{Stab}_G(x_1)$ and $\operatorname{Stab}_G(x_2)$ are conjugate in $G$: \[\operatorname{Stab}_G(g \cdot x_1) = g \operatorname{Stab}_G(x_1) g^{-1}.\] (2) Note that if $x_1, x_2$ lie in different orbits, their stabilizers need not be conjugate.

Solution

For $h\in G$, \[ \begin{aligned} h\in\operatorname{Stab}_G(gx) &\iff h(gx)=gx\\ &\iff (g^{-1}hg)x=x\\ &\iff g^{-1}hg\in\operatorname{Stab}_G(x)\\ &\iff h\in g\operatorname{Stab}_G(x)g^{-1}. \end{aligned} \] Hence \[ \operatorname{Stab}_G(gx)=g\operatorname{Stab}_G(x)g^{-1}. \] Thus points in the same orbit have conjugate stabilizers.

For different orbits this can fail. For example, let $S_3$ act on \[ \{1,2,3\}\sqcup\{*\} \] in the usual way on the first orbit and trivially on $*$. Then \[ \operatorname{Stab}(1)\cong C_2, \qquad \operatorname{Stab}(*)=S_3, \] so the stabilizers are not conjugate.

A subgroup of index equal to the smallest prime dividing $|G|$ is normal (Tag P-BQVDL)

Let $G$ be a finite group, and let $H \le G$ be a subgroup of index $[G : H] = p$, where $p$ is the smallest prime dividing $|G|$. Prove that $H$ is a normal subgroup of $G$ ($H \trianglelefteq G$).

Solution

Let $G$ act on the $p$ left cosets of $H$. This gives \[ \rho:G\to S_p, \] and let $K=\ker\rho$. Then $K\trianglelefteq G$, $K\le H$, and \[ [G:K]=|\operatorname{im}\rho|\mid p!. \] Since $[G:H]=p$, \[ [G:K]=p[H:K], \] so $[H:K]\mid (p-1)!$.

On the other hand, $[H:K]\mid |G|$. If $[H:K]>1$, choose a prime $q\mid[H:K]$. Then $q\mid|G|$, so by minimality of $p$ we have $q\ge p$. But $q\mid(p-1)!$, so $q<p$, a contradiction. Hence $[H:K]=1$, so $H=K\trianglelefteq G$.

Infinite simple groups have no finite-index subgroups (Tag P-KOQPF)

Show that if $G$ is an infinite simple group, then $G$ cannot have a proper subgroup of finite index.

Solution

Suppose $H<G$ has finite index $n$. The action of $G$ on the left cosets $G/H$ gives a homomorphism \[ \rho:G\to S_n. \] Its kernel is the core \[ \ker\rho=\bigcap_{g\in G}gHg^{-1}, \] which is a normal subgroup of $G$ contained in $H$.

If $G$ is simple, then $\ker\rho$ is either $1$ or $G$. Since $H$ is proper, $\ker\rho\ne G$, so $\ker\rho=1$. Hence $\rho$ is injective and embeds $G$ into the finite group $S_n$, forcing $G$ to be finite. This contradicts the hypothesis that $G$ is infinite.

Therefore an infinite simple group has no proper subgroup of finite index.

Subgroups of order 5 in $S_5$ are transitive (Tag P-Q44SQ)

Show that every subgroup of order $5$ in $S_5$ acts transitively on $\{1,2,3,4,5\}$.

Solution

Let $H\le S_5$ have order $5$. Then $H$ is cyclic, say \[ H=\langle\sigma\rangle \] with $\sigma\ne e$. Since $\sigma$ has order $5$, its cycle decomposition in $S_5$ must consist of a single $5$-cycle. Therefore the orbit of any point under $\langle\sigma\rangle$ has size $5$ and is the whole set.

Hence $H$ is transitive.

Euler totient of primes and composites (Tag P-BXBHJ)

How do you compute the totient $\phi(p)$ for $p$ prime? Or $\phi(n)$ for $n$ composite?

Solution

<1>1. If $p$ is prime, then \[ \varphi(p)=p-1. \]

Every nonzero residue modulo $p$ is relatively prime to $p$, so the invertible residue classes are exactly $1,2,\ldots,p-1$.

<1>2. For a prime power $p^a$ with $a\ge1$, \[ \varphi(p^a)=p^a-p^{a-1}=p^{a-1}(p-1). \]

Among the $p^a$ residue classes modulo $p^a$, exactly the $p^{a-1}$ multiples of $p$ fail to be relatively prime to $p^a$.

<1>3. Euler’s totient is multiplicative on coprime integers.

If $\gcd(m,n)=1$, the Chinese remainder theorem gives \[ \ZZ/mn\ZZ\cong \ZZ/m\ZZ\times\ZZ/n\ZZ. \] Taking unit groups gives \[ (\ZZ/mn\ZZ)^\times\cong(\ZZ/m\ZZ)^\times\times(\ZZ/n\ZZ)^\times, \] so \[ \varphi(mn)=\varphi(m)\varphi(n). \]

<1>4. Therefore, if \[ n=\prod_{i=1}^r p_i^{a_i} \] is the prime factorization, then \[ \varphi(n)=\prod_{i=1}^r p_i^{a_i-1}(p_i-1) =n\prod_{p\mid n}\left(1-\frac1p\right). \]

Apply <1>2 to each prime-power factor and then <1>3.

Order of $\GL_n(\FF_q)$ (Tag P-BJDIE)

Show that the order of the group $\mathrm{GL}_n(\mathbb{F}_q)$ of invertible $n\times n$ matrices over the field $\mathbb{F}_q$ of $q$ elements is $(q^n-1)(q^n-q)\cdots(q^n-q^{n-1})$.

Solution

An invertible $n\times n$ matrix over $\FF_q$ is the same thing as an ordered basis of the vector space $\FF_q^n$, namely its ordered list of columns.

<1>1. The first column can be any nonzero vector, giving $q^n-1$ choices.

There are $q^n$ vectors in $\FF_q^n$, and only the zero vector is forbidden.

<1>2. After choosing $k$ linearly independent columns, the next column has $q^n-q^k$ choices.

The span of $k$ independent vectors is a $k$-dimensional subspace, hence has $q^k$ elements. The next column must lie outside that span.

<1>3. Therefore \[ |\GL_n(\FF_q)|=(q^n-1)(q^n-q)(q^n-q^2)\cdots(q^n-q^{n-1}). \]

Choose the columns successively and multiply the numbers of choices from <1>1 and <1>2.

$\Aut(\ZZ/p)$, $\Aut((\ZZ/p)^n)$, and $\Aut(\ZZ/n)$ (Tag P-TRZ3O)

Identify \[ \operatorname{Aut}(\ZZ/p\ZZ), \qquad \operatorname{Aut}((\ZZ/p\ZZ)^n), \qquad \operatorname{Aut}(\ZZ/n\ZZ). \]

Solution

For the additive cyclic group $C_p=\ZZ/p\ZZ$, every automorphism is determined by the image of $1$, which may be any nonzero residue class. Thus \[ \operatorname{Aut}(\ZZ/p\ZZ)\cong(\ZZ/p\ZZ)^\times. \] Since $p$ is prime, this is a cyclic group of order $p-1$.

The group \[ (\ZZ/p\ZZ)^n \] is the additive group of the $n$-dimensional vector space $\FF_p^n$. Its group automorphisms are exactly the invertible $\FF_p$-linear maps, so \[ \operatorname{Aut}((\ZZ/p\ZZ)^n)\cong GL_n(\FF_p). \]

Finally, for the additive cyclic group $\ZZ/n\ZZ$, an endomorphism is determined by \[ 1\longmapsto a\pmod n. \] It is invertible exactly when $a$ is a unit modulo $n$. Therefore \[ \operatorname{Aut}(\ZZ/n\ZZ)\cong(\ZZ/n\ZZ)^\times. \]

Number of elements of order $4$ in $\Aut(\ZZ/20)$ (Tag P-SUU6D)

How many elements in the automorphism group $\operatorname{Aut}(\mathbb{Z}/20\mathbb{Z})$ have order 4?

Solution

Since $\mathbb Z/20\mathbb Z$ is cyclic, \[ \operatorname{Aut}(\mathbb Z/20\mathbb Z)\cong(\mathbb Z/20\mathbb Z)^\times. \] By the Chinese remainder theorem, \[ (\mathbb Z/20\mathbb Z)^\times \cong(\mathbb Z/4\mathbb Z)^\times\times(\mathbb Z/5\mathbb Z)^\times \cong C_2\times C_4. \] An element $(a,b)\in C_2\times C_4$ has order $4$ exactly when $b$ has order $4$. There are two choices for $a$ and two elements of order $4$ in $C_4$, hence \[ 2\cdot2=4 \] such elements.

Explicitly, they correspond to \[ 3,7,13,17\pmod{20}. \] Therefore the answer is \[ \boxed{4}. \]

Non-isomorphic groups can have isomorphic automorphism groups (Tag P-BIFDZ)

Find two groups $G\not\cong H$ where $\Aut G\cong \Aut H$.

Solution

Take \[ G=C_3, \qquad H=C_4. \] Then $G\not\cong H$ because they have different orders. On the other hand, \[ \Aut(C_n)\cong (\ZZ/n\ZZ)^\times. \] Therefore \[ \Aut(C_3)\cong (\ZZ/3\ZZ)^\times\cong C_2 \] and \[ \Aut(C_4)\cong (\ZZ/4\ZZ)^\times\cong C_2. \] Hence \[ \Aut(G)\cong\Aut(H) \] even though $G\not\cong H$.

Isomorphic subgroups need not have isomorphic quotients (Tag P-3OUVI)

Let $H, K \leq G$ be subgroups with $H\cong K$. Is it true that $G/H \cong G/K$?

Hint: consider a group with distinct subgroups of order 2 whose quotients have order 4.

Solution

No.

<1>1. Let \[ G=C_4\oplus C_2, \] and write elements additively. Define \[ H=\langle(2,0)\rangle, \qquad K=\langle(0,1)\rangle. \] Then $H\cong K\cong C_2$.

Both displayed generators have order $2$. Since $G$ is abelian, both subgroups are normal, so both quotients are defined.

<1>2. The quotient $G/H$ is isomorphic to $C_2\oplus C_2$.

Quotienting the $C_4$ factor by its subgroup of order $2$ gives $C_2$, while the second factor survives unchanged: \[ G/H\cong (C_4/\langle2\rangle)\oplus C_2\cong C_2\oplus C_2. \]

<1>3. The quotient $G/K$ is isomorphic to $C_4$.

The subgroup $K$ is exactly the second direct factor, so \[ G/K\cong C_4. \]

<1>4. Hence $G/H\not\cong G/K$.

The group $C_4$ contains an element of order $4$, whereas every nonzero element of $C_2\oplus C_2$ has order $2$. Thus the quotients are not isomorphic even though $H\cong K$.

Inner automorphisms preserve conjugacy of subgroups (Tag P-HPSTB)

Show that any automorphism (and in particular, every inner automorphism) sends conjugate subgroups to conjugate subgroups.

Solution

Suppose $H_2=gH_1g^{-1}$ and let $\varphi\in\operatorname{Aut}(G)$. Then \[ \varphi(H_2) =\varphi(gH_1g^{-1}) =\varphi(g)\,\varphi(H_1)\,\varphi(g)^{-1}. \] Thus $\varphi(H_1)$ and $\varphi(H_2)$ are conjugate subgroups of $G$. In particular, this holds for every inner automorphism.

$\Aut(S_n)=\Inn(S_n)$ for $n\neq 6$ (Tag P-WBK5B)

Show that for $n\ne6$, \[ \operatorname{Aut}(S_n)=\operatorname{Inn}(S_n). \]

Solution

For $n=1,2$ the statement is immediate, so assume $n\ge3$.

<1>1. Every automorphism sends transpositions to transpositions when $n\ne6$.

An involution in $S_n$ has cycle type $2^k1^{n-2k}$. Its conjugacy class has size \[ \frac{n!}{2^k k!(n-2k)!}. \] A transposition has class size $\binom n2$. Automorphisms preserve element order and conjugacy-class size, so the image of a transposition must be an involution satisfying \[ \frac{n!}{2^k k!(n-2k)!}=\binom n2. \] For $k\ge2$ this is equivalent to \[ \frac{(n-2)!}{2^{k-1}k!(n-2k)!}=1. \] For $k=2$ there is no integer solution. For $k=3$ the unique solution is $n=6$. For $k\ge4$, already at the smallest possible $n=2k$ the left side is \[ \frac{(2k-2)!}{2^{k-1}k!}>1, \] and it increases with $n$. Thus, when $n\ne6$, necessarily $k=1$.

<1>2. An automorphism preserving transpositions is inner.

Let \[ s_i=(i\ i+1),\qquad 1\le i<n. \] These generate $S_n$. Put $t_i=\phi(s_i)$. Each $t_i$ is a transposition. Moreover, \[ \operatorname{ord}(t_it_j)= \operatorname{ord}(s_is_j) = \begin{cases} 3,&|i-j|=1,\\ 2,&|i-j|>1. \end{cases} \] Two distinct transpositions have product of order $3$ exactly when their supports meet in one point, and they commute exactly when their supports are disjoint.

Write $t_1=(a_1a_2)$. Since $t_2$ meets $t_1$ in exactly one point, after interchanging $a_1,a_2$ if necessary write $t_2=(a_2a_3)$. Inductively, $t_i$ must meet $t_{i-1}$ but be disjoint from $t_1,\ldots,t_{i-2}$, so \[ t_i=(a_i a_{i+1}) \] for distinct $a_1,\ldots,a_n$.

Let $\pi\in S_n$ be defined by $\pi(i)=a_i$. Then \[ \pi s_i\pi^{-1}=t_i=\phi(s_i) \] for every $i$. Since the $s_i$ generate $S_n$, \[ \phi(g)=\pi g\pi^{-1} \qquad(g\in S_n). \] Hence $\phi$ is inner.

Therefore \[ \operatorname{Aut}(S_n)=\operatorname{Inn}(S_n) \] for all $n\ne6$.

Pairs $(n,p)$ for which $\SL_n(\FF_p)$ is solvable (Tag P-KPEPM)

Let $p$ be prime and $n\ge1$. Determine all pairs $(n,p)$ for which \[ \SL_n(\FF_p) \] is solvable.

Solution

The solvable cases are \[ n=1\quad\text{for every prime }p, \] and \[ (n,p)=(2,2),(2,3). \]

<1>1. If $n=1$, then $\SL_1(\FF_p)=1$, so it is solvable.

<1>2. The groups $\SL_2(\FF_2)$ and $\SL_2(\FF_3)$ are solvable.

We have \[ \SL_2(\FF_2)\cong S_3, \] which is solvable.

<1>3. All other cases with $n\ge2$ are nonsolvable.

Groups of order $pq$ need not be nilpotent (Tag P-NLMMW)

If $|G| = pq$ (distinct primes), is $G$ necessarily nilpotent?

Solution

No. For example, $S_3$ has order $6=2\cdot3$ but is not nilpotent: its center is trivial, so its upper central series cannot reach $S_3$.

More generally, for a finite group, nilpotence is equivalent to normality of every Sylow subgroup. If $|G|=pq$ with primes $p<q$, the Sylow $q$-subgroup is always normal. A nonabelian group of order $pq$ exists exactly when $p\mid(q-1)$; it is a semidirect product \[ C_q\rtimes C_p \] with nontrivial action, and its Sylow $p$-subgroups are not normal. Hence such a group is not nilpotent. The case $p=2,q=3$ is $S_3$.

Solvable groups have a nontrivial normal subgroup (Tag P-RPSTO)

Show that if $G$ is solvable, then $G$ contains a nontrivial normal subroup.
- What does this mean on the Galois theory side?

Hint: consider the derived series.

Solution

Taken literally, every nontrivial group $G$ contains a nontrivial normal subgroup, namely $G$ itself. The solvability hypothesis becomes meaningful if one asks for a proper nontrivial normal subgroup.

If $G$ is nonabelian and solvable, then \[ 1\ne [G,G]\ne G. \] Indeed, $[G,G]\ne1$ because $G$ is nonabelian, while $[G,G]\ne G$ because otherwise the derived series would be stationary at $G$ and could not reach $1$. Since $[G,G]$ is characteristic, it is normal. Thus every nonabelian solvable group has a proper nontrivial normal subgroup.

For abelian solvable groups this stronger assertion can fail: $C_p$ is simple. More generally, a nontrivial abelian group has a proper nontrivial subgroup unless it has prime order.

For a finite Galois extension $L/F$ with solvable Galois group $G$, a normal subgroup $H\trianglelefteq G$ corresponds to an intermediate field $L^H$ that is Galois over $F$. A proper nontrivial normal subgroup gives a proper nontrivial Galois intermediate field. The derived series gives the usual tower whose successive Galois groups are abelian.

$\Aut((\ZZ/p)^n)$ as a matrix group (Tag P-ZHTDW)

Identify $\Aut_\Grp(\bigoplus_{i=1}^n \ZZ/p)$ as a matrix group and determine its size.

Solution

Write \[ V=(\ZZ/p)^n\cong \FF_p^n. \] Every group endomorphism of the elementary abelian $p$-group $V$ is automatically $\FF_p$-linear, because for $a\in\FF_p$ and $v\in V$, \[ \phi(av)=\phi(v+\cdots+v)=a\phi(v). \] Hence \[ \Endo_{\Grp}(V)=\Endo_{\FF_p}(V)\cong M_n(\FF_p), \] and the automorphisms are exactly the invertible linear maps: \[ \Aut_{\Grp}(V)\cong \GL_n(\FF_p). \]

To count them, choose the columns of an invertible matrix successively. The first column can be any nonzero vector, giving $p^n-1$ choices. Once $j$ linearly independent columns have been chosen, their span has $p^j$ elements, so the next column has $p^n-p^j$ choices. Therefore \[ \left|\Aut_{\Grp}\bigl((\ZZ/p)^n\bigr)\right| =|\GL_n(\FF_p)| =\prod_{j=0}^{n-1}(p^n-p^j). \]

The Sylow theorems (Tag P-ZBTCU)

State the three Sylow theorems.

Solution

Let $G$ be a finite group and let \[ |G|=p^n m, \qquad p\nmid m. \] A subgroup of order $p^n$ is called a Sylow $p$-subgroup.

The Sylow theorems are:

<1>1. Existence. There exists a subgroup \[ P\le G \] of order $p^n$.

<1>2. Conjugacy and containment. Every $p$-subgroup of $G$ is contained in a Sylow $p$-subgroup, and any two Sylow $p$-subgroups are conjugate in $G$.

<1>3. Number of Sylow subgroups. If $n_p$ denotes the number of Sylow $p$-subgroups, then \[ n_p\equiv1\pmod p \] and \[ n_p\mid m. \]

In particular, \[ n_p=1 \iff \text{the Sylow $p$-subgroup is normal in $G$}. \]

Groups of order $pq$ with $p>q$ have a proper nontrivial normal subgroup (Tag P-PKJVI)

Let $G$ be a group of order $|G| = p q$, where $p$ and $q$ are distinct primes with $p > q$. Prove that $G$ possesses a unique (and hence normal) Sylow $p$-subgroup of order $p$.

Solution

Let $n_p$ be the number of Sylow $p$-subgroups. Sylow’s theorem gives \[ n_p\mid q,\qquad n_p\equiv1\pmod p. \] Since $q$ is prime, $n_p$ is either $1$ or $q$. But $p>q$, so $q\not\equiv1\pmod p$. Hence $n_p=1$.

Thus the Sylow $p$-subgroup is unique and therefore normal. Its order is $p$, so it is nontrivial and proper in a group of order $pq$.

A finite abelian group is the product of its Sylow subgroups (Tag P-USPU6)

Show that any finite abelian group is isomorphic to the direct product of its Sylow subgroups

Solution

Goal: Prove that every finite abelian group $G$ is isomorphic to the direct product of its Sylow subgroups.

<1>1. Definition of the Sylow subgroups: Proof: <2>1. Let $G$ be a finite abelian group of order $|G| = p_1^{a_1} p_2^{a_2} \cdots p_k^{a_k}$, where $p_1, \dots, p_k$ are distinct primes and $a_i \ge 1$. <2>2. For each $i \in \{1, \dots, k\}$, define \[P_i = \left\{ x \in G : x^{p_i^{a_i}} = e \right\}.\] <2>3. Because $G$ is abelian, $(xy)^{p_i^{a_i}} = x^{p_i^{a_i}} y^{p_i^{a_i}} = e$ for all $x, y \in P_i$, and $(x^{-1})^{p_i^{a_i}} = e$, so $P_i$ is a subgroup of $G$. <2>4. The order of every element in $P_i$ is a power of $p_i$, so $P_i$ is a $p_i$-subgroup. By Lagrange’s Theorem and Sylow’s Theorems (or Cauchy’s Theorem), $|P_i| = p_i^{a_i}$, so $P_i$ is the unique Sylow $p_i$-subgroup of $G$. <2>5. Since $G$ is abelian, $P_i \trianglelefteq G$ for all $i$.

<1>2. Linear independence / trivial intersections of Sylow subgroups: Proof: <2>1. For each $j \in \{1, \dots, k\}$, define $Q_j = P_1 P_2 \cdots P_{j-1} P_{j+1} \cdots P_k$. <2>2. Every element $y \in Q_j$ has order dividing $\prod_{i \neq j} p_i^{a_i} = |G| / p_j^{a_j}$, which is relatively prime to $p_j$. <2>3. Every element $x \in P_j$ has order dividing $p_j^{a_j}$. <2>4. If $z \in P_j \cap Q_j$, then the order $o(z)$ divides both $p_j^{a_j}$ and $|G|/p_j^{a_j}$. <2>5. Since $\gcd(p_j^{a_j}, |G|/p_j^{a_j}) = 1$, $o(z) = 1$, which forces $z = e$. <2>6. Thus $P_j \cap Q_j = \{e\}$ for all $j \in \{1, \dots, k\}$.

<1>3. Generation of $G$ via Bézout’s identity: Proof: <2>1. For each $i \in \{1, \dots, k\}$, let $m_i = |G| / p_i^{a_i}$. <2>2. Since $p_1, \dots, p_k$ are distinct primes, $\gcd(m_1, m_2, \dots, m_k) = 1$. <2>3. By Bézout’s identity in $\mathbb{Z}$, there exist integers $u_1, u_2, \dots, u_k \in \mathbb{Z}$ such that \[\sum_{i=1}^k u_i m_i = 1.\] <2>4. For any $g \in G$: \[g = g^1 = g^{\sum_{i=1}^k u_i m_i} = \prod_{i=1}^k g^{u_i m_i}.\] <2>5. For each $i$, let $g_i = g^{u_i m_i}$. Then $g_i^{p_i^{a_i}} = g^{u_i m_i p_i^{a_i}} = g^{u_i |G|} = (g^{|G|})^{u_i} = e^{u_i} = e$. <2>6. Thus $g_i \in P_i$ for each $i$, which proves that $g \in P_1 P_2 \cdots P_k$. <2>7. Therefore $G = P_1 P_2 \cdots P_k$.

<1>4. Construction of the direct product isomorphism: Proof: <2>1. Define the map $\Phi: P_1 \times P_2 \times \cdots \times P_k \to G$ by \[\Phi(g_1, g_2, \dots, g_k) = g_1 g_2 \cdots g_k.\] <2>2. Since $G$ is abelian, $\Phi$ is a group homomorphism. <2>3. By <1>3, $\Phi$ is surjective. <2>4. If $\Phi(g_1, \dots, g_k) = e$, then for each $j$, $g_j^{-1} = \prod_{i \neq j} g_i \in P_j \cap Q_j$. <2>5. By <1>2, $P_j \cap Q_j = \{e\}$, so $g_j = e$ for all $j$. <2>6. Thus $\ker \Phi = \{(e, \dots, e)\}$, so $\Phi$ is injective. <2>7. Therefore $\Phi$ is an isomorphism.

<1>5. Conclusion: Proof: The finite abelian group $G$ is isomorphic to the direct product of its Sylow subgroups $P_1 \times P_2 \times \cdots \times P_k$.

A group of order $231$ has a normal Sylow $7$-subgroup and a central Sylow $11$-subgroup (Tag P-K2U6D)

Let $G$ be a group of order \[ 231=3\cdot7\cdot11. \] Show that $G$ has a normal Sylow $7$-subgroup and that its Sylow $11$-subgroup is central.

Solution

<1>1. The Sylow $7$-subgroup is normal.

Let $n_7$ be the number of Sylow $7$-subgroups. Sylow gives \[ n_7\equiv1\pmod7, \qquad n_7\mid 33. \] The divisors of $33$ are $1,3,11,33$, and only $1$ is congruent to $1$ modulo $7$. Hence \[ n_7=1. \] So the Sylow $7$-subgroup is unique and therefore normal.

<1>2. The Sylow $11$-subgroup is unique.

Similarly, \[ n_{11}\equiv1\pmod{11}, \qquad n_{11}\mid21. \] Among $1,3,7,21$, only $1$ is congruent to $1$ modulo $11$. Thus $n_{11}=1$. Let $Q$ denote this unique Sylow $11$-subgroup; then $Q\trianglelefteq G$ and $Q\cong C_{11}$.

<1>3. The subgroup $Q$ is central.

Conjugation gives a homomorphism \[ G\longrightarrow\Aut(Q)\cong C_{10}. \] Because $Q$ is abelian, $Q$ lies in the kernel, so the image has order dividing \[ |G/Q|=21. \] It also has order dividing $|\Aut(Q)|=10$. Since \[ \gcd(21,10)=1, \] the image is trivial. Thus every element of $G$ centralizes every element of $Q$, so \[ Q\subseteq Z(G). \]

Possible Galois groups of an abelian extension of degree $540$, and quadratic intermediate extensions $L/E$ (Tag P-4ZDQQ)

If $L/k$ is an abelian Galois extension of degree $540 = 2^2 \times 3^3\times 5$, what are the possible Galois groups $\Gal(L/k)$?
- Are there any intermediate fields $E$ for which $L/E$ is a quadratic extension?

Solution

Let \[ G=\operatorname{Gal}(L/k). \] Since $L/k$ is abelian Galois and $[L:k]=540$, $G$ is a finite abelian group of order \[ 540=2^2 3^3 5. \]

<1>1. The possible primary components are \[ G_{(2)}\cong C_4\quad\text{or}\quad C_2\oplus C_2, \] \[ G_{(3)}\cong C_{27},\quad C_9\oplus C_3,\quad\text{or}\quad C_3\oplus C_3\oplus C_3, \] and \[ G_{(5)}\cong C_5. \]

These are exactly the abelian groups corresponding to the partitions of the exponents $2$, $3$, and $1$ in the primary decomposition theorem.

<1>2. Hence there are exactly six possible Galois groups.

The $2$-primary and $3$-primary choices are independent, while the $5$-primary part is forced. Thus there are \[ 2\cdot3\cdot1=6 \] isomorphism classes. In invariant-factor form they are \[ C_{540}, \qquad C_3\oplus C_{180}, \qquad C_3\oplus C_3\oplus C_{60}, \] \[ C_2\oplus C_{270}, \qquad C_6\oplus C_{90}, \qquad C_3\oplus C_6\oplus C_{30}. \]

<1>3. An intermediate field $E$ satisfies $[L:E]=2$ exactly when the corresponding subgroup $\operatorname{Gal}(L/E)\le G$ has order $2$.

By the fundamental theorem of Galois theory, \[ [L:E]=|\operatorname{Gal}(L/E)|. \] Thus quadratic extensions $L/E$ correspond precisely to order-$2$ subgroups of $G$.

<1>4. Such intermediate fields always exist.

Since $2\mid |G|$, Cauchy’s theorem gives an element of order $2$, hence a subgroup of order $2$. By <1>3 this subgroup corresponds to an intermediate field $E$ with $[L:E]=2$.

<1>5. The number of such fields is determined by the $2$-primary part.

If $G_{(2)}\cong C_4$, there is a unique subgroup of order $2$, hence exactly one such $E$. If $G_{(2)}\cong C_2\oplus C_2$, its three nonzero elements generate three distinct order-$2$ subgroups, hence there are exactly three such fields. In every case \[ [E:k]=[L:k]/2=270. \]

A commutative unital ring is a field iff its only ideals are $\{0\}$ and $R$ (Tag P-3PTP5)

Prove that a commutative ring with unit is a field if and only if its only ideals are {0} and the whole ring

Solution

<1>1. If $R$ is a field, then its only ideals are $\{0\}$ and $R$.

Let $I\subseteq R$ be a nonzero ideal and choose $0\ne a\in I$. Since $R$ is a field, $a^{-1}\in R$, so \[ 1=a^{-1}a\in I. \] Hence $I=R$. Thus the only ideals are $0$ and $R$.

<1>2. Conversely, suppose the only ideals of the commutative unital ring $R$ are $\{0\}$ and $R$. Then every nonzero element is a unit.

Take $0\ne a\in R$. The principal ideal $(a)$ is nonzero, so by hypothesis \[ (a)=R. \] Hence $1\in(a)$, so there exists $b\in R$ with \[ ab=1. \] Thus $a$ is a unit.

<1>3. Therefore $R$ is a field.

By <1>2, every nonzero element of $R$ is invertible. Since $R$ is commutative with $1$, this is exactly the definition of a field.

A polynomial of degree $2$ or $3$ in $k[x]$ is irreducible iff it has no root in $k$ (Tag P-HWBXR)

Let $k$ be a field and let $f(x) \in k[x]$ be a polynomial of degree $\deg(f) \in \{2, 3\}$. Prove that $f(x)$ is irreducible in $k[x]$ if and only if $f(x)$ has no roots in $k$.

Solution

If $f$ is irreducible and had a root $a\in k$, then the factor theorem would give \[ f(x)=(x-a)g(x) \] with $\deg g=\deg f-1\ge1$, a nontrivial factorization. Hence an irreducible polynomial of degree $2$ or $3$ has no root in $k$.

Conversely, suppose $f$ is reducible. Then \[ f=gh \] with $\deg g,\deg h\ge1$ and \[ \deg g+\deg h=\deg f\in\{2,3\}. \] Therefore at least one factor has degree $1$. Write that factor as $ax+b$ with $a\ne0$. It vanishes at $-b/a\in k$, so $f$ has a root in $k$.

Thus, for degrees $2$ and $3$, \[ f\text{ irreducible}\quad\Longleftrightarrow\quad f\text{ has no root in }k. \]