Topics
Subgroups
The one-step subgroup test
- A nonempty subset \(H\subseteq G\) with \(x\inverseof{y} \in H\) for all \(x,y\in H\) is a subgroup
Cosets
- \(xH \da \ts{xh\st h\in H}\), \(G/H \da \ts{xH \st x\in G}\), and the left cosets partition \(G\)
The index of a subgroup
Normal subgroups
Quotients
The normalizer of a subgroup
Maximal and proper subgroups
Characteristic subgroup
Cauchy’s theorem
Lagrange’s theorem
Definitions and properties of common special families of groups:
Cyclic groups \(C_n\)
Symmetric groups \(S_n\)
Alternating groups \(A_n\)
Dihedral groups \(D_{n}\)
The quaternion group \(Q_8\)
Matrix groups \(\GL_n(k), \Orth_n(k), \SL_n(k), \SO_n(k)\)
\(p\dash\)groups
Free groups \(F_n\) (and presentations/relations)
The four isomorphism theorems
Finite groups of order at most \(20\)
Implications among classes of groups:
- cyclic \(\implies\) abelian \(\implies\) nilpotent \(\implies\) solvable
Review exercises
State the definitions of the following:
Group homomorphism
Centralizer
Normalizer
Conjugacy class
Center
Inner automorphism
Commutator
\(p\dash\)group
Write definitions or presentations for the families of groups in the topic list.
State what it means for a permutation to be even or odd.
Find a counterexample for the converse of Lagrange’s theorem.
State the four isomorphism theorems.
Problems by topic
For everything that follows, assume \(G\) is a finite group.
\(H\leq G\) denotes that \(H\) is a subgroup of \(G\).
\(\size G\) denotes the order of \(G\).
\(e\) or \(e_G\) denotes the identity element of \(G\).
The group operation is written multiplicatively.
\(G/H\) is the set of left cosets of \(H\) in \(G\).
Cosets
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.
Orders
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 \(g^{|G|}=e\).
Normal subgroups
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.
Symmetric groups
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 \(n\ge2\).
Matrix groups
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 \[ \boxed{|\operatorname{SL}_n(\mathbb F_p)|= \frac1{p-1}\prod_{k=0}^{n-1}(p^n-p^k).} \]
Warmup problems
$G/Z(G)$ 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.
Let \([G:H]=p\) and let \(G\) act on the left cosets \(G/H\). This gives \[ \rho:G\to S_p. \] The image is transitive, so \(p\mid |\rho(G)|\). Also \(|\rho(G)|\) divides both \(|G|\) and \(p!\).
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.
Qual problems
A normal subgroup of smallest prime order lies in the center (Tag P-4XOCE)
Let \(H \normal G\) be a normal subgroup of a finite group \(G\), where the order of \(H\) is the smallest prime \(p\) dividing \(\abs{G}\). Prove that \(H\) is contained in the center of \(G\).
\(x\in Z(G)\) iff \(\size C_x = 1\), i.e. the size of its conjugacy class is one.
Normal subgroups are disjoint unions of (some) conjugacy classes in \(G\).
In fact, this is a characterization of normal subgroups (i.e. \(H\) is normal iff \(H\) is a union of conjugacy classes in \(G\)).
Why: if \(H\normal G\) then \(gh\inverseof{g} \in H\) for all \(g\), so \(C_h \subseteq H\) and \(\Union_h C_h = H\). Conversely, if \(H = \Union_{h\in H} C_h\), then \(gh\inverseof{g} \in C_h \subseteq H\) and thus \(gH\inverseof{g} = H\).
Orbit stabilizer theorem: \(\size C_g = \size G/ \size K_g\) where \(C_g\) is the centralizer and \(K_g\) is the conjugacy class of \(g\).
- In particular, \(\size C_g\) divides \(\size G\).
Show an element \(x\) is central by showing \(\size C_x = 1\).
Solution due to Swaroop Hegde, typed up + modifications added by DZG.
Let \(p \da \size H\).
Let \(\ts{ C_i }_{i\leq n}\) be the conjugacy classes in \(G\), then \(G = \disjoint_{i\leq n} C_i\)
By the first fact, there is a sub-collection \(\ts{ C_{i_j}}_{j\leq k }\) such that \[ H = \disjoint_{j\leq k} C_{i_j} .\]
The identity is always in a single conjugacy class, so \(C_e = \ts{ e }\).
Since \(e\in H\), without loss of generality, label \(C_{i_1} = \ts{ e }\).
So \[ H = \Disjoint_{j\leq k} C_{i_j} = C_{i_1}{\textstyle \coprod} \displaystyle\Disjoint_{\substack{ j\leq k \\ j\neq 1} } C_{i_j} .\]
Take cardinality in the above equation \[ p = 1 + \sum_{\substack{ j\leq k \\ j\neq 1 }} \size C_{i_j} .\]
So \(\size C_{i_j} \leq p-1\) for all \(j\neq 1\).
Every \(\size C_{i_j}\) divides \(\size G\), but \(p\) was the minimal prime dividing \(\size G\), forcing \(\size C_{i_j} = 1\) for all \(j \neq 1\).
- This rules out \(\size C_{i_j}\) being a prime less than \(p\), but also rules out composites: if a prime \(q\divides \size C_{i_j}\), then \(q<p\) and \(q\divides \size G\), a contradiction.
By fact 3, each \(x\in C_{i_j}\) satisfies \(x\in Z(G)\).
\(\union C_{i_j} = H\), so \(H \subseteq Z(G)\).
Solution
Because \(|H|=p\) is prime, \(H\cong C_p\). Conjugation by \(G\) on \(H\) gives a homomorphism \[ \varphi:G\longrightarrow \operatorname{Aut}(H)\cong C_{p-1}. \] Hence \(|\operatorname{im}\varphi|\) divides both \(|G|\) and \(p-1\). Every prime divisor of \(|\operatorname{im}\varphi|\) therefore divides \(|G|\), so by minimality of \(p\) it is at least \(p\); but it also divides \(p-1\). Thus \(|\operatorname{im}\varphi|=1\).
Therefore conjugation by every \(g\in G\) acts trivially on \(H\). Equivalently, \(ghg^{-1}=h\) for all \(g\in G\) and \(h\in H\), so \(H\subseteq Z(G)\).