Groups and subgroups · Qual Corpus

Two left cosets of a subgroup $H\le G$ are equal or disjoint, so for finite $G$, $\abs{G}=[G:H]\,\abs{H}$ (Lagrange’s theorem). A subgroup $N\le G$ is normal if and only if $(aN)(bN)=abN$ is well defined on cosets, and then $G/N$ is a group. The isomorphism theorems: a homomorphism $\varphi\colon G\to G'$ induces $G/\ker\varphi\cong\im\varphi$; for $H\le G$ and $N\trianglelefteq G$, $HN/N\cong H/(H\cap N)$; for normal subgroups $N\le M$ of $G$, $(G/N)/(M/N)\cong G/M$; and $M\mapsto M/N$ is a bijection from the subgroups of $G$ containing $N$ to the subgroups of $G/N$. Every subgroup of a cyclic group is cyclic, and a cyclic group of order $n$ has exactly one subgroup of each order dividing $n$. Cauchy’s theorem: if a prime $p$ divides $\abs{G}$, then $G$ has an element of order $p$.

Subgroup (Tag D-IQ4OX)

Let $G$ be a group with identity $e$. A subset $H\subseteq G$ is a subgroup of $G$ if

$gh\in H$ for all $g,h\in H$;
$e\in H$;
$g^{-1} \in H$ for all $g\in H$.

One-step subgroup test (Tag PR-U5CHT)

A nonempty subset is a subgroup once it is closed under $ab^{-1}$.Suggest a slogan

Let $G$ be a group and $H \subseteq G$ a nonempty subset. If $ab^{-1}\in H$ for all $a,b\in H$, then $H$ is a subgroup of $G$.

Choose $a\in H$; then $1=aa^{-1}\in H$. For $b\in H$, $b^{-1}=1\cdot b^{-1}\in H$. For $a,b\in H$, $ab=a(b^{-1})^{-1}\in H$.

One-step subgroup test (Tag FF-QVQXM)

Nonempty and closed under $ab^{-1}$ already means subgroup.Suggest a slogan

Let $G$ be a group and let $H\subseteq G$ be a nonempty subset. If $ab^{-1}\in H$ for all $a,b\in H$, then $H$ is a subgroup of $G$.

Choose $a\in H$; then $e=aa^{-1}\in H$. For $b\in H$, $b^{-1}=eb^{-1}\in H$. For $a,b\in H$, $ab=a(b^{-1})^{-1}\in H$.

State the one-step subgroup test.

Subgroup generated by a subset (Tag D-D36IY)

Let $G$ be a group and let $H\subseteq G$ be a subset. The subgroup generated by $H$ is the intersection of all subgroups of $G$ containing $H$: \[ \gens{H} \coloneqq \bigcap \theset{M \suchthat H\subseteq M \leq G}. \]

Let $G$ be a group and let $H\subseteq G$ be a subset. Then $\gens{H}$ is the smallest subgroup of $G$ containing $H$, and \[ \gens{H} = \theset{ h_1^{\varepsilon_1} \cdots h_n^{\varepsilon_n} \suchthat n\geq 0,\ h_i \in H,\ \varepsilon_i\in\theset{1,-1}}. \]

Index of a subgroup (Tag D-VJGH5)

Let $G$ be a group and $H\leq G$ a subgroup. The index of $H$ in $G$, written $[G:H]$, is the cardinality of the set $G/H=\theset{gH \st g\in G}$ of left cosets of $H$.

The map $gH\mapsto Hg^{-1}$ is a well-defined bijection from left cosets to right cosets, since $g_1H=g_2H$ if and only if $g_2^{-1}g_1\in H$ if and only if $Hg_1^{-1}=Hg_2^{-1}$. So $[G:H]$ is also the number of right cosets of $H$.

Cosets are equal or disjoint (Tag PR-VUKHO)

Two cosets of the same subgroup either coincide or do not meet.Suggest a slogan

Let $G$ be a group, $H\leq G$ a subgroup, and $x,y\in G$. Then the left cosets $xH$ and $yH$ are either equal or disjoint.

Suppose $g\in xH\cap yH$, say $g=xh=yh'$ with $h,h'\in H$. Then $x=yh'h^{-1}\in yH$, so $xH\subseteq yHH=yH$; by symmetry $yH\subseteq xH$.

Equality of left cosets (Tag T-5KGVV)

Two left cosets are equal exactly when their representatives differ by an element of the subgroup.Suggest a slogan

Let $G$ be a group, $H\leq G$ a subgroup, and $a,b\in G$. Then \[ aH = bH \iff a^{-1} b \in H \iff b^{-1} a\in H. \]

If $aH=bH$, then $b=b\cdot e\in aH$, so $b=ah$ with $h\in H$ and $a^{-1}b=h\in H$. If $a^{-1}b=h\in H$, then $bH=ahH=aH$. Finally, $b^{-1}a=(a^{-1}b)^{-1}$, and $H$ is closed under inverses.

Lagrange's theorem (Tag T-SZRXI)

A finite group is partitioned into equal-sized cosets, so subgroup order times index equals group order.Suggest a slogan

If $H$ is a subgroup of a finite group $G$ with index $[G:H]$, then \[ \abs G = [G:H]\abs H. \] In particular, $\abs H$ divides $\abs G$, and the order of every element of $G$ divides $\abs G$.

The left cosets of $H$ partition $G$. Multiplication by a coset representative is a bijection $H\to gH$, so every coset has $\abs H$ elements. There are $[G:H]$ cosets, and summing their equal sizes gives the formula.

Lagrange's theorem (Tag T-GJNT5)

A subgroup order divides the finite group order, with quotient equal to its index.Suggest a slogan

Let $G$ be a finite group and $H \leq G$ a subgroup. Then $\abs H$ divides $\abs G$, and $[G:H] = \abs G/ \abs H$ [@DF04].

The left cosets of $H$ partition $G$ (PR-VUKHO), and each coset $gH$ has $\abs H$ elements because $h\mapsto gh$ is a bijection $H\to gH$. Hence $\abs G=[G:H]\,\abs H$.

Coset counting formula $\# G = [G:H]\,\#H$ (Tag C-HWX2P)

A finite group is $[G:H]$ equal-sized cosets of $H$.Suggest a slogan

Let $G$ be a finite group, let $H\leq G$ be a subgroup, and let $G/H$ be the set of left cosets of $H$ in $G$, so that $[G:H] = \size (G/H)$ is the index of $H$. Then \[ \size G = [G:H]\, \size H, \qquad\text{equivalently}\qquad [G:H] = \frac{\size G}{\size H}. \]

Tower law for subgroups (Tag PR-SF6ZE)

Subgroup indices multiply in towers.Suggest a slogan

Let $G$ be a group and $K\leq H \leq G$ subgroups. Then \[ [G: K] = [G:H]\, [H: K] \] as cardinals; in particular $[G:K]$ is finite if and only if $[G:H]$ and $[H:K]$ are both finite.

Write $G=\coprod_{i\in I}g_iH$ and $H=\coprod_{j\in J}h_jK$ as disjoint unions of left cosets, with $\abs{I}=[G:H]$ and $\abs{J}=[H:K]$. Then $G=\bigcup_{i,j}g_ih_jK$. If $g_ih_jK=g_{i'}h_{j'}K$, then $g_ih_j\in g_{i'}h_{j'}K\subseteq g_{i'}H$, so $g_iH=g_{i'}H$ and $i=i'$; cancelling $g_i$ gives $h_jK=h_{j'}K$ and $j=j'$. Hence the cosets $g_ih_jK$ are pairwise distinct, and $[G:K]=\abs{I\times J}$.

Larger subgroups have smaller index (Tag PR-YY3JG)

Subgroup inclusion reverses index size.Suggest a slogan

Let $G$ be a group and $H\leq K \leq G$ subgroups. Then \[ \abs{G} = [G:1] \geq [G:H] \geq [G:K] \geq [G:G] = 1 \] as cardinals. In particular, for arbitrary subgroups $H, K\leq G$, applying this to $H \cap K \leq H\leq G$ and $H\cap K\leq K\leq G$ gives $[G: H \cap K] \geq [G:H]$ and $[G:H\cap K]\geq[G:K]$.

By the tower law, $[G:H]=[G:K]\,[K:H]\geq[G:K]$ since $[K:H]\geq1$; the outer inequalities are the cases $H=1$ and $K=G$.

Order of a group element (Tag D-ZTPHF)

Let $G$ be a group with identity $e$ and $g\in G$. If $g^n=e$ for some integer $n\geq 1$, the order of $g$, written $o(g)$, is the smallest such $n$. If no such $n$ exists, $g$ has infinite order.

Cauchy's theorem (Tag T-XDMK2)

Every prime dividing a finite group order occurs as an element order.Suggest a slogan

Let $G$ be a finite group and let $p$ be prime. If $p$ divides $\abs G$, then $G$ contains an element of order $p$, and therefore a subgroup of order $p$.

By Lagrange's theorem, the order of every element of $G$ divides $\abs G$; Cauchy’s theorem is the converse for prime divisors of $\abs G$. Sylow's first theorem strengthens the existence statement from order $p$ to order $p^a$, where $p^a$ is the largest power of $p$ dividing $\abs G$.

Subgroups of cyclic groups (Tag T-QMKXQ)

A finite group is cyclic exactly when every divisor of its order determines a unique subgroup.Suggest a slogan

Let $G$ be a finite group of order $n$. Then $G$ is cyclic if and only if, for each positive divisor $d$ of $n$, $G$ has a unique subgroup of order $d$.

Normal subgroup (Tag D-EKE4Q)

Let $G$ be a group. A subgroup $N\leq G$ is normal, written $N\normal G$, if $gNg^{-1} = N$ for every $g\in G$.

Let $G$ be a group and $N\leq G$. The following are equivalent:

$N\normal G$.
$gN = Ng$ for every $g\in G$.
$\psi(N) = N$ for every inner automorphism $\psi \in \Inn(G)$.
$N$ is the only subgroup of $G$ conjugate to $N$.

Normal closure of a subgroup (Tag D-BPRD3)

Let $G$ be a group and $H\leq G$ a subgroup. The normal closure of $H$ in $G$ is the subgroup generated by the conjugates of $H$: \[ H^G \coloneqq \gens{ \Union_{g\in G} gHg^{-1} }. \]

Let $G$ be a group and $H\leq G$. Then $H^G$ is the smallest normal subgroup of $G$ containing $H$: \[ H^G = \bigcap \theset{N \st H \leq N \normal G }. \]

Normal core of a subgroup (Tag D-QMVEB)

Let $G$ be a group and $H\leq G$ a subgroup. The normal core of $H$ in $G$ is \[ H_G \coloneqq \Intersect_{g\in G} gHg^{-1}. \]

Let $H\leq G$, and let $\psi\colon G \to \Sym(G/H)$, $g \mapsto (xH\mapsto gxH)$, be the action of $G$ on the left cosets of $H$. Then $H_G$ is the largest normal subgroup of $G$ contained in $H$, and \[ H_G = \gens{ N \st N \normal G,\ N \leq H} = \ker \psi. \]

For $g\in G$, $\psi(g)=\id$ if and only if $gxH=xH$ for all $x\in G$, that is, $x^{-1}gx\in H$ for all $x$, that is, $g\in\Intersect_{x\in G}xHx^{-1}=H_G$. Hence $H_G=\ker\psi$ is a normal subgroup of $G$, and $H_G\subseteq 1\cdot H\cdot 1=H$. If $N\normal G$ and $N\leq H$, then $N=gNg^{-1}\subseteq gHg^{-1}$ for every $g$, so $N\subseteq H_G$. Therefore the subgroup generated by all such $N$ lies in $H_G$, and it contains $H_G$, which is one of them.

Characteristic subgroups (Tag D-53LTN)

Let $G$ be a group. A subgroup $H\leq G$ is characteristic in $G$, written $H \ch G$, if $\phi(H) \leq H$ for every $\phi \in \Aut(G)$.

If $H \ch G$, then $\phi(H) = H$ for every $\phi\in\Aut(G)$: applying the definition to $\phi^{-1}$ gives $\phi^{-1}(H)\leq H$, so $H\leq\phi(H)$. The automorphisms need not fix $H$ pointwise.

The center is a characteristic subgroup (Tag PR-GJLQP)

Every automorphism preserves the center.Suggest a slogan

For every group $G$, the center $Z(G)$ is a characteristic subgroup of $G$, $Z(G) \ch G$.

Let $\phi \in \Aut(G)$, $z \in Z(G)$, and $g \in G$. Then $\phi(z)\,g = \phi\big(z\,\inverseof{\phi}(g)\big) = \phi\big(\inverseof{\phi}(g)\,z\big) = g\,\phi(z)$, so $\phi(z) \in Z(G)$.

Characteristic subgroups of normal subgroups are normal (Tag PR-IOMVN)

Characteristic inside normal is normal in the ambient group.Suggest a slogan

Let $G$ be a group, $H \normal G$ a normal subgroup, and $N \ch H$ a characteristic subgroup of $H$. Then $N \normal G$.

Order of a quotient group (Tag T-WR7CD)

For a finite group, quotient order is subgroup index: $\abs{G/H}=\abs G/\abs H$.Suggest a slogan

Let $G$ be a finite group and $H\normal G$ a normal subgroup. Then \[ [G: H] = \abs{G/H} = \frac{\abs G}{\abs H}. \]

First isomorphism theorem for groups (Tag T-I5N43)

Quotient the domain by the kernel and what remains is exactly the image.Suggest a slogan

Let $\phi\colon G\to H$ be a group homomorphism. Then $\ker \phi$ is a normal subgroup of $G$, $\im \phi$ is a subgroup of $H$, and \[ G/\ker \phi \to \im \phi,\qquad g\ker\phi\mapsto\phi(g), \] is a well-defined group isomorphism.

For $x\in\ker\phi$ and $g\in G$, $\phi(gxg^{-1})=\phi(g)\phi(g)^{-1}=e$, so $\ker\phi\normal G$; $\im\phi$ is the image of a group under a homomorphism, hence a subgroup of $H$. The map is well defined and injective because $\phi(g)=\phi(g')$ if and only if $g^{-1}g'\in\ker\phi$, it is surjective by construction, and it is a homomorphism because $\phi$ is.

Second isomorphism theorem for groups (Tag T-NFQBO)

Quotienting $SN$ by $N$ is the same as quotienting $S$ by its overlap with $N$.Suggest a slogan

Let $G$ be a group, $S \leq G$ a subgroup, and $N \normal G$ a normal subgroup. Then \[ \frac{SN}{N} \cong \frac{S}{S\intersect N}. \] If $S$ and $N$ are finite, then \[ \abs{SN} = \frac{\abs S \abs N}{\abs{S\intersect N}}. \]

Diamond diagram for the second isomorphism theorem

Correspondence and third isomorphism theorems (Tag T-6CQEB)

Subgroups above $N$ are exactly subgroups of $G/N$, and quotienting twice collapses to quotienting by the larger subgroup.Suggest a slogan

Let $G$ be a group and $N\normal G$ a normal subgroup.

If $N\leq K\leq G$, then $K/N$ is a subgroup of $G/N$.
For $N\leq K\leq G$, $K\normal G$ if and only if $K/N\normal G/N$.
The map $K\mapsto K/N$ is a bijection from the subgroups $K$ of $G$ containing $N$ to the subgroups of $G/N$, and it restricts to a bijection between normal subgroups.
If $N\leq K$ and $K\normal G$, then \[ \frac{G/N}{K/N} \cong \frac{G}{K}. \]

Let $\pi\colon G\to G/N$ be the quotient map. (a) $K/N=\pi(K)$ is the image of a subgroup. (c) The inverse of $K\mapsto\pi(K)$ is $\bar K\mapsto\pi^{-1}(\bar K)$: $\pi^{-1}(\bar K)$ is a subgroup containing $N$, $\pi(\pi^{-1}(\bar K))=\bar K$ since $\pi$ is surjective, and $\pi^{-1}(\pi(K))=KN=K$ since $N\leq K$. (b) Since $\pi(gKg^{-1})=\pi(g)\pi(K)\pi(g)^{-1}$ and $\pi$ is a bijection on subgroups containing $N$, $gKg^{-1}=K$ for all $g\in G$ if and only if $\bar g(K/N)\bar g^{-1}=K/N$ for all $\bar g\in G/N$; this also gives the second half of (c). (d) The map $G/N\to G/K$, $gN\mapsto gK$, is well defined because $N\leq K$, and it is a surjective homomorphism with kernel $K/N$; apply the first isomorphism theorem.

Correspondence theorem for groups (Tag T-RLVA4)

Subgroups of $G/N$ are exactly the subgroups of $G$ that contain $N$.Suggest a slogan

Let $G$ be a group, $N \normal G$ a normal subgroup, and $\pi\colon G\to G/N$ the quotient map. Then \[ \theset{H \leq G \suchthat N \subseteq H} \mapstofrom \theset{\overline{H} \suchthat \overline{H} \leq G/N}, \qquad H \mapsto H/N, \qquad \pi^{-1}(\overline{H}) \mapsfrom \overline{H}, \] are mutually inverse bijections between the subgroups of $G$ containing $N$ and the subgroups of the quotient group $G/N$.

Isomorphism theorems for modules (Tag PR-DJZLY)

Module quotients follow four laws: kernel–image, diamond, quotient tower, and submodule correspondence.Suggest a slogan

Let $R$ be a ring and $M$ an $R$-module.

For an $R$-module homomorphism $\phi\colon M \to M'$, there is an isomorphism $M/\ker\phi \cong \im\phi$.
For submodules $A, B \subseteq M$, \[ \frac{A+B}{B} \cong \frac{A}{A \intersect B}. \]
For submodules $A \subseteq B \subseteq M$, \[ \frac{M/A}{B/A} \cong \frac{M}{B}. \]
For a submodule $N \subseteq M$, there is an inclusion-preserving bijection \[ \begin{aligned} \correspond{\text{Submodules of } M \\ \text{containing } N} &\mapstofrom \correspond{\text{Submodules of } M/N} \\ A &\mapstofrom A/N. \end{aligned} \]

The bijection in (d) preserves sums and intersections: for submodules $A, A' \subseteq M$ containing $N$, $(A+A')/N = A/N + A'/N$ and $(A \intersect A')/N = A/N \intersect A'/N$.

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