ALG2_L9

Orbit-stabilizer theorem

Theorem 1(Theorem).

Let $G$ be a group. Let $\varphi :G\times X\to X$ be a group action. The orbits of $x\in X$ partition $X$.

This was expected, since we saw in the previous lecture that the orbits of elements of $G$ under the conjugation action were conjugacy classes, and we know that the conjugacy classes of $G$ partition $G$.

Theorem 2(Orbit-stabilizer theorem).

Let $\varphi :G\times X\to X$ and $|G|<\infty$. Then, $|G_x||{\theta }_x|=|G|$ for all $x\in X$.

Proof
Let $x\in X$. We will construct a bijection between $G/G_x$ and ${\theta }_x$, where the former is the set of all left cosets of $G_x$ in $G$ (may not be a group, since $G_x$ is not necessarily normal). Define the map $\varphi :G/G_x\to {\theta }_x$ by $a{G}_x\mapsto ax$. $\varphi$ is well defined, because $a{G}_x=b{G}_x$ implies $a=bg$ for some $g\in G_x$, so $ax=bgx=bx$. $\varphi$ is injective, since $\varphi (a{G}_x)=\varphi (b{G}_x)$ implies $b^{-1}ax=x$, which implies $b^{-1}a\in G_x$, which implies $a{G}_x=b{G}_x$. $\varphi$ is surjective since $y\in {\theta }_x$ implies $y=ax$ for some $a\in G$, so $a{G}_x\mapsto y$.

The action of $G$ on $A$ is called transitive if there is only one orbit.

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The class equation

Definition 3(Definition).

Two elements $a$ and $b$ of $G$ are said to be conjugate in $G$ if there is some $g\in G$ such that $b=ga{g}^{-1}$ (i.e, iff there are in the same orbit of $G$ acting on itself by conjugation). The orbits of $G$ acting on itself by conjugation are called the conjugacy classes of $G$.

Notes:

• If $|G|>1$ then, unlike the action by left multiplication, $G$ does not act transitively on itself by conjugation because $\{e\}$ is always a conjugacy class.
• $\{a\}$ is a conjugacy class iff $a\in Z(G)$.

$G$ acting on itself by conjugation can be generalized to $G$ acting on its power set by conjugation: ${\varphi }_g(S)=gS{g}^{-1}$, $S\in \mathcal{P}(G)$, $g\in G$. Note that the action of $G$ acting on itself by conjugation is embedded in this action, since ${\varphi }_g(\{s\})=\{gs{g}^{-1}\}$, $s,g\in G$. As expected, we call two subsets $S$ and $T$ of $G$ to be conjugates if they are in the same orbit.

Given $S\in \mathcal{P}(G)$, the orbit stabilizer theorem allows us to compute the size of its conjugacy class to be the index $[G:G_S]$ of the stabilizer of $S$. Now, $G_S=\{g\in G|gS{g}^{-1}=S\}=N_G(S)$. Thus, the size of the conjugacy class of $S$ is $[G:N_G(S)]$.

Theorem 4(Proposition).

The size of the conjugacy class of a subset $S$ of a group $G$ is the index of the normalizer of $S$ in $G$. In particular, the size of the conjugacy class of $s\in G$ is the index of the centralizer of $s$ in $G$.

The second statement follows from the fact that $N_G(\{s\})=C_G(s)$.

The fact that the orbits partition the set being acted upon yields the following equation:

Theorem 5(The class equation).

Let $G$ be a finite group and let $g_1,g_2,\dots ,g_r$ be representatives of the distinct conjugacy classes of $G$ not contained in the center $Z(G)$. Then,

$$|G|=|Z(G)|+\sum _{i=1}^r[G:C_G(g_i)].$$

Note that all summands on the RHS of the class equation are divisors of $|G|$.

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p-groups

Groups of order $p^{\alpha }$, $\alpha \ge 1$, where $p$ is prime, are called p-groups.

Theorem 6(Theorem).

If $P$ is a p-group, $Z(P)$ is non-trivial.

Proof
Follows directly from the class equation: $p^{\alpha }=|Z(G)|+kp$ (each non-trivial conjugacy class must be a multiple of $p$)

Theorem 7(Theorem).

If $|P|=p^n$, then $|Z(P)|\ne p^{n-1}$.

Proof
FTSOC, assume $|P|=p^n$ and $|Z(P)|=p^{n-1}$. Then, $|P/Z(P)|=p$, so $P/Z(P)$ is cyclic. Thus, $P$ is abelian. But this implies $|Z(P)|=p^n$, a contradiction.

Theorem 8(Theorem).

If $|P|=p^2$ for some prime $p$, then $P$ is abelian. More precisely, either $P\cong {\mathbb{Z}}_{p^2}$ or $P\cong {\mathbb{Z}}_p\times {\mathbb{Z}}_p$.

Proof
It follows directly from the two preceding theorems that $P$ is abelian. Let $x\in P$. If $P$ has an element $x$ of order $p^2$, $P=⟨x⟩$, and $P\cong {\mathbb{Z}}_{p^2}$. Else, $P$ must have two elements $x$ and $y$ such that $|x|=|y|=p$ and $⟨x⟩\ne ⟨y⟩$.
Lagrange’s theorem forces $⟨x⟩\cap ⟨y⟩=\{1\}$. Thus, $|⟨x⟩⟨y⟩|=p^2$, so $⟨x⟩⟨y⟩=P$. These facts along with $⟨x⟩◃P$ and $⟨y⟩◃P$ allow us to write $P=⟨x⟩\times ⟨y⟩$. Thus, $P\cong {\mathbb{Z}}_p\times {\mathbb{Z}}_p$.

Sylow’s theorems have a lot more to say about p-groups and subgroups.

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Conjugacy in Sn

Cycle decomposition of conjugates

Definition 9(Definition).

Define an action $\ast$ of $S_n$ on the set $\mathcal{A}$ of all $n$-tuples with distinct entries form $[n]$, $\{(a_1,\dots ,a_n)|a_i\in [n],a_i\ne a_j\text{ for }i\ne j\}$ as

$$\mathbf{a}=(a_1,\dots ,a_n)\stackrel{\sigma }{\mapsto }\sigma \ast \mathbf{a}=(\sigma (a_1),\dots ,\sigma (a_n)).$$

$\sigma \ast \mathbf{a}$ is called an entry permutation of $\mathbf{a}$.

Define a map $\star :S_n\times \mathcal{A}\to \mathcal{A}$ by

$$\sigma \star (a_1,\dots ,a_n)=(a_{\sigma (1)},\dots ,a_{\sigma (n)}).$$

$\sigma \star \mathbf{a}$ is called a place permutation of $\mathbf{a}$. Note that $\star$ is NOT an action of $S_n$ on $\mathcal{A}$.

Since the action $\ast$ is transitive, it makes sense to ask: given $\sigma \in S_n$ and $\mathbf{a}\in \mathcal{A}$, which $\pi \in S_n$ satisfies $\pi \ast \mathbf{a}=\sigma \star \mathbf{a}$?

Let $\mathbf{e}$ denote the tuple $(1,2,\dots ,n)\in \mathcal{A}$. Observe that $\omega \ast \mathbf{e}=\omega \star \mathbf{e}$ for all $\omega \in S_n$. Thus, there exists $\tau \in S_n$ such that $\mathbf{a}=\tau \ast \mathbf{e}=\tau \star \mathbf{e}$. So, the equation reduces to $\pi \ast (\tau \ast \mathbf{e})=\sigma \star (\tau \star \mathbf{e})$, and since $\ast$ is a group action, it reduces further to $\pi \tau \ast \mathbf{e}=\sigma \star (\tau \star \mathbf{e})$. Now, let $\mathbf{b}=\tau \star \mathbf{e}$. Since $\tau \star \mathbf{e}=\tau \ast \mathbf{e}$, $b_i=\tau (e_i)=\tau (i)$.

$$\begin{array}{rl}\sigma \star \mathbf{b}& =(b_{\sigma (1)},\dots ,b_{\sigma (n)})\\ & =(\tau (\sigma (1)),\dots ,\tau (\sigma (n)))\\ & =\tau \sigma \ast \mathbf{e}.\end{array}$$

Therefore, we have

$$\begin{array}{rl}& \pi \tau \ast \mathbf{e}=\tau \sigma \ast \mathbf{e}\\ ⟹& \pi \tau =\tau \sigma \\ ⟹& \pi =\tau \sigma {\tau }^{-1}.\end{array}$$

Therefore, given $\sigma \in S_n$ and $\mathbf{a}=\tau \ast \mathbf{e}\in \mathcal{A}$, conjugating $\sigma$ by $\tau$ gives us the entry permutation equivalent to the place permutation $\sigma \star \mathbf{a}$. The following theorem follows.

Theorem 10(Theorem).

Let $\sigma ,\tau \in S_n$. Let the cycle decomposition of $\sigma$ be (what follow are NOT tuples)

$$(a_1{a}_2\dots a_n)(b_1{b}_2\dots b_n)\dots .$$

Then $\tau \sigma {\tau }^{-1}$ has cycle decomposition

$$(\tau (a_1)\tau (a_2)\dots \tau (a_n))(\tau (b_1)\tau (b_2)\dots \tau (b_n))\dots .$$

This should be reminiscent of how change of basis works in linear algebra: If $A\mathbf{a}=\mathbf{b}$, then $PA{P}^{-1}(P\mathbf{a})=P\mathbf{b}$.

Conjugacy classes of Sn

Definition 11(Definition).

If $\sigma \in S_n$ is the product of disjoint cycles of lengths $n_1,n_2,\dots ,n_r$ with $n_1\le n_2\le \cdots \le n_r$ (including its 1-cycles) then the integers $n_1,n_2,\dots ,n_r$ are called the cycle type of $\sigma$.

Theorem 12(Theorem).

Two elements of $S_n$ are conjugate in $S_n$ iff they have the same cycle type. The number of conjugacy classes of $S_n$ equals the number of partitions of $n$.

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More on Automorphisms

Theorem 13(Theorem).

Let $H◃G$. Then, $G$ acts by conjugation on $H$ as automorphisms of $H$. The permutation representation of this action is a homomorphism of $G$ into $\text{Aut}(H)$ with kernel $C_G(H)$. In particular, $G/C_G(H)$ is isomorphic to a subgroup of $\text{Aut}(H)$.

It follows that for any subgroup $H$ of $G$, the quotient group $N_G(H)/C_G(H)$ is isomorphic to a subgroup of $\text{Aut}(H)$ (take $G=N_G(H)$ in the above theorem).
Thus, any information we have about $\text{Aut}(H)$ for $H\le G$ translates to information about $N_G(H)/C_G(H)$.

Definition 14(Definition).

Let $G$ be a group and let $g\in G$. Conjugation by $g$ is called an inner automorphism of $G$ and the subgroup of $\text{Aut}(G)$ consisting of all inner automorphisms is denoted by $\text{Inn}(G)$.

Note that if $H=G$ in the above theorem, we get that $G/C_G(G)=G/Z(G)$ is isomorphic to $\text{Inn}(G)$.

Theorem 15(Corollary).

$G/Z(G)\cong \text{Inn}(G)$.

So, a group $G$ is abelian iff every inner automorphism is trivial.

Theorem 16(Corollary).

If $H$ is an abelian normal subgroup of $G$ and $H$ is not contained in $Z(G)$, then there is some $g\in G$ such that conjugation by $g$ restricted to $H$ is not an inner automorphism of $H$.

Proof
If for all $g\in G$, conjugation by $g$ restricted to $H$ is an inner automorphism of $H$ (note that since $H$ is abelian, all inner automorphisms of $H$ are trivial), $gh{g}^{-1}=h$ for all $h\in H$. This would imply $H\le Z(G)$, a contradiction.

Automorphism groups

Theorem 17(Theorem).

$\text{Aut}({\mathbb{Z}}_n)\cong U_n$, where $U_n$ is the group of units modulo $n$.

Proof
Let $x$ be a generator of ${\mathbb{Z}}_n$. If $\varphi \in \text{Aut}({\mathbb{Z}}_n)$, then $\varphi (x)=x^a$ for some $a\in \mathbb{Z}$, and $a$ uniquely determines $\varphi$. Denote this automorphism by ${\varphi }_a$. Since ${\varphi }_a$ is an automorphism, $x$ and $x^a$ must have the same order, so $(a,n)=1$. Furthermore, for every $a$ relatively prime to $n$, the map $x\mapsto x^a$ is an automorphism of ${\mathbb{Z}}_n$. Hence we have a bijective map

$$\begin{array}{rl}\Phi :\text{Aut}({\mathbb{Z}}_n)& \to U_n\\ {\varphi }_a& \mapsto a\mathrm{mod}n\end{array}$$

which is a homomorphism because

$${\varphi }_a\circ {\varphi }_b(x)={\varphi }_a(x^b)=x^{ab}={\varphi }_{ab}(x)$$

for all ${\varphi }_a,{\varphi }_b\in \text{Aut}({\mathbb{Z}}_n)$, so that

$$\Phi ({\varphi }_a\circ {\varphi }_b)=\Phi ({\varphi }_{ab})=ab\mathrm{mod}n=\Phi ({\varphi }_a)\Phi ({\varphi }_b).$$

Theorem 18(Theorem).

If $G$ and $H$ are two groups whose orders are relatively prime, then $\text{Aut}(G\times H)\cong \text{Aut}(G)\times \text{Aut}(H)$.

Characteristic groups

Definition 19(Definition).

A subgroup $H$ of a group $G$ is called characteristic in $G$, denoted $H◀B$, if every automorphism of $G$ maps $H$ to itself.

Note that this definition is stronger than that of a normal subgroup, which only requires every inner automorphism of $G$ to map $H$ to itself.

It should be obvious that characteristic subgroups are normal. Also, if $H$ is the unique subgroup of $G$ of a given order, then $H◀G$, since every $\sigma \in \text{Aut}(G)$ must map $H$ to another subgroup of $G$ that is isomorphic to $H$, and only one such group exists, namely $H$ itself.

Theorem 20(Theorem).

If $K◀H$ and $H◃G$, then $K◃G$.

Proof
Let $g\in G$. Let ${\varphi }_g\in \text{Aut}(H)$ be the conjugation $x\mapsto gx{g}^{-1}$ restricted to $H$. Note that this may not be an inner automorphism of $H$; this is why $K◃H$ does not suffice. Since $K◀H$, ${\varphi }_g(K)=K$. Thus, $gK{g}^{-1}=K$ for all $g\in G$, and $K◃G$.