Conjugation Group Action

Back to Algebra Page

On this page:

Introduction

One of the most important group actions in group theory is the action of a group on itself by conjugation. That is, if $G$ is a group, then we define

\[g\cdot a \coloneqq gag^{-1}\]

for every $g, a\in G$. It’s pretty immediate that this is a group action so we skip verification. This definition is frequently generalized to an action on the set $2^G$ by defining

\[g\cdot S \coloneqq gSg^{-1} = \{gsg^{-1} \mid s\in S\}.\]

This is also pretty clearly a group action and so we skip the verification that it is actually a group action.

Basic Definitions

[definition] [deftitle]Definition (conjugate elements).[/deftitle]

We say that $a, b\in G$ are conjugate in $G$ if

\[b = gag^{-1}\]

for some $g \in G$. Equivalently, $a$ and $b$ are in the same orbit of $G$ acting on itself by conjugation. We call the orbits of $G$ acting on itself by conjugation the congjugacy classes of $G$. [/definition]

[definition] [deftitle]Definition (conjugate subsets).[/deftitle]

We say that $S, T \subseteq G$ are conjugate in $G$ if

\[T = gSg^{-1}\]

for some $g \in G$. Equivalently, $S$ and $T$ are in the same orbit of $G$ acting on its subsets by conjugation. [/definition]

It should be fairly clear that the conjugaction action is strongly related to normalizers. If we are given a subset $S$ of $G$, then $G/N_G(S)$ annihilates all $g \in G$ for which $gSg^{-1} = S$. Accordingly, the index of $G/N_G(S)$ counts the number of conjugates of $S$. Let’s state this more formally.

[theorem] [thmtitle]Proposition.[/thmtitle]

Let $G$ be a group and $S \subseteq G$. Then the number of conjugates of $S$ is $[G:N_G(S)]$. In particular, the number of conjugates of $s \in G$ is $[G:C_G(s)]$. [/theorem]

[proof] We proceed by exhibiting a bijection between the conjugates of $S$ and $G/N_G(S)$. We claim that the map

\[gSg^{-1} \mapsto gN_G(S)\]

gives the desired bijection.

(Well-definedness) Let $gSg^{-1} = hSh^{-1}$. Then

\[S = h^{-1}gSg^{-1}h = (h^{-1}g)S(h^{-1}g)^{-1}.\]

In other words, $h^{-1}g \in N_G(S)$. So then it follows that

\[h^{-1}g N_G(S) = N_G(S) \quad\iff\quad gN_G(S) = hN_G(S).\]

(Injectivity) Suppose $gN_G(S) = hN_G(S)$. Then it follows that $h^{-1}g \in N_G(S)$ which means that

\[S = (h^{-1}g)S(h^{-1}g)^{-1} = h^{-1}g S g^{-1}h.\]

So it follows that $gSg^{-1} = hSh^{-1}$ and we obtain injectivity.

(Surjectivity) Obvious.

To prove the final claim, we let $S = {s}$ and note that $N_G(S) = C_G(s)$ and the claim follows pretty much immediately. [/proof]

Another to also realize this result is that this is really the bijection in the orbit-stabilizer theorem in disguise.

Examples of Conjugacy Classes

[example] [extitle]Conjugacy classes of abelian group.[/extitle]

Let $G$ be an abelian group. Then it follows immediately that the conjugaction action of $G$ on itself is trivial and the conjugacy classes of $G$ are of the form $\{a\}$ where $a \in G$. [/example]

[example] [extitle]Transitivity of the conjugacy action.[/extitle]

Let $G$ be a group such that $|G| > 1$. Then the conjugaction action fails to be transitive since $\{1\}$ is one of the conjugacy classes of $G$. Furthermore, $\{a\} \subseteq G$ is conjugacy class of $G$ provided that $a \in Z(G)$. [/example]

[example] [extitle]Conjugacy classes of $\Sym_n$.[/extitle]

The conjugacy classes of $\Sym_n$ are indexed by the integer partitions of $n$. We will discuss this result in more depth later down the page. [/example]

The Class Equation

The conjugacy action of $G$ on itself is a group action and we know that group actions partition $G$-sets into orbits. Applying this to the special case of the conjugaction action, $G$ partitions itself into conjugacy classes. Combining the proposition above that puts a bijection between the conjugates of $a \in G$ and $G/C_G(a)$ along with the partitioning-property of the orbits leads to what one calls the class equation.

[theorem] [thmtitle]Theorem (Class equation).[/thmtitle]

Let $G$ be a finite group and let $a_1, \ldots, a_r$ be representatives of the distinct conjugacy classes of $G$ not contained in $Z(G)$. Then

\[|G| = |Z(G)| + \sum_{1 \le i \le r} [G:C_G(a_i)]\]

[/theorem]

[proof] Since the conjugacy classes of $G$ are the orbits of the conjugaction action, they partition $G$. Let us call the conjugacy classes of $G$ that contain more than one element $C_1, \ldots, C_r$. Since ${z}$ is a conjugacy class if and only if $z \in Z(G)$, it follows that the remaining conjugacy classes are those given by ${z}$ where $z \in Z(G)$. Therefore,

\[G = \left(\bigsqcup_{z \in Z(G)}\{z\}\right) \sqcup \left(\bigsqcup_{1\le i \le r} C_i\right).\]

Thus,

\[|G| = |Z(G)| + \sum_{1 \le i \le r} |C_i|.\]

Note $C_i$ is in bijection with $G/C_G(a_i)$ where $a_i$ is some element of $C_i$. So it follows that

\[|G| = |Z(G)| + \sum_{1 \le i \le r} [G : C_G(a_i)]\]

and we are done. [/proof]

While the class equation looks like a stupid result (it did when *I* first learned about it), it gives significant restrictions on “what can happen in a group.” For instance, the bijection between each conjugcy class $C$ and $G/C_G(a)$ (where $a\in C$) tells us that $|C|$ is a divisor of $|G|$ by Lagrange.

[theorem] [thmtitle]Corollary.[/thmtitle]

Let $p$ be a prime and $G$ a group of order $p^n$ for some $n \ge 1$. Then $G$ has a nontrivial center $Z(G)$. [/theorem]

[proof] By the class equation,

\[|G| = |Z(G)| + \sum_{1 \le i \le r} [G:C_G(a_i)].\]

By Lagrange’s theorem, each $[G:C_G(a_i)]$ is a divisor of $p^n$ and is therefore divisible by $p$. Taking the class equation modulo $p$ gives

\[0 \equiv |Z(G)| \pmod{p}.\]

Since $Z(G)$ contains at least one element (the identity), it follows that $|Z(G)|$ must be a nontrivial multiple of $p$ and the claim follows. [/proof]

One final classic result that follows nicely from the class equation is that any group whose order is a square of a prime is abelian.

[theorem] [thmtitle]Corollary.[/thmtitle]

Let $p$ be a prime and $G$ a group of order $p^2$. Then $G$ is abelian and $G \cong \ZZ/p^2\ZZ$ or $G \cong (\ZZ/p\ZZ) \oplus (\ZZ/p\ZZ)$. [/theorem]

[proof] By the previous result, $Z(G)$ is nontrivial and so $Z(G)$ has order $p$ or $p^2$. In either case, $G/Z(G)$ is a cyclic group and we can show that this means $G$ is abelian. Indeed, if $a,b \in G$. then

\[aZ(G) = g^rZ(G) \quad\text{ and }\quad bZ(G) = g^sZ(G)\]

which means

\[a = g^rz_1 \quad\text{ and }\quad b = g^sz_2\]

for some $z_1, z_2 \in Z(G)$. It follows that

\[ab = (g^rz_1)(g^sz_2) = (g^sz_2)(g^rz_1) = ba.\]

Thus, $G$ is abelian. At this point, there are two ways to proceed: apply the classification of finitely-generated abelian groups or use the classic ad-hoc approach. We will go with the ad-hoc approach. If $G$ has an element of order $p^2$, then the final claim follows immediately. If not, then $G$ has some nonidentity element $x$ of order $p$ (otherwise all elements would have order 1). Then pick out some nonidentity element $y\in G\setminus\generator{x}$. Then $\generator{x, y} = G$. Now we claim that

\[\begin{align*} G &\to (\ZZ/p\ZZ) \oplus (\ZZ/p\ZZ) \\ x^ay^b &\mapsto (a, b) \pmod{p} \end{align*}\]

is an isomorphism. The fact that this map is a surjective homomorphism is immediate. To see the injectivity, we note that $a = b = 0$ implies that $a$ and $b$ are multiples of $p$. But since $x$ and $y$ are order $p$, $x^ay^b = 1$. So the claim follows. [/proof]

Examples of the Class Equation

[example] [extitle]Abelian groups.[/extitle]

One place where the class equation is actually totally useless is abelian groups. All conjugacy classes in an abelian group have cardinality 1 and so no new information is actually gained. [/example]

[example] [extitle]Groups of order 6.[/extitle]

The case where $G$ is abelian is fully taken care of by the classification of finitely-generated modules over a PID. In particular, $G \cong \ZZ/6\ZZ$ or $G \cong (\ZZ/2\ZZ) \oplus (\ZZ/3\ZZ)$.

In the case where $G$ fails to be abelian, $Z(G)$ has order $1, 2$, or $3$. If the order is either $2$ or $3$, then $G/Z(G)$ is a cyclic group. We claim that this implies that $G$ is actually abelian. Suppose that $gZ(G)$ generates $G/Z(G)$ and let $a, b \in G$. Then

\[aZ(G) = g^{r}Z(G) \quad\text{ and }\quad bZ(G) = g^{s}Z(G).\]

Thus, there are $z_1, z_2 \in Z(G)$ for which $a = g^r z_1$ and $b = g^s z_2$. It follows that

\[ab = (g^r z_1)(g^s z_2) = (g^s z_2)(g^r z_1) = ba.\]

Thus, $G$ is abelian which is a contradiction. Therefore, $Z(G)$ is trivial. Since the remark we made above forces the conjugacy classes to have cardinality a divisor of $6$, it follows that the only possibility for the class equation is

\[6 = 1 + 2 + 3.\]

With some ad-hoc arguments, one can show that $Z(G)$ being trivial implies $G \cong \Sym_3$. [/example]

Conjugacy Classes of $\Sym_n$

One of the examples of conjugacy classes that anyone in abstract algebra studies is conjugation in $\Sym_n$. In linear algebra, we know that given any $P \in \GL_n(\FF)$, conjugating $A$ by $P$ gives the matrix

\[PAP^{-1}.\]

That is, $P$ is realized as a change-of-basis transformation. In the symmetric group, conjugation is a change-of-labeling that is completely analogous to change-of-basis in matrices.

[theorem] [thmtitle]Theorem (“Change-of-labeling”).[/thmtitle]

Let $\sigma, \tau \in \Sym_n$ and suppose $\sigma$ has cycle decomposition

\[(a_1 \; a_2 \; \ldots \; a_{k_1})(b_1 \; b_2 \; \ldots \; b_{k_2}) \cdots.\]

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

\[(\tau(a_1) \; \tau(a_2) \; \ldots \; \tau(a_{k_1}))(\tau(b_1) \; \tau(b_2) \; \ldots \; \tau(b_{k_2})) \cdots.\]

[/theorem]

[proof] Suppose that $\sigma(i) = j$. Then

\[\tau\sigma\tau^{-1}\tau(i) = \tau\sigma(i) = \tau(j).\]

Thus, if $i \mapsto j$ under $\sigma$, we have $\tau(i) \mapsto \tau(j)$ under $\tau\sigma\tau^{-1}$. Therefore, in the cycle decomposition, we just replace each occurence of $i$ with $\tau(i)$. [/proof]

[example] [extitle]Example (from Dummit and Foote).[/extitle]

Let $\sigma = (1\;2)(3\;4\;5)(6\;7\;8\;9)$ and $\tau = (1\;3\;5\;7)(2\;4\;6\;8)$. Then

\[\tau\sigma\tau^{-1} = (3\;4)(5\;6\;7)(8\;1\;2\;9).\]

[/example]

Since conjugation in $\Sym_n$ is just a change-of-labeling, the “shapes” of the cycles is preserved under conjugation. This observation trivializes finding conjugacy classes in $\Sym_n$. Let us quickly formalize this:

[definition] [deftitle]Definition (Cycle type).[/deftitle]

Let $\sigma \in \Sym_n$. The cycle type of $\sigma$ is the integer partition of $n$ given by the sizes of the cycles of $\sigma$. [/definition]

[theorem] [thmtitle]Theorem (Conjugate elements in $\Sym_n$)[/thmtitle]

Two elements of $\Sym_n$ are conjugate if and only if they have the same cycle type. [/theorem]

[proof] ($\impliedby$) Follows immediately by the change-of-labeling theorem we proved above.

($\impliedby$) Say $\sigma$ and $\sigma’$ have the same cycle shape. Reorder the cycle decomposition so that both products “decrease” in cycle length as we read from left to right:

\[\begin{align*} \sigma &= (a_1 \; a_2 \; \ldots \; a_{k_1})(b_1 \; b_2 \; \ldots \; b_{k_2}) \cdots \\ \sigma' &= (a_1' \; a_2' \; \ldots \; a_{k_1}')(b_1' \; b_2' \; \ldots \; b_{k_2}') \cdots. \end{align*}\]

Now we define $\tau$ by sending

\[\begin{align*} a_1 &\mapsto a_1', \\ a_2 &\mapsto a_2', \\ a_{k_1} &\mapsto a_{k_1}', \\ &\;\;\vdots \\ b_1 &\mapsto b_1', \\ b_2 &\mapsto b_2', \\ b_{k_2} &\mapsto b_{k_2}', \\ &\;\;\vdots. \end{align*}\]

By the change-of-labeling theorem, it follows that $\sigma’ = \tau\sigma\tau^{-1}$. So the claim follows. [/proof]

[theorem] [thmtitle]Theorem (Conjugacy classes of $\Sym_n$).[/thmtitle]

The conjugacy classes of $\Sym_n$ are in bijection with the integer partitions of $n$. [/theorem]

[proof] The previous theorem implies that that each conjugacy class is uniquely determined by the cycle shape that all of its elements have so the claim follows. [/proof]

Worked Exercises

[example] [extitle]TODO: Cite[/extitle]

Let $S \subseteq G$ and $g \in G$. Then

\[gN_G(S)g^{-1} = N_G(gSg^{-1}) \quad\text{ and }\quad gC_G(S)g^{-1} = C_G(gSg^{-1}).\]

[/example]

[proof] We will start by proving the first claim. Suppose $gxg^{-1} \in gN_G(S)g^{-1}$. Noting that

\[(gxg^{-1})gSg^{-1}(gxg^{-1})^{-1} = (gx)S(x^{-1}g^{-1}) = gSg^{-1},\]

it follows that $gxg^{-1} \in N_G(gSg^{-1})$. Similarly, if $y \in N_G(gSg^{-1})$, then define $x \coloneq g^{-1}yg$. Then $gxg^{-1} = y$ and

\[xSx^{-1} = (g^{-1}yg)S(g^{-1}yg)^{-1} = S.\]

So $x \in N_G(S)$ which implies $y \in gN_G(S)g^{-1}$ as desired. A very similar argument will show the second claim so we omit the proof of the second claim. [/proof]

[example] [extitle]TODO: Cite[/extitle]

Let $G$ be a finite group. Then $[G:Z(G)] \ge |C|$ for every conjugacy class $C$ of $G$. [/example]

[proof] Recall that a given conjugacy class $C$ with representative $a$ is in bijection with $G/C_G(a)$. Furthermore, we know that

\[Z(G) \le C_G(a) \le G.\]

So, by Lagrange’s theorem, we have

\[[G:Z(G)] \ge [G:C_G(a)] = |C|\]

as desired. [/proof]

[example] [extitle]TODO: Cite[/extitle]

Let $G$ be a finite group with exactly two conjugacy classes. Then $G$ is isomorphic to $\ZZ/2\ZZ$. [/example]

[proof] Clearly $\ZZ/2\ZZ$ is one such group. We will show that there are no other such finite groups. One of the conjugacy classes is necessarily $\{1\}$ so the other conjugacy class $C$ must have $|G| - 1$ elements by the class equation. Since we showed that conjugacy classes have size that is a factor of $|G|$, it follows that

\[\frac{|G|}{|C|} = \frac{|G|}{|G|-1}\]

is an integer. But this can never be the case unless $|G| = 2$ and there is only one such group: $\ZZ/2\ZZ$. [/proof]

[example] [extitle]TODO: Cite[/extitle]

Let $G$ be a group of odd order and let $x$ be a non-identity element of $G$. Then $x$ and $x^{-1}$ are not conjugate in $G$. [/example]

[proof] For the sake of contradiction, suppose that $x$ and $x^{-1}$ are actually conjugate to each other. If $C$ is the conjugacy class that contains $x$, then $x^{-1} \in C$ as well. Note that this implies that $C$ is closed under inversion since any $a = hxh^{-1} \in C$ implies $a^{-1} = hx^{-1}h^{-1}$, and $x^{-1}\in C$ means $a^{-1} \in C$. Consider the map

\[\begin{align*} C &\to C \\ a &\mapsto a^{-1}. \end{align*}\]

Since $a \neq a^{-1}$ (because $|G|$ is odd and $a\neq 1$), this map has no fixed points and is therefore a fixed-point-free involution. Thus, $C$ can be partitioned up into pairs of sets of the form $\{a, a^{-1}\}$. But, since $C$ is in bijection with $[G:C_G(x)]$, $|G|$ is even by Lagrange’s theorem. Contradiction! [/proof]

[example] [extitle]TODO: Cite[/extitle]

Let $\sigma \in \Alt_n$. The conjugacy class of $\sigma \in \Sym_n$ splits into two conjugacy classes in $\Alt_n$ if and only if the cycle type consists of distinct odd integers. [/example]

($\implies$) Let $C$ be the conjugacy class of $\sigma \in \Sym_n$ and suppose $C$ splits into the conjugacy classes $C_1, C_2$ of $\sigma \in \Alt_n$.

($\impliedby$)