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On this page:

• Introduction
• Basic Definitions
• Examples of Conjugacy Classes
• The Class Equation
• Examples of the Class Equation
• Conjugacy Classes of $\Sym_n$
• Worked Exercises

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

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

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

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

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

gives the desired bijection.

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

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

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

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

[/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,

Thus,

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

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,

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

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

which means

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

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

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

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

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

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

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

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

[/theorem]

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

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

[/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:

Now we define $\tau$ by sending

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

[/example]

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

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

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

So, by Lagrange’s theorem, we have

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

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

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]KU Algebra Qual August 2019 Problem 4[/extitle]

Let $[x]$ denote the conjugacy class of an element $x \in G$.

1. Prove that for $x \in G$, $[x]$ has exactly one element if and only if $x \in Z(G)$.
2. More generally, prove that the number of elements in $[x]$ is $[G:C_G(x)]$.
3. Suppose that $G$ is a finite group of odd order, and suppose that $N$ is normal subgroup of $G$ of order $3$. Prove that $N \le Z(G)$. [/example]

[proof] (1) If $[x]$ has exactly one element, then $x$ by definition is fixed under the conjugation action which implies that $x \in Z(G)$. Conversely, if $x \in Z(G)$, then $xg = gx$ for all $g \in G$ which implies that $x = gxg^{-1}$. Thus, $[x] = \{x\}$ as desired.

(2) By Lagrange, we know that

The orbit-stabilizer theorem, applied to the conjugation action of $G$ on itself says that

But, by definition, $\Stab_G(x)$ is just $C_G(x)$. So it follows that $|[x]| = [G:C_G(x)]$ as desired.

Alternatively, if we want to do an argument from scratch, let $f:[x] \to G/C_G(x)$ be given by sending $h \mapsto hC_G(x)$. For well-definedness, consider $gxg^{-1} = hxh^{-1}$. Then $h^{-1}g x (h^{-1}g)^{-1} = x$. Thus, $h^{-1}g \in C_G(x)$ which implies that $hC_G(x) = gC_G(x)$. Surjectivity is immediate since we may send $hxh^{-1}$ to $hC_G(x)$. For injectivity, note that $hC_G(x) = gC_G(x)$ implies that $h^{-1}g \in C_G(x)$. Thus, $h^{-1}g x (h^{-1}g)^{-1} = x$ and so $hxh^{-1} = gxg^{-1}$. So $f$ is a bijection and the claim follows.

(3) Let $N = \{1, a, b\}$. Then the normality of $N$ tells us that $[a] = \{a\}$ or $[a] = \{a, b\}$. We claim the latter is impossible. Indeed, by Lagrange’s theorem,

where the second equality follows from part 2. If this is the case, then $G$ has even order which is a contradiction. On the other hand, if $[a] = \{a\}$, then $[b] = \{b\}$. So then it follows that each element of $N$ is in $Z(G)$ by part 1 and we are done. [/proof]

[example] [extitle]KU Algebra Qual January 2017 Problem 1[/extitle]

Let $G$ be a finite group with $|G| = p^n$, $p$ prime.

1. Prove that $Z(G)$, the center of $G$, is non-trivial.
2. Prove that if $N\subseteq G$ is a normal subgroup of order $p$, then $N\subseteq Z(G)$.
3. Give an example of a non-abelian group of order $p^n$ whose center contains more than one normal subgroup of order $p$. [/example]

[proof] (1) By the class equation, we know that

where the summation is taken over representors of distinct conjugacy classes of $G$ and $[G:C_G(x)]$ is the index of $C_G(x)$ in $G$. Then taking both sides modulo $p$ gives

where the summation vanishes due to each index being a multiple of $p$ by Lagrange’s theorem. Thus, $Z(G)$ has order that is a multiple of $p$. Since $1 \in Z(G)$, it follows that $Z(G)$ is nontrivial.

(2) Let

Now as $N$ is normal, it follows that each conjugacy class $[a_i]$ of $a_i$ is a subset of $N$. Furthermore, notice that $1\notin [a_i]$ since $1$ constitutes its own conjugacy class (which is a singleton). Now since we know that $|[a_i]| = [G:C_G(a)]$, it follows that $|[a_i]|$ is a factor of $p^n$. But since $N$ has $p$ elements, this implies that $|[a_i]| = 1$. So each element has a trivial conjugacy class which means that $a_i \in Z(G)$ as desired. [/proof]

[example] [extitle]KU Algebra Qual January 2024 Problem 5[/extitle]

How many distinct conjugacy classes are there in the symmetric group $\Sym_5$. [/example]

[proof] Given any permutation $\sigma \in \Sym_5$, consider its disjoint cycle decomposition

arranged so that $n_1 \ge n_2 \ge \cdots \ge 0$. We call the sequence $(n_1, n_2, \ldots)$ the cycle type of $\sigma$ and we claim that another permutation $\sigma’\in \Sym_5$ is conjugate to $\sigma$ if and only if $\sigma’$ has the same cycle type as $\sigma$. Suppose that $\sigma’ = \tau\sigma\tau^{-1}$. Then we know that given any $i \in [5] \coloneqq \{1, 2, 3, 4, 5\}$ with $\sigma(i) = j$, it follows that

So $\tau(i)$ gets sent to $\tau(j)$. This tells us that

So $\sigma’$ has the same cycle shape as $\sigma$. Conversely, if we start with an arrangement of the cycles of $\sigma’$ to match the shapes of $\sigma$

then we may construct bijection $\tau \in \Sym_5$ by setting $\tau(a_i) = a_i’$, $\tau(b_i) = b_i’$, and so on. Running through the same argument above shows that $\tau\sigma\tau^{-1} = \sigma’$ which gives the desired claim.

Notice that this implies that result we proved above implies that the conjugacy classes of $\Sym_5$ are in bijection with the integer partitions of $5$. To see this, we note that any partition of $5$ defines a conjugacy class where the cycle type corresponds to that partition. For example, $5 = 2 + 2 + 1$ corresponds to the cycle type $(2, 2, 1)$. Conversely, any cycle type defines an integer partition of $5$ as the disjoint cycles (including the $1$-cycles) lengths must add to $5$. As the integer partitions of $5$ are

it follows that there are 7 distinct conjugacy classes of $\Sym_5$. [/proof]

[example] [extitle]KU Algebra Qual August 2016 Problem 3[/extitle]

Let $G$ be a group and $\Aut(G)$ be the group of automorphisms of $G$. Recall that for any $c\in G$, we have an automorphism of $G$ given by $f_c(x) = c^{-1}xc$ ( such an automorphism is called an inner automorphism of $G$). It is not hard to see that $\Inn(G)$, the collection of inner automorphisms of $G$, is a subgroup of $\Aut(G)$.

1. Show that $G/Z(G) \cong \Inn(G)$. Here $Z(G)$ is the center of $G$.
2. Show that if $\Aut(G)$ is cyclic, then $G$ is abelian.
3. Find all prime numbers $p$ such that there is a finite group $G$ with $\Aut(G) \cong \ZZ/p\ZZ$. [/example]

[proof] (1) Let $f:G \to \Inn(G)$ be given by $c\mapsto f_c$. Surjectivity is immediate and so the First Isomorphism Theorem implies that

We claim that $\ker f = Z(G)$. If $c \in \ker f$, then we know that $f_c$ is the identity mapping on $G$. So $c^{-1}xc = x$ for all $x \in G$. This implies that $c \in Z(G)$. Conversely, if $c \in Z(G)$, then we know that $c^{-1}xc = x$ for all $x\in G$ and so $f_c$ is the identity mapping on $G$. Thus, $\ker f = Z(G)$ and we are done.

(2) Any subgroup of a cyclic group is again cyclic. This implies that $G/Z(G)$ is cyclic. So if $g, h \in G$, then there is some $r \in G$ for which

This implies that there is some $s,t\in Z(G)$ for which

Thus,

So $G$ is abelian.

(3) If we let $G = \ZZ/3\ZZ$, then it is straightforward to show that $\Aut(G) \cong \ZZ/2\ZZ$. We claim that this is the only $p$ for which this is true (that is, $p = 2$).

Let $p > 2$ and consider the mapping $f:G \to G$ that sends $g$ to $g^{-1}$. It is straightforward to see that this is self-inverse and that this is a homomorphism (as $G$ is abelian by part 2). This implies that $\Aut(G)$ has an element of order $2$ assuming that $f$ is not the identity mapping. In particular, $f$ being the identity mapping would imply that $g^2 = 1$ for ALL $g\in G$. By the classification of finite abelian groups, this occurs nontrivially only when $G \cong (\ZZ/2\ZZ)^{\oplus n}$ as there being other types of summands would imply the existence of an element that has order greater than 2. So $\Aut(G)$ is isomorphic to $\Aut((\ZZ/2\ZZ)^{\oplus n})$. If $n = 1$, then we just get $(\ZZ/2\ZZ)^\times$. If $n > 1$, then we can just take the map defined by

This map is clearly involutive so $\Aut(G)$ contains an element of order $2$ which implies that $\Aut(G)$ cannot be isomorphic to $\ZZ/p\ZZ$.

REMARK. It was pointed out to me that you can also do this by noting that $\Aut(G)$ is isomorphic to $\GL_n(\ZZ/2\ZZ)$ as the classification theorem tells us that $G$ has a natural $\ZZ/2\ZZ$-vector space structure. [/proof]