Conjugacy Classes

A conjugacy class of a group is a set of elements that are connected by an operation called conjugation. This operation is defined in the following way: in a group $G$, the elements $a$ and $b$ are conjugates of each other if there is another element $g\in G$ such that $a=gbg^{-1}$.

Conjugacy classes partition the elements of a group into disjoint subsets, which are the orbits of the group acting on itself by conjugation. Conjugacy classes of a group can be used to classify groups; they can be used to show that two groups are not isomorphic, or to discover properties of an isomorphism between groups if it exists. In general, the sizes of conjugacy classes in a group give information about its structure.

Here is the definition of conjugacy, repeated for clarity.

An element $b$ in a group $G$ is conjugate to an element $a$ if there is a $g\in G$ such that $a = gbg^{-1}.$ $($Alternatively, one says that $a$ is a conjugate of $b.)$

To frame this in the language of group actions, consider the function $\varphi \colon G \times G \to G$ defined by

$$\varphi(g,b) = gbg^{-1}.$$

Then $\varphi(e,b) = ebe^{-1} = b$ for all $b,$ and

$$\varphi\big(g,\varphi(h,b)\big) = \varphi\big(g,hbh^{-1}\big) = g\big(hbh^{-1}\big)g^{-1} = (gh)b\big(h^{-1}g^{-1}\big) = (gh)b(gh)^{-1} = \varphi(gh,b).$$

This shows that $\varphi$ defines a group action of $G$ on itself.

Then the conjugates of $b$ are precisely the members of the orbit of $b$ under the action. The stabilizer of $b$ is the subgroup of elements $g$ such that $gbg^{-1}=b,$ or $gb=bg.$ This is called the centralizer of $b,$ the subgroup of elements of $G$ which commute with $b.$

As mentioned above, elements $a$ and $b$ of a group are conjugate if there is another element $g$ such that $a=gbg^{-1}$. In fact, this is an equivalence relation. This follows immediately from the general theory of group actions, but it is instructive to prove it directly as well.

Conjugation is an equivalence relation.

There are three properties to check:

• Reflexivity, that $a$ is conjugate to $a$. This follows from $a=eae^{-1}$, where $e$ is the identity element of the group.
• Symmetry, that $a$ being conjugate to $b$ implies $b$ being conjugate to $a$. This follows from $$a=gbg^{-1}\implies b=g^{-1}a\big(g^{-1}\big)^{-1}.$$
• Transitivity, that $a$ being conjugate to $b$ and $b$ being conjugate to $c$ implies that $a$ is conjugate to $c$. This follows from $$a=gbg^{-1}, b=hch^{-1}\implies a=(gh)c(gh)^{-1}.$$

It follows now that conjugation partitions $G$ into equivalence classes, and this gives the formal definition of conjugacy classes.

The conjugacy classes of $G$ are the equivalence classes produced by the relation of conjugation. So a conjugacy class in $G$ is a subset of $G$ consisting of elements which are all conjugate to one another.

Equivalently, the conjugacy classes are the orbits of $G$ acting on itself by conjugation.

In linear algebra, the matrix group $\text{GL}_n(\mathbb{R}$) contains all $n\times n$ invertible matrices with real entries. In this group, two matrices $A$ and $B$ are conjugates if there is a matrix $P$ such that $A=PBP^{-1}$, which corresponds to matrix similarity. The conjugacy classes of this group then are the sets of matrices that represent the same linear transformation in different bases.

In the symmetry group $S_3$, there are 3 conjugacy classes:

• There is the identity permutation, which does nothing and is in its own class.
• There are the cyclic permutations, which take $abc$ to $bca$ or $abc$ to $cab$.
• Finally, there are the permutations that swap two elements—these are $abc$ to $bac$, $abc$ to $acb$, and $abc$ to $cba$.

Thus there are 3 conjugacy classes of $S_3$.

The final example of the previous section illustrates that the sum of the sizes of the conjugacy classes must be equal to the size of the group. This fact, along with the orbit-stabilizer theorem, can be used to derive an important equation known as the class equation.

Recall from the above discussion that the centralizer $C_G(b)$ of an element $b\in G$ is the subgroup of $G$ consisting of elements which commute with $b;$ it is the stabilizer of $b$ with respect to the action of conjugation.

Let $G$ be a finite group with $k$ conjugacy classes, and pick representatives $g_1,\ldots,g_k$ of each class. Then $$|G| = \sum_{i=1}^k \big[G:C_G(g_i)\big].$$ Alternatively, suppose $g_1,\ldots,g_\ell$ are representatives of the $\ell$ conjugacy classes with more than one element. Then $$|G|=|Z(G)|+\sum_{i=1}^\ell \big[G:C_G(g_i)\big],$$ where $Z(G)$ is the center of $G,$ consisting of all the elements which commute with every element of $G.$

$|G|$ equals the sum of the sizes of its conjugacy classes. The orbit-stabilizer theorem says that the size of the conjugacy class of an element equals the index of its stabilizer, and the stabilizer of $g_k$ is $C_G(g_k)$ as discussed above. Putting these facts together gives the first formula immediately.

The second formula is an immediate consequence of the first, because the conjugacy classes with one element correspond precisely to the elements of the center: $gbg^{-1}=b$ for all $g\in G$ if and only if $gb=bg$ for all $g\in G,$ which happens if and only if $b \in Z(G).$ $_\square$

This formula has quite a few well-known applications. One of them is in the $p$-group wiki: the center of a group of prime power order is always non-trivial. Here is another.

Write down the class equation of $A_5.$ Use it to show that $A_5$ is a simple group.

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The elements of $A_5$ come in four types: the identity, products of two disjoint transpositions, 3-cycles, and 5-cycles.

Any product of two disjoint transpositions is conjugate to $(12)(34)$ in $A_5.$ Proof: let $h$ be such a product; then $h$ is conjugate to $(12)(34)$ in $S_5.$ So $g(12)(34)g^{-1} = h$ for some $g\in S_5.$ If $g\in A_5,$ then we are done; otherwise, if $g$ is odd, let $g' = g(12)$ and note that $g'(12)(34)(g')^{-1} =h$ as well, and $g'$ is even. Note that there are $10 \cdot \frac32 = 15$ elements of this form.

Any 3-cycle is conjugate to $(123)$ in $A_5.$ Proof: same idea as above; if $g(123)g^{-1} = h,$ then either $g$ or $g'=g(45)$ is even, and both of them conjugate $(123)$ to $h.$ Note that there are $\binom{5}{3} \cdot 2 = 20$ elements of this form.

On the other hand, two 5-cycles that are conjugate in $S_5$ may not be conjugate in $A_5.$ In fact, it is impossible for all 5-cycles to be conjugate in $A_5,$ because there are $24$ of them, and the size of a conjugacy class must divide $60.$ Since there are 24 5-cycles in $S_5,$ the centralizer of a 5-cycle in $S_5$ has size $5,$ and thus must consist only of the powers of the 5-cycle. This must also be true in $A_5,$ so the conjugacy class of a 5-cycle must have size $\frac{60}5 = 12.$

Putting this all together, the class equation of $A_5$ is

$$60 = 1 + 15 + 20 + 12 + 12.$$

Any proper normal subgroup of $A_5$ must be a union of some of these conjugacy classes from $A_5.$ So its class equation will give its order as the sum of some subset of the above terms, including $1$ (since it must contain the identity). The order of a proper normal subgroup is a proper nontrivial divisor of $60.$ But there is no way to add up any subset of the other four terms, plus $1,$ to get a proper nontrivial divisor of $60.$ So $A_5$ has no nontrivial proper normal subgroup; so it is simple. $_\square$