Conjugacy

We now return from an investigation of permutation groups to the general situation where $G$ is an arbitrary group. In this section, we will consider two special subgroups of $G$. Also we investigate a partition of $G$ into equivalence classes with respect to a certain equivalence relation (conjugacy) on $G$. Another partition of $G$ into equivalence classes (cosets) will be considered in the final section. If $a \in G$, we define the centralizer of $a$ in $G$, $C_G(a)$, as follows:

$$C_G(a)=\{g\in G\ \vert\ ag=ga\}.$$

Thus the centralizer of $a$ in $G$ consists of those elements in $G$ which commute with $a$. In exercise 3 for Chapter 2, it was shown that $C_G(a)$ is a subgroup of $G$. Next we consider the set

$$Z(G)=\cap_{a\in G} C_G(a).$$

Recall from exercise 7 for Chapter 2, that $Z(G) \subset G$ is called the center of $G$. Clearly, $g\in Z(G)$ if and only if $ga = ag$, for all $a \in G$, i.e., the center of a group consists precisely of those elements which commute with all elements of the group. Thus $G$ is abelian if and only if $G = Z(G)$. Since $Z(G)$ is a subgroup of $G$, the identity element $e\in Z(G)$. It is entirely possible that for a given group $G$, $Z(G) = \{e\}$ (e.g., see exercise 1 of this section). In this case, one says that the group $G$ has a trivial center; otherwise, one says the center of G is non-trivial. Again let $G$ be an arbitrary group. We introduce a relation on $G$ as follows: for $a,b \in G$, define $a \sim b$ if and only if there is a $g\in G$ such that

$$a = gbg^{-1}.$$ (4.1)

Elements $a,b \in G$ related as in (4.1) are called conjugate. We claim that (4.1) is an equivalence relation on $G$; viz,

(1)

(Reflexivity) $a \sim a$ since $a = eae^{-1}$.

(2)

(Symmetry) $a \sim b$ implies that there exists a $g\in G$ such that $a = gbg^{-1}$. Solving for $b$, $b = g^{- 1}ag = (g^{- 1})a(g^{- 1})^{- 1}$. Thus $b \sim a$.

(3)

(Transitivity) If $a \sim b$ and $b \sim c$, then there exist elements $g, h\in G$ such that $a = gbg^{-1}$ and $b = hch^{- 1}$. Therefore $a = ghch^{- 1}g^{- 1} = (gh)c(gh)^{- 1}$. So $a \sim c$.

Now $G$ is therefore partitioned according to Theorem 1.1.4 into disjoint equivalence classes $[a]$. For this particular equivalence relation, that of (4.1), we call the equivalence classes conjugacy classes and write $Cl(a)$ instead of $[a]$. Theorem 1.1.4 also yields

$$G=\coprod Cl(a),$$

where this disjoint union ($\coprod$ has the same meaning as $\cup$, except that the union is disjoint) is taken over certain $a \in G$. Let us see next what it means that $Cl(a) = \{a\}$. This is equivalent to the fact that $gag^{- 1} = a$ for all $g\in G$; i.e., $a\in Z(G)$. Thus $Cl(a) = \{a\}$ if and only if $a\in Z(G)$. If we collect all such one element conjugacy classes together, then we can write

$$G=Z(G)\cup (\coprod Cl(a)),$$ (4.2)

where the union is taken over certain $a \in G$ such that $\vert Cl(a)\vert\geq 2$. We turn to the case of $S_n$ to illustrate in a specific example the concept of conjugate elements. Let $f \in S_n$, then using the 2-row form of (2.1), we can display the effect of $f$ on the set $A = \{1, 2, ..., n\}$ by writing

$$f= \left( \begin{array}{cccc} 1 & 2 & ... & n\\ f(1)& f(2)& ...& f(n) \end{array} \right)$$

Now if $g\in S_n$ also, then $gfg^{- 1}$ maps $g(i)$ into $gf(i)$. So $gfg^{- 1}$ can be displayed by

$$gfg^{-1} = \left( \begin{array}{cccc} g(1) & g(2) & ... & g(n)\\ gf(1)& gf(2)& ...& gf(n) \end{array} \right).$$

(Note: since $g$ is 1-1 and onto $g[A] = \{g(1), g(2), ..., g(n)\} = A$.) Thus, using the cycle representation of $f$ from Theorem 3.1.2, we may write $f$ as in (3.2), i.e., $f=(a_1,...,a_k)(b_1,...,b_\ell)...(h_1,...,h_t)$, Then since e.g. $f(a_i) = a_{i+1}$ ($1 \leq i < k$) and $f(a_k) = a_1$ then $gfg^{- 1}(g(a_i)) = g(a_{i+1})$ ($1 \leq i < k$) and $gfg^{- 1} (g(a_k)) = g(a_1)$, we can write

$$gfg^{-1}=(g(a_1),...,g(a_k))(g(b_1),...,g(b_\ell))... (g(h_1),...,g(h_t)).$$ (4.3)

As an illustration, consider $S_5$ and let $f = (1,5) (2,3,4)$, and $g = (1,2,3,4,5)$. To obtain $gfg^{- 1}$, we must just see what $g$ does to the elements occurring in $f$, e.g., $g(1) = 2$, $g(5) = 1$, etc.; hence, $gfg^{- 1} = (2,1) (3,4,5)$. Going back to the more general situation, let us assume now that the cycle representation of $f$ given in (3.2) above is such that $k \geq \ell \geq ... \geq t$ and also let's assume that all cycles, including even the 1-cycles, i.e., elements left fixed by $f$, are present. Then

$$M=k+\ell +...+t.$$

This is called a partition of $n$. If $f$ and $\hat{f}$ are two permutations of $\{1,2,...,n\}$ which are conjugate by an element of $S_n$, then observing that the cycle structure of (4.3) is the same as that of (3.2), we see that the same partition of $n$ is associated with them. Conversely, it is easy to see that permutations having the same cycle structure must be conjugate (see exercise 2 of this section). We have thus proven the following result.

Theorem 4.1.1   Two permutations of degree n are conjugate (in $S_n$) if and only if they have the same cycle structure, i.e., if and only if they induce the same partition of n. Moreover, the number of conjugacy classes in $S_n$ is equal to the number of partitions of n.