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Let $G$ be a group with center $Z(G)$ and let $\sim$ denote the conjugancy equivalence relation on $G$ (that is, $a \sim b$ iff $\exists g \in G$ s.t. $a = gbg^{-1}$). Finally let $[a]$ denote the conjugacy class which contains $a \in G$.

Now if $a \in Z(G)$ it is clear that $[a] = \{a\}$, for if $b = gag^{-1}$, then $b = a$ so that $[a] = \{a\}$ as desired.

Yet it is not clear to me why the converse is true (that is if $[a] = \{a\}$ that we necessarily have $a \in Z(G)$). If we assume that $[a] = \{a\}$, then clearly there exists a $g \in G$ s.t. $a = gag^{-1}$ (since $\sim$ is an equivalence relation which means $a \sim a$ $\forall a \in G$). But we cannot know that $a \in Z(G)$ until all such elements $g \in G$ satisfy this property, not merely a single one of them. How do I show this?

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2 Answers 2

up vote 3 down vote accepted

Note that $a=gbg^{-1}$ if and only if $b=(g^{-1})a(g^{-1})^{-1}$. Hence, if $b\in[a]$, then $b\in\{hah^{-1}:h\in G\},$ and the converse holds, as well. In other words, $[a]=\{hah^{-1}:h\in G\}$. If $[a]=\{a\}$, then $hah^{-1}=a$ for all $h\in G$, so $a\in Z(G)$.

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The conjugacy class of an element is its orbit under conjugation by the whole group. In other words, the conjugacy class of $x$ is the set $\{g^{-1}xg : g \in G\}$. If this set only consists of $x$, then necessarily $g^{-1}xg =x$ for every $g\in G$. So $x\in Z(G)$.

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I'm not sure how we get from $Orb(a) = \{gag^{-1} : g \in G\}$ to $Orb(a) = [a] = \{b \in G$ : s.t. $\exists g \in G$ whereby $gbg^{-1} = a\}$. These two sets are for some reason not obviously the same to me. –  user1770201 Nov 24 '12 at 10:09
@user1770201 $b=g^{-1}ag$ if and only if $a=(g^{-1})^{-1}b(g^{-1})$. –  Alexander Gruber Nov 24 '12 at 16:47

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