TITLE: Prove the centralizer of an element in group $G$ is a subgroup of $G$
QUESTION [5 upvotes]: We have a group $G$ where $a$ is an element of $G$. Then we have a set $Z(a) = \{g\in G : ga = ag\}$ called the centralizer of $a$. If I have an $x\in Z(a)$, how do I go about proving that the inverse of $x$, $x^{-1}$, is also an element of $Z(a)$? I have already proved step 1, the subgroup test: I just need step 2, described above, and I have no idea how to start.

REPLY [2 votes]: Let ${C_a} = \left\{ {g \in G:ag = ga} \right\}$. Since 
$\left( {gh} \right)a = g\left( {ha} \right) = g\left( {ah} \right) = \left( {ga} \right)h = \left( {ag} \right)h = a\left( {gh} \right) \Rightarrow gh \in {C_a}$ and 
$ga = ag \Rightarrow {g^{ - 1}}\left( {ga} \right){g^{ - 1}} = {g^{ - 1}}\left( {ag} \right){g^{ - 1}} \Rightarrow a{g^{ - 1}} = {g^{ - 1}}a \Rightarrow {g^{ - 1}} \in {C_a}$ for every 
$g,h \in {C_a}$, we get ${C_a} \le G$.