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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.

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suppose $ga = ag$ then $gag^{-1} = agg^{-1} = a$ hence $g^{-1}gag^{-1} = g^{-1}a$ or $ag^{-1} = g^{-1}a$. – Deven Ware Dec 7 '11 at 0:22

Deven has given you a fine answer in the comments. Here is a slightly different way of thinking about the problem which foreshadows group actions, which you will surely learn about soon.

For $x \in G$, to say that $x \in Z(a)$ is equivalent to saying that forming the conjugate $xax^{-1}$ of $a$ by $x$ yields $a$ again. Now, what happens when you conjugate both sides of the equality $xax^{-1} = a$ by $x^{-1}$?

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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 ${C_a} \le G$.

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