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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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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