Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
2  
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

2 Answers 2

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 $x^{-1}$?

share|improve this answer

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 ${C_a} \le G$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.