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Let $a$ be an element of a group G. Prove that $a$ commutes with each of its conjugates in $G$ iff $a$ belongs to an abelian normal subgroup of $G$.

My Try:

I proved the backward direction. For forward one, given $g\in G, a(gag^{-1})=(gag^{-1})a$. I was trying to prove that $a\in Z(G)$. In order to prove that, I must show that $ag=ga$. But failed. Am I on the right track? Can anybody please give me a hint? I want to try it myself.

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  • $\begingroup$ The statement $a \in Z(G)$ is too strong $\endgroup$ Aug 6, 2015 at 14:58

2 Answers 2

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Let $H$ be generated by all conjugates of $a$. Then $H\lhd G$ because it is conjugation-invariant. Use the given condition that $H$ is abelian.

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Consider the group $$A =\left \langle a^g|g\in G \right \rangle$$ It is straightforward to check that it is a normal subgroup of $G$ and it is clear that $a\in A.$

Let us check that it is an abelian subgroup

$$a^ga^h = g^{-1}agh^{-1}ah = g^{-1}a(gh^{-1}ahg^{-1})g = g^{-1}aa^{hg^{-1}}g = g^{-1}a^{hg^{-1}}ag = a^ha^g.$$

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