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Let $a\in G$, where $G$ is a group. Prove that $a$ commutes with each of its conjugates in $G$ if and only if $a$ belongs to an abelian normal subgroup of $G$.

This is what I did:

$"⟹"$

First, I thought of $<a>$ to be that normal subgroup containing a. But if it is true, I am not sure how to prove it. Then I thought of another subgroup, namely $C_G(a)$. I conclude from the hypothesis given that $Cl_G(a) ⊂ C_G(a)$ And I know that $|Cl_G(a)| = |G : C_G(a)|$ by G acting transitively on $Cl(a)$ by conjugation. But I don't know how to get that $ C_G(a)$ is normal in $G$.

I am thinking also of a normal subgroup of $G$ , say $K$ containing $a$, so $<a>$ $\cap$ $K$ and $ C_G(a)$ $\cap$ $K$ are subgroups of G inheriting abelianity from $<a>$ and $ C_G(a)$ respectively, and normality from $K$.

$"\Leftarrow"$

Let $a ∈ H$ s.t. $H$ is an abelian normal subgroup of G. So $ah=ha$, ∀ $h∈H$.

Since $H$ is normal in $G$, hence $g^{-1}hg ∈ H$, ∀ $g∈G$

So $g^{-1}ag ∈ H$, ∀ $g∈G$

Which means $Cl_G(a)$ $\subseteq$ $ H$

Therefore, a commutes with each of its conjugates in G.

Could someone correct me?

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marked as duplicate by Derek Holt group-theory Aug 17 '15 at 16:45

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  • $\begingroup$ Your implication "$\Leftarrow$" is correct. $\endgroup$ – Crostul Aug 17 '15 at 15:15
  • $\begingroup$ @Crostul thanks. $\endgroup$ – Fabian Aug 17 '15 at 15:32
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For the implication "$\Rightarrow $", you're on the right track. Consider the subgroup $H$ generated by all conjugates of powers of $a$. It's easy to see that this is a normal subgroup, since the generating set is closed under conjugation.

To see that the subgroup is abelian, it suffices to prove that generators commute (since the generating set is closed under taking inverses). So consider two generators $ha^mh^{-1}$ and $g a^n g^{-1}$ of $H$, where $h,g \in G$.

$$ (g a^n g^{-1})*( h a^m h^{-1}) = g a^n *(g^{-1} h a^m h^{-1}g)*g^{-1} = g * (g^{-1} h a^m h^{-1}g) *a^n g^{-1},$$ where the last equality comes from the commutativity assumption. Reassociating and canceling, we get:

$$ (g a^n g^{-1})*( h a^m h^{-1}) = ( h a^m h^{-1}) *(g a^n g^{-1}).$$

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  • $\begingroup$ Great! Thank you so much MPO. $\endgroup$ – Fabian Aug 17 '15 at 15:41

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