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That is, the set of elements of a group that commute with all elements of the group is a subgroup. What is the question asking? From what I'm interpreting it as is that for all elements of $a$ that can commute with $x$ is in a subgroup of $G$. However, if that's the case, how would I go about that? I believe this subgroup would be an abelian group.

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I think this is nothing but the center (Z) of the group G and center of any group is always a subgroup of the group G. So, question is, prove that center of the group is always a subgroup of the group G.

I think this might be helpful for you

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  • $\begingroup$ If you need proof, then i would like to prove it in my free time, thanks $\endgroup$ – Krishna Srivastav May 13 '16 at 7:11
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Denote $Z=\{a\in G,\,\forall x\in G\space ax=xa\}$. To prove that this sunset we need to prove the following

$$\begin{align} &e\in G\\&\forall a,b\in Z,\, ab^{-1}\in Z\end{align}$$

We have $\forall x\space ex=xe=x$ by definition of the neutral element and we have the first one.

For the second $ab^{-1}x=a(b^{-1}x)=(b^{-1})xa$ because $a\in Z$. On the other hand $bx^{-1}=x^{-1}b$ because $b\in Z$ and so by taking the inverse $xb^{-1}=b^{-1}x$ (remember $(xy)^{-1}=y^{-1}x^{-1}$. We get $ab^{-1}x=xb^{-1}a=xab^{-1}$ because $a\in Z$ commutes with $b^{-1}$

And we're done

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