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Let $(G,∗)$ be a group and $a\in G$ then so that set of elements $x$ of $G$ such that $a∗x = x∗a$ is a subgroup of $G$.

I have tried by using theorem that $H$ is as subgroup of $G$ if and only if for any $a,b\in H$ ,

$a∗b^{-1}\in H$

but didn’t get the result

Please help me

Martin Sleziak
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kalpeshmpopat
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    Please stop giving your posts useless titles. – Alexander Gruber Mar 14 '13 at 07:42
  • @kalpeshmpopat Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. In this case I've changed the title. Please check, whether this title corresponds with what you want to ask. – Martin Sleziak Mar 14 '13 at 07:58

2 Answers2

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This subset is called centralizer of $x$ and it is often denoted $$C_G(x)=\{a\in G; xa=ax\}.$$

You can verify that this is a subgroup like this:

  • $C_G(x)\ne\emptyset$ since $e\in C_G(x)$
  • $a,b\in C_G(x)$ $\Rightarrow$ $(ax=xa) \land (bx=xb)$ $\Rightarrow$ $(ab)x=a(bx)=a(xb)=(ax)b=(xa)b=x(ab)$ $\Rightarrow$ $ab\in C_G(x)$
  • $a\in C_G(x)$ $\Rightarrow$ $ax=xa$ $\Rightarrow$ $axa^{-1}=x$ $\Rightarrow$ $xa^{-1}=a^{-1}x$ $\Rightarrow$ $a^{-1}\in C_G(x)$

You can find a proof that it is a subgroup also at ProofWiki.

Another useful thing to notice is that $x\in C_G(x)$ and thus the subgroup generated by $x$ is in centralizer, $\langle x \rangle \subseteq C_G(x)$.

Martin Sleziak
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  • I've forgotten to show that $C_G(x)\ne\emptyset$, now I've added this (after I have seen the other answer, where this was shown, too). – Martin Sleziak Mar 14 '13 at 07:34
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Hint: let denote by $H$ the set of elements $x$ of $G$ such that $a∗x = x∗a$. You must first show $H$ is non-empty, which is easy because it's easy to prove that $e\in H$, ($e$ is the identity element of $G$) and $H$ is closed under multiplication.

Let $b\in G$ s.t $ba=ab$ then multiplying the equality on the LHS and RHS by $b^{-1}$ gives $ab^{-1}=b^{-1}a$ so $H$ is closed under taking inverse. Now you can conclude.

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