Let $G$ be a group and let $a \in G$ be a fixed element. Let $$H= \{\text{$g \in G$ | $g^{-1}ag=a$}\}.$$

Prove that $H$ is a subgroup of $G$.

So I know I need to show:

(i) $H$ is closed under the operation on $G$.

(ii) $H$ is closed under inversion.

(iii) $H\neq \emptyset$

For closure:

Suppose $x,y \in H.$ Then $x^{-1}ax=a$ and $y^{-1}ay=a.$ So $$\begin{align} (xy)^{-1}a(xy)& = y^{-1}x^{-1}axy \\ & = y^{-1}(x^{-1}ax)y \\ & = y^{-1}ay \\ & = a \\ \end{align}$$ Thus $xy \in H$.

Now how do I show it's closed under inversion? I'm not sure what the inverse is...

And to show $H \neq \emptyset$ all I need to do is show that the identity is in the set. And I'm not sure how to do that either.


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    $\begingroup$ $H\ni g\iff g^{-1}ag=a\iff a=gag^{-1}\iff g^{-1}\in H$ $\endgroup$ – blue Mar 10 '14 at 5:12
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    $\begingroup$ $H$ is known as "centralizer of $a$ in $G$". You can look at that condition $g^{-1}ag=a$ as $ag=ga$. $\endgroup$ – Woria Mar 10 '14 at 5:18
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    $\begingroup$ Of course solutions have been given, but it seems to me that an answer to the actual question would have been: You have to show that if $g\in H$, then also $g^{-1}\in H$. $\endgroup$ – Carsten S Mar 10 '14 at 10:44


$$g^{-1}ag=a\iff a=gag^{-1}$$



For more understanding:

The subgroup $H$ is very famous in group theory. For every subset $X\subseteq G$, we can define $C_G(X)=\{g \in G\,|\, g^{-1}xg=x\,$ for every $x\in X\}$ and $C_G(X)$ is called the centralizer of $X$. Now, for this case, put $X=\{a\}$.

Theorem: Let $G$ be group and $X\subseteq G$. Then $C_G(X)$ is a subgroup of $G$.

Proof: Assume that $g_1,g_2\in C_G(X)$. So for every $x\in X$, $g_1x=g_1x$ and $g_2x=xg_2$. Thus, $x=g_2^{-1}xg_2$ and so $xg_2^{-1}=g_2^{-1}x$. We claim that $g_1g_2^{-1}\in C_G(X)$. Suppose that $x\in X$ is an arbitrary element of $X$.

$g_1g_2^{-1}x=g_1xg_2^{-1}=xg_1g_2^{-1}$. Hence, $g_1g_2^{-1}\in C_G(X)$ and the claim is proved.


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