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$\begingroup$

Let $G$ be a group. We define

$$H=\{h\in G\mid \forall g\in G: hg=gh\},$$

the center of $G$.

Prove that $H$ is a (normal) subgroup of $G$.

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    No, it was my exam question. Just finnished it. I think I managed it up to part to prove that it is normal. But I couldn't prove that it is a subgroup.2011-09-20
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    http://www.proofwiki.org/wiki/Center_is_a_Normal_Subgroup2011-09-20
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    I think this answers everything. Thanks!2011-09-20
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    Just curious: what does it mean to prove that it is normal without proving that it is a subgroup?2011-09-20
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    @ShreevatsaR Let $S$ be a subset of a group $G$. We say that $S$ is a **normal subset** of $G$ if for all $s\in S$ and $g\in G$, we have $g^{-1}sg\in S$. In particular, a subset $S$ of $G$ is a normal subgroup of $G$ if and only if $S$ is a subgroup of $G$ and $S$ is a normal subset of $G$.2011-09-20
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    @AmiteshDatta: Thanks.2011-09-20

3 Answers 3

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As you said in a comment you already showed that it is normal. So I will only show that it is a subgroup.

Clearly it contains $e$, since $eg = ge$.

Now, we will show that it is closed. Let $a,b \in H$, we know that $\forall g: ag = ga$ and $gb = gb$. Thus, $gab = agb = abg$ and thus $ab \in H$.

Now we only have to show that every $h \in H$ has an inverse and we are done. Let $h \in H$, we know that $\forall g \in G: gh = hg$, thus $$\begin{align*}h^{-1}(gh)h^{-1} &= h^{-1}(hg)h^{-1}\\ h^{-1}g(hh^{-1}) &= (h^{-1}h)gh^{-1}\\ h^{-1}(ge) &= (eg)h^{-1}\\ h^{-1}g &= gh^{-1} \end{align*}$$

Which implies that $h^{-1} \in H$.

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sxd has shown that it is a subgroup.

That it is normal follows from here:

Let $x\in Z(G)$ (center of $G$).

Then for any $g\in G$, $gxg^{-1}=gg^{-1}x=x\in Z(G)$.

This proves $Z(G)$ is a normal subgroup.

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i think the kernel argument (Mikko's no 1) is the most elegant approach. but sxd shows with thoroughness that the centre is a group, and it is perhaps worth a passing mention that once we know it is a group its normality is implied by the fact that it consists of complete conjugacy classes, albeit they are all singletons.