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If $G$ is a group, then the center of $G$, assume its $Z(G)$, is the set of all elements of $G$ which commute with everything. So $Z(G)$ $=$ {$x$ $\in$ $G$: $xy$ $=$ $yx$ for all $y$ $\in$ $G$}. Prove that $Z(G)$ is a normal subgroup of $G$ for any group $G$.

First I need to show $G$ is a subgroup; and then show it's normal.

To show its a subgroup I know I need to prove that it has an identity, all elements contain inverses and it is closed under the operation

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  • $\begingroup$ Can you prove it has the identity? That it's closed under inverses and the operation? Have you tried? $\endgroup$
    – blue
    Mar 25 '14 at 23:37
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It should be clear that $1_G \in Z(G)$, but even if not, showing inverses and closure is enough to show $Z(G)$ contains the identity. For inverses: take arbitrary $x \in Z(G)$ and $g \in G$. Then we have:

$$x^{-1}g = (g^{-1}x)^{-1} = (xg^{-1})^{-1} = gx^{-1} \Rightarrow x^{-1} \in Z(G).$$

Can you do a similar thing to show that for $x,y \in Z(G)$, we have $xy \in Z(G)$? Showing it's normal is very easy once you use the commutativity property.

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    $\begingroup$ Actually, you also need to prove that $Z(G)$ is non-empty (and generally the easiest way to show that a subset of a group is non-empty when we want to show that it is a subgroup is to show that it contains the identity). $\endgroup$
    – fkraiem
    Mar 26 '14 at 0:16
  • $\begingroup$ @fkraiem thank you, that was sloppiness on my part. $\endgroup$
    – ah11950
    Mar 26 '14 at 0:18
  • $\begingroup$ Hmm, how do I show its normal though? $\endgroup$
    – atherton
    Mar 26 '14 at 4:19
  • $\begingroup$ @RoyM. Did you try? Do you know what it means to say that it's normal? $\endgroup$
    – blue
    Mar 26 '14 at 8:10

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