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Let G be a group and $g \in G$. Show that $Z(G) = \{ x \in G : gx = xg$ for all $g \in G$} is a subgroup of $G$. This subgroup is called the center of $G$.

I know that associativity can be inherited from G, but I'm not 100% sure on how to show it's closed, that an identity exists, and that an inverse exists.

Claire
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  • Closure: if $g_1, g_2 \in Z$ then is $g_1g_2$? Identity element: what do you think it is? Inverse: show if $g \in Z$ then $g^{-1} \in Z$ – J. W. Tanner Feb 14 '19 at 02:45
  • I understand what each is asking, I'm just unsure of how I should go about it in this format. – Claire Feb 14 '19 at 02:46
  • it's very similar to that post, but they already know that it is a subgroup, they're just proving it's abelian. I don't have to prove it's abelian in this question @YuiToCheng – Claire Feb 14 '19 at 02:47
  • Can you prove that if $gg_1=g_1g$ and $gg_2=g_2g$ for all $g \in G$ then $gg_1g_2=g_1g_2g$ for all $g \in G$ ? – J. W. Tanner Feb 14 '19 at 02:50

1 Answers1

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Well, if

$x, y \in Z(G), \tag 1$

then

$\forall g \in G, \; xg = gx, \; yg = gy$ $\Longrightarrow \forall g \in G, \; (xy)g = x(yg) = x(gy) = (xg)y = (gx)y = g(xy), \tag 2$

which shows that $Z(G)$ is closed under the group operation; clearly the identity element $e \in G$ satisfies

$eg = ge, \forall g \in G; \tag 3$

also,

$x \in Z(G) \Longrightarrow xg = gx, \forall g \in G \Longrightarrow gx^{-1} = x^{-1}g, \; \forall g \in G \Longrightarrow x^{-1} \in Z(G); \tag 4$

thus $Z(G)$ contains the group identity and also the inverse of each of its elements; since we have agreed that the associativity of the group operation is in fact manifestly inherited by $Z(G)$ from $G$, we conclude that $Z(G)$ is a subgroup of $G$. $OE\Delta$.

Robert Lewis
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