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Show that $Z(G)$ is subgroup of G. $Z(G)$ is the center of the group

Proof: First we show $Z(G)$ is a group being a subgroup

if $Z(G)$ is a group then

1) $1_G \in Z(G)$
2) $Z(G)$ is closed
3) if $a \in$ then $a^{-1}\in Z(G)$

1) $1_G*x=1_Gx$ $\forall x\in G$, so $1_G \in Z(G)$
2) Let $a,b \in Z(G)$, so $\forall$ $x \in G$ $ax=xa$ and $bx=xb$, how do you proceed from here?
3)if $a\in Z(G)$ then $ax=xa$ $\forall x \in G$ $$ax=xa$$ $$axa^{-1}=xaa^{-1}$$ $$axa^{-1}=x$$ $$a^{-1}axa^{-1}=a^{-1}x$$ $$xa^{-1}=a^{-1}x$$ so $a^{-1}\in Z(G)$

How does that look? I just

daniel
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1 Answers1

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Yes, just as Nex has mentioned:

For any $a,b\in Z(G)$ and $x\in G$, we know that $ax=xa$ and $bx=xb$. Using these with the fact that the multiplication is associative we see that for $ab$ we have $$(ab)x=a(bx)=a(xb)=(ax)b=(xa)b=x(ab).$$ Therefore $ab$ is also in the center.

MattW
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