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Let $G$ be any group , $Z(G)$ is the center of the group $G$ , prove that : $\forall \tau \in Aut(G) , \tau [(Z(G)] = Z(G)$

My first trial was to prove that the center of any group is the unique subgroup of its order hence is a characteristic . but i found a counterexample easily which is $D_8$ the Dihedrial group of order $8$ ( i use the notation $D_{2n}$ )

So , any hints ?

FNH
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3 Answers3

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Can you show that for any $x\in Z(G)$ and $y\in G$, $\tau(x)\tau(y)=\tau(y)\tau(x)$? This will say $\tau(x)$ commutes with every element of $G$ (since $y$ was arbitrary), thus $\tau(x)\in Z(G)$. This shows containment in one direction, can you show containment in the other direction now?

Clayton
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  • Your answer shows $\tau(Z) \subset Z$? – user5826 Dec 05 '17 at 21:51
  • @Al: yes.... that is exactly what my answer says based on the fact that it says $x\in Z(G)\Rightarrow\tau(x)\in Z(G)$. – Clayton Dec 06 '17 at 00:10
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    Hmm... Actually we don't have to show the reverse containment. This is because $\tau$ is an automorphism, so it is certainly injective, hence $|\tau(Z(G))| = |Z(G)|$. Thus, together with the forward containment, we are done – Tan Yong Boon Apr 28 '24 at 10:39
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  • Is every element of $\tau(Z(G))$ central in $G$? (Translation: is $\tau(Z(G)) \subseteq Z(G)$? The answer should be "yes".)
  • Is anything else central in $G$ that isn't already in $\tau(Z(G))$? (Translation: is $\tau(Z(G)) \subsetneq Z(G)$? The answer should be "no". Otherwise, what happens when you hit it with $\tau^{-1}$?)
Billy
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    Of course, having proved that $\sigma(Z(G)) \subseteq Z(G)$ for any automorphism $\sigma$, you can just let $\sigma = \tau$ for one direction and $\sigma = \tau^{-1}$ for the other. – Billy Jun 27 '13 at 15:11
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consider that any Automorphism preserves the order, means order of Z(G) = order of τ(Z(G)).

MathGuy
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