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Let $G$ be a group and $x, a \in G$ where $x=a^2$. Then $x\in Z(G) ⟹ G≇ Inn(G)$.

I have already shown that if $x=b^{-1}a$, then $G≇Inn(G)$ since $\phi_{a}=\phi_{b}$. However, I don't think this argument applies for the cause that $x=a^2$.

Does anyone have any hints on proving this using properties of isomorphisms?

Alexander Gruber
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luke
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    By $\to$ do you want $\implies$ (logical implication)? It is not clear to me what your question is. For one thing, in $S_3$ we have $e=(12)^2$ and $e\in Z(S_3)=1$ so $G\cong G/Z(G)\cong {\rm Inn}(G)$ (they are isomorphic). Similarly, set $a=(12)=b$ and the same argument works for $x=b^{-1}a$. And if $Z(G)$ contains any nontrivial element of a finite group $G$, then $|G/Z(G)|<|G|$ so $G\not\cong{\rm Inn}(G)$ - this holds true regardless of whether $x$ is the square of some $a$ or not. Are you not telling us everything? – anon Mar 13 '13 at 03:17
  • Isn't there the condition that $,a\notin Z(G),$? – DonAntonio Mar 13 '13 at 03:22
  • @zach I do not understand how this differs from your earlier, apparently identical, question http://math.stackexchange.com/questions/324398/prove-that-g-cong-mathrminng-if-and-only-if-zg-is-trivial/ – user1729 Mar 14 '13 at 13:31

2 Answers2

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Hint. Let $\phi:G\rightarrow \operatorname{Aut}(G)$ be defined by $\phi(g)=\theta_g$, where $\theta_g$ is the automorphism $\theta_g(x)=g^{-1}xg$.

  1. What is the image of $G$ under $\phi$?

  2. What is $\operatorname{ker}\phi$?

Is $\phi$ an isomorphism?

Alexander Gruber
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Presumably we are assuming that $1 \ne x$, but I still don't believe this is correct.

Let $H$ be the (infinite) multiplicative group whose elements are the complex $2^r$ th roots of 1 for all $r \ge 0$, and let $G$ be a semidirect product $G = H \rtimes \langle z \rangle$ where $z^2=1$ and $z^{-1}hz=h^{-1}$ for all $h \in H$. Let $a=i$ be a primitive fourth root of 1. Then $Z(G) = \{\pm 1\}$, and $-1 = a^2 \in Z(G)$ with $G \cong {\rm Inn}(G) \cong G/Z(G)$.

Derek Holt
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