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Assume that $G$ is a finite group such that $|G| \gt 1$ and there exists $x \in G$ such that $x^2 \neq e$. Prove that there exists a non-trivial automorphism on $G$.

Note: This question has been asked and answered here. But, I can't understand that answers. Both of them use the concept of inner-automorphism. We haven't studied this in the class, So i'm not allowed to use the term 'Inner-Automorphism'. Also, I don't understand that what is their automorphism doing. We are given a group. Maybe its abelian. Maybe it's not. We should provide a function which takes an element of $G$ and maps it to another element of $G$. Which element is that another element?

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If $G$ is abelian, then $\phi:g \mapsto g^{-1}$ is a non-trivial automorphism because $\phi(x)\ne x$.

If $G$ is not abelian, there are elements $a,b \in G$ such that $ab\ne ba$. Then $\phi:g \mapsto a^{-1}ga$ is a non-trivial automorphism because $\phi(b)\ne b$.

Thus, the existence of $x \in G$ such that $x^2 \neq e$ in only used in the abelian case, but as the original answer notes, it is not really needed.

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  • $\begingroup$ @llrf If G is not abelian , in your example just take $x \notin Z(G)$. $\endgroup$ – Riju May 12 '17 at 14:06
  • $\begingroup$ This answer is just a distillation of math.stackexchange.com/a/8401/589. $\endgroup$ – lhf May 12 '17 at 14:12
  • $\begingroup$ This may be a distillation... But i can understand it :) That's the point :) and i appreciate that :) $\endgroup$ – Arman Malekzadeh May 12 '17 at 14:26

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