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Let $G$ be a finite group with no element of order $p^2$ for each prime $p$. Also suppose that $\vert G\vert\neq p$, for each prime $p$. Does there always exist an automorphism $\phi$ of order 2 such that for at least one subgroup of $G$ say $H$, we have $\phi(H)\neq H$?

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    $\begingroup$ $\{1\}$ is a counter example. $\endgroup$
    – user583416
    Apr 8 '17 at 7:54
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A series of counterexamples is given by the cyclic groups $C_n$ where $n$ is a squarefree composite integer. Indeed $C_n$ can be written as $C_{p_1} \times C_{p_2} \times \ldots \times C_{p_n}$. Where the $C_{p_i}$ are minimal subgroups of $C_n$. Each automorphism of $C_n$ leaves $C_{p_i}$ invariant. I presume that these are the only counterexamples but I found no immediate proof.

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  • $\begingroup$ They are not the only counterexamples. See YCor's comments on mathoverflow.net/questions/266649 $\endgroup$
    – Derek Holt
    Apr 9 '17 at 12:58
  • $\begingroup$ @DerekHolt That's the advantage of "presuming" things instead of taking them for granted :-) $\endgroup$ Apr 9 '17 at 13:04
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Okay so look at $Z_2 \oplus Z_2$. The automorphisms of this group form $GL_2(F_2)$. Now look at $\phi$ which sends $(0,1)$ to $(1,0)$ and vice versa and keeps the others fixed. Take $H=(0,0),(1,0)$. Here $\phi^2$ is identity. This should do. There is no element of order $4$ in G.

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  • $\begingroup$ @ Ronit Debnath. You gave an example which satisfies the conditions. But I am looking for a proof in generel case( if it is true). $\endgroup$ Apr 8 '17 at 7:42
  • $\begingroup$ O okay I missed the always...ok I will see $\endgroup$
    – CoffeeCCD
    Apr 8 '17 at 7:50

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