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I'm working on the classification on the group of order 75. Part of its question is to decide the homomorphism between $\Bbb Z_3$ and $\operatorname{Aut}(\Bbb Z_5\times \Bbb Z_5)$, which is to find the elements in $\operatorname{Aut}(\Bbb Z_5\times \Bbb Z_5)$ with order 3.

Since $\lvert\operatorname{Aut}(\Bbb Z_5\times \Bbb Z_5)\rvert=24\times 20$, it is messy to discuss the Sylow $3$-subgroups based on the order, also, direct computation could do, but it does not seem a good way. I wonder if there's any easier way to determine it.

Shaun
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  • The symbol $\ast$ is usually reserved for the free product of groups in the context of group theory. Do you mean $\times$ for $\times$ instead? – Shaun Sep 24 '19 at 16:02
  • You could use GAP, I suppose. – Shaun Sep 24 '19 at 16:24
  • It appears that $$\operatorname{Aut}(\Bbb Z_5\times \Bbb Z_5)\cong GL_2(\Bbb Z_{5}).$$ – Shaun Sep 24 '19 at 16:36
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    All subgroups of order $3$ are conjugate, so you really only need to find one. I would recommend using the fact that the automorphism group in question is naturally isomorphic to ${\rm GL}(2,5)$, the multiplicative group of non-singular $2 \times 2$-matrices over the field of order $5$. So you just need to write down a matrix of order $3$ in this group. All such matrices lie in ${\rm SL}(2,5)$ (determinant $1$), and they have trace $-1$. – Derek Holt Sep 24 '19 at 16:37
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    Yes thank you for editting, also the GAP thing is pretty useful and thx for mentioning that! @Shaun –  Sep 24 '19 at 20:24
  • I think the trace thing is wrong since the identity matrix dose not have trace 1, but this method could work out. @DerekHolt –  Sep 24 '19 at 20:41
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    I did not say that the identity had trace $1$, I said that elements of order $3$ have trace $-1$. That's because their eigenvalues are $\omega$ and $\omega^2$, where $\omega$ is a cube root of $1$. – Derek Holt Sep 24 '19 at 21:14
  • Ohh I got it wrong, that is correct.@DerekHolt –  Sep 24 '19 at 21:36

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