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I'm trying to find the number of the subgroups whose order $5$ in the $Aut(G)$. [Here $G \simeq \mathbb{Z}_3 \times \mathbb{Z}_{11}\times\mathbb{Z}_{11} $]. Definitely, If the $G$ is a cyclic, $Aut(G)$ is $U(G)$. But, like the case I suggested, How can I find the isomorphic group?

Does $Aut(G \times H) \simeq Aut(G) \times Aut(H)$ ? (Under the hypothesis, order of the G and H are relatively prime)

So back to the first question, $Aut(G) \simeq Aut(\mathbb{Z}_3)\times Aut(\mathbb{Z}_{11}\times\mathbb{Z}_{11}) $ What should I do next?

Is there any theorem realted with Aut( $\mathbb{Z}_n \times \mathbb{Z}_{m}$)? Plus what if the $gcd(n,m) \not = 1$ ?

Arturo Magidin
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han fei
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  • Since $\mathbb{Z}{11}$ is a field ${\rm Aut}({\mathbb{Z}{11}\times\mathbb{Z}{11}})={\rm GL}_2(\mathbb{Z}{11})$. – Andrea Mori Mar 27 '21 at 14:08
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    In general, $\mathrm{Aut}(G\times H)$ does not equal $\mathrm{Aut}(G)\times\mathrm{Aut}(H)$. They do, however, in the special case in which $G$ and $H$ are both abelian of relatively prime orders. See here – Arturo Magidin Mar 27 '21 at 14:22
  • What concepts are you expected to use? Do you know Sylow's theorems? Do you know that $Aut(\mathbb{Z}_p \times \mathbb{Z}_p)$ is isomorphic to $GL_2(\mathbb{Z}_p \times \mathbb{Z}_p)$ (this can be proved easily regardless). Do you know how to compute the order of $GL_2(\mathbb{Z}_p \times \mathbb{Z}_p)$? This question is not "that hard" if you know the right theorems,but not much context is given. – Dionel Jaime Mar 28 '21 at 04:23

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I will go ahead and solve your particular example, and note that the tools are presented in the comments.

Since $(3,121)=1$, we have that $\rm {Aut}(\mathbb Z_3×\mathbb Z_{11}×\mathbb Z_{11})\cong\rm {Aut}(\mathbb Z_3)×\rm {Aut}(\mathbb Z_{11}×\mathbb Z_{11}) $.

And this will be $\mathbb Z_2×\rm {GL_2 (\mathbb Z_{11})} $.