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What is the order of the automorphism group of the finite group $G=\mathbb{Z}/5\mathbb{Z}\oplus \mathbb{Z}/25\mathbb{Z}.$ Is the group $Aut(G)$ Abelian?

My attempt: If $\phi \in Aut(G)$, then $\phi$ sends generator to the generator. For this question, we need to count the total number of generator in $G$. There are $4$ generator in $\mathbb{Z}/5\mathbb{Z}$; and $20$ generator in $\mathbb{Z}/25\mathbb{Z}$. This implies $|Aut(G)|=80$.

I think $Aut(G)$ is not abelian, because a finite group $G$ has an abelian automorphism group iff $G$ is cyclic. In this case, $G$ is abelian, not cyclic.

Any one please suggest me whether this idea is correct for this question?

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  • $\begingroup$ By your argument $Aut(C_2\times C_4)$ consists of 2 elements, hence Abelian, hence $C_2\times C_4$ is cyclic which is absurd. $\endgroup$ – Mark Sapir Jul 4 '20 at 2:03
  • $\begingroup$ The automorphism group has order 2000. What you miss is e.g. a map $a\mapsto a$, $b\mapsto ab$. $\endgroup$ – ahulpke Jul 4 '20 at 2:10
  • $\begingroup$ $\DeclareMathOperator{\Aut}{Aut}$Your two ideas are not consistent. The subgroup $\Aut(C_5)\times \Aut(C_{25})\leq \Aut(G)$ actually is abelian. $\endgroup$ – tomasz Jul 4 '20 at 2:16
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As you say there are 20 generators of $\mathbb{Z}/25\mathbb{Z}$. Thus there are 100 elements of order 25 in $G$. Therefore an automorphism of $G$ can map $(0,1)$ to 100 different places. Suppose it gets mapped to some element $x$. Now consider the subgroup $$\mathbb{Z}/5\mathbb{Z}\times\mathbb{Z}/5\mathbb{Z}\subseteq \mathbb{Z}/5\mathbb{Z}\times\mathbb{Z}/25\mathbb{Z}.$$

We know that $(1,0)$ has order $5$, so must map to an element $y$ in this subgroup. We know $5x$ is in this subgroup, generating a subgroup of size 5. In order for our map to be surjective, $(1,0)$ must map to one of the 20 other elements in $\mathbb{Z}/5\mathbb{Z}\times\mathbb{Z}/5$.

Thus there are $20\times 100=2000$ automorphisms.

If you believe this group is not commutative, then try writing down two elements which do not commute. Start simple.

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Let $G=Z_p\times Z_{p^2}$ where $p$ is prime and $Z_n$ is cyclic of order $n$.

Then $G$ has $p^3-p^2$ elements of order $p^2$. Let $a$ be any of them. Then $G$ has $p^2-p$ elements of order $p$ outside $\left<a\right>$, call one of them $b$.

There is a unique automorphism sending the standard generators of $G$ to $a$ and $b$,. Therefore $G$ has $$(p^3-p^2)(p^2-p)=p^3(p-1)^2$$ automorphisms.

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