# What is the order of the automorphism group of the finite group $G=\mathbb{Z}/5\mathbb{Z}\oplus \mathbb{Z}/25\mathbb{Z}.$

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?

• 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. – Mark Sapir Jul 4 '20 at 2:03
• The automorphism group has order 2000. What you miss is e.g. a map $a\mapsto a$, $b\mapsto ab$. – ahulpke Jul 4 '20 at 2:10
• $\DeclareMathOperator{\Aut}{Aut}$Your two ideas are not consistent. The subgroup $\Aut(C_5)\times \Aut(C_{25})\leq \Aut(G)$ actually is abelian. – tomasz Jul 4 '20 at 2:16

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.

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 $$\left$$, 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.