A question about the automorphism group of $\mathbb{Z}_{2} \times \mathbb{Z}_{4}$

Question

I wanted to clarify some confusion I was having on the automorphism group of $\mathbb{Z}_{2} \times \mathbb{Z}_{4}$, which I call $\mathrm{Aut}(\mathbb{Z}_{2} \times \mathbb{Z}_{4})$.

I considered the following as a presentation of this group $\mathbb{Z}_{2} \times \mathbb{Z}_{4} = \langle r,s : r^{2}=1=s^{4}, sr=rs \rangle$. Looking at this presentation, an element $\alpha \in \mathrm{Aut}(\mathbb{Z}_{2} \times \mathbb{Z}_{4})$ will send $r$ to $r$ or $s^{2}r$ and will send $s$ to $s, s^{3}, sr$, or $s^{3}r$.

Using this, I was able to list $8$ possible automorphisms. I did not check this carefully, but the autmorphisms that I listed each had order $2$ and I may not be remembering this correctly but a group of order $8$ where all the non-identity elements are of order $2$ is abelian.

I turned to looking at $\mathrm{Aut}(\mathbb{Z}_{5} \times \mathbb{Z}_{25})$ where I found this question:

Properties of automorphism group of $G={Z_5}\times Z_{25}$

The answer uses the following proposition the result of which is found in the paper below(which I haven't finished reading yet to verify):

Christopher J. Hillar, Darren Rhea, Automorphisms of finite Abelian groups, arXiv

For example, if $p$ is a prime, then $$\mathrm{End}(\mathbb{Z}/p \times \mathbb{Z}/p^2) \cong \begin{pmatrix} \hom(\mathbb{Z}/p,\mathbb{Z}/p) & \hom(\mathbb{Z}/p^2,\mathbb{Z}/p) \\ \hom(\mathbb{Z}/p,\mathbb{Z}/p^2) & \hom(\mathbb{Z}/p^2,\mathbb{Z}/p^2) \end{pmatrix} \cong \begin{pmatrix} \mathbb{Z}/p & \mathbb{Z}/p \\ \mathbb{Z}/p & \mathbb{Z}/p^2 \end{pmatrix}$$

But I think based on that result, my conclusion that $\mathrm{Aut}(\mathbb{Z}_{2} \times \mathbb{Z}_{4})$ is abelian looks to be false.

I am essentially wondering if I did something wrong

$r\mapsto sr$ can't be happen in an automorphism, since $sr$ is not of order $2$. Maybe you meant $r\mapsto s^2r$. — Thomas Andrews, Dec 21 '14 at 22:08
Endomorphisms are not the same as automorphisms, BTW. You'd have to figure out which of those are invertible. — Thomas Andrews, Dec 21 '14 at 22:18
The automorphism $r \mapsto rs^2$, $s \mapsto rs$ has order $4$, not $2$, and the full automorphism group is nonabelian and isomorphic to the dihedral group of order $8$. — Derek Holt, Dec 21 '14 at 22:39
@DerekHolt I think, you pointed me to my mistake. I thank you with great enthusiasm. — user135520, Dec 21 '14 at 22:48
@DerekHolt Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. — Julian Kuelshammer, Apr 11 '15 at 08:55

score 2 · Accepted Answer · answered Nov 07 '15 at 08:41

This question was answered by a comment:

The automorphism $r↦rs^2$, $s↦rs$ has order 4, not 2, and the full automorphism group is nonabelian and isomorphic to the dihedral group of order 8. – Derek Holt Dec 21 '14 at 22:39

Also helpful:

Endomorphisms are not the same as automorphisms, BTW. You'd have to figure out which of those are invertible. – Thomas Andrews Dec 21 '14 at 22:18

A question about the automorphism group of $\mathbb{Z}_{2} \times \mathbb{Z}_{4}$

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