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

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 $r\mapsto s^2r$. Dec 21, 2014 at 22:08
• Endomorphisms are not the same as automorphisms, BTW. You'd have to figure out which of those are invertible. Dec 21, 2014 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 $8$. Dec 21, 2014 at 22:39
• @ThomasAndrews Sorry, I made the correction Dec 21, 2014 at 22:46
• @DerekHolt I think, you pointed me to my mistake. I thank you with great enthusiasm. Dec 21, 2014 at 22:48

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