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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

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    $\begingroup$ $r\mapsto sr$ can't be happen in an automorphism, since $sr$ is not of order $2$. Maybe you meant $r\mapsto s^2r$. $\endgroup$ Dec 21, 2014 at 22:08
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    $\begingroup$ Endomorphisms are not the same as automorphisms, BTW. You'd have to figure out which of those are invertible. $\endgroup$ Dec 21, 2014 at 22:18
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    $\begingroup$ 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 $8$. $\endgroup$
    – Derek Holt
    Dec 21, 2014 at 22:39
  • $\begingroup$ @ThomasAndrews Sorry, I made the correction $\endgroup$
    – user135520
    Dec 21, 2014 at 22:46
  • $\begingroup$ @DerekHolt I think, you pointed me to my mistake. I thank you with great enthusiasm. $\endgroup$
    – user135520
    Dec 21, 2014 at 22:48

1 Answer 1

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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

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