-1

1) Let $G=\mathrm{Aut}(\mathbb {Z}_4\times\mathbb {Z}_2)$. Show that $|G|\leq8$ and then $G \cong D_4$.

2) Show that $\mathrm{Aut}(S_3\times \mathbb {Z}_3) \cong S_3\times\mathbb {Z}_2$.

At the moment I was able to do the first part of the first problem. I noticed that $\mathbb {Z}_4\times\mathbb {Z}_2$ has two generators, $(1,0),(0,1)$, so an automorphism $ϕ$ will be fixed if I choose their images. Both $(1,0)$ and $(0,1)$ have order $4$, and I can see that there are only $4$ elements of order $4$, so I only have $4$ elements that can be their images. Because of this facts I have $4$ possibilities for $ϕ((1,0))$, but only $2$ for $ϕ((0,1))$, because it can't be an element of order $4$ that is in the subgroup generated by $ϕ((1,0))$. So $|G|=8$. Is it correct?

user26857
  • 52,094
GRE
  • 103

1 Answers1

0

Hint: To see that $G\simeq D_4$, observe that there are four elements of $(\mathbb{Z}/4)\times(\mathbb{Z}/2)$ which are of order $4$. Namely, these elements are $\{(1,0),(3,0),(1,1),(3,1)\}$. Observe that these form two pairs of inverses ($(1,0)+(3,0)=(0,0)$ and $(1,1)+(3,1)=(0,0)$). These four points can be represented as a square $$ \begin{array}{ccc} (1,0)&\cdots&(1,1)\\ \vdots&&\vdots\\ (3,1)&\cdots&(3,0) \end{array} $$ Observe that inverses are shown as opposite corners of the square. $G$ acts by symmetry on that square (requires proof).

Michael Burr
  • 32,867
  • Sorry, I am not of sure I have understood what you said.. what do you mean when you say "acts by symmetry"? – GRE Nov 01 '16 at 18:21
  • @GRE $D_4$ is the symmetry group of the square. The four elements above form a square. The automorphism group acts as the symmetry group on those four elements. – Michael Burr Nov 01 '16 at 21:56