Automorphisms and Mappings

Question

Consider $Aut(Z_4 ⊕Z_2)$. Any automorphism φ is determined by where we send the two generators (1, 0) and (0, 1) of $Z_4 ⊕ Z_2$. Also, in any automorphism an element must be sent to an element with the same order.

My first question is what is $Aut(Z_4 ⊕Z_2)$. Like what are the elements. I understand it is the group of Automorphisms such that $ghg^{-1}=h$ and what not. But What is that in this case.

a) How many choices are there for $φ(1, 0)$? How many choices are there for $φ(0, 1)$? Why can you not send $φ(0, 1) = (2, 0)$? What is the order of the group $Aut(Z_4 ⊕ Z_2)$?

There are three choices for $φ(1, 0)$ because there are three generators of $Z_4 ⊕ Z_2$ right? Mainly (1,0), (0,1) and (1,1). The same would go for $φ(0, 1)$. Also it seems to me that you can send $φ(0, 1) = (2, 0)$ because (0,1) has order 1 and (2,0) has order 1. However if $φ(0, 1) = (2, 0)$, then $φ(1,0)=(?,?)$. That I believe is where the contradiction would occur. The order of $Aut(Z_4 ⊕ Z_2)$ is 8 right?

b) Let $φ_1$ :$(1, 0) → (1, 1)$; $(0, 1) → (0, 1)$ and $φ_2$ : $(1, 0) → (1, 0)$; $(0, 1) → (2, 1)$.

Show that $φ_1φ_2$ is not equal to $φ_2φ_1$ so that $Aut(Z_4 ⊕ Z_2)$ is non-abelian.

I have done this, but included it for full information.

c) The two non-abelian groups of this order are D4 and he quaternions Q. Show $Aut(Z_4 ⊕ Z_2)$ is the dihedral group. Essentially Find two automorphisms that have order 2.

This I would like some help on.

Answer 1

The hint that part (c) of the question is trying to give is that there are only two options, now: either the automorphism group is $Q_8$ or it is $D_4$.

But $Q_8$ contains only one involution, so if you can show there are two different involutions then you are done.

[ I also believe there is something wrong with your solution to part (a). For $\phi(1,0)$, not e that this is an element of order 4. It has to be mapped to another element of order $4$ so there are $4$ options: $(1,0), (1,1), (3,0), (3,1)$. For $\phi(0,1)$, note that $(0,1)$ is an involution so it must be mapped to an involution. A priori, there are $3$ options namely $(2,0)$, $(2,1)$ and $(0,1)$. However the first element is actually the square of any of the order $4$ elements while $(2,1)$ and $(0,1)$ are not. So at most these options remain. This gives at most $4 \times 2$ possible automorphims. ]