How many automorphisms of $\Bbb Z \oplus \Bbb Z_2$

The question is how many isomorphisms (of groups) exist from $\Bbb Z \oplus \Bbb Z_2$ to itself.

I tried to define isomorphism by taking generators to generators, but I want to see convincing proof for that.

Thanks

• By Z2 do you mean $\mathbb{Z}/2\mathbb{Z}$? If so, (0,1) is the unique element of order 2 so it must get sent to itself. Then you can figure out how many places you can send (1,0) to make the map surjective. – Nate Feb 3 '14 at 21:00
• What's not convincing about the argument you have? Tell us so that we can help you fix it. – Ayman Hourieh Feb 3 '14 at 21:16
• en.wikipedia.org/wiki/… – Robert Wolfe Feb 3 '14 at 21:20

First, let's suppose that $f$ is an automorphism of $\Bbb Z\oplus\Bbb Z_2.$ Now, we know that $f(0,1)\ne(0,0).$ (Why?) However, $(0,1)+(0,1)=(0,0),$ so we know that $f(0,1)+f(0,1)=f(0,0)=(0,0).$ Thus, $f(0,1)$ has order $2,$ but $(0,1)$ is the unique element of $\Bbb Z\oplus\Bbb Z_2$ having order $2.$ (Why?) Hence, we must have $f(0,1)=(0,1),$ and so for all $k\in\Bbb Z,$ we have $$f(k,1)=f(k,0)+f(0,1)=f(k,0)+(0,1)=k f(1,0)+(0,1).$$ (Why?) Now, suppose that $f(1,0)=(x,y).$ Since $(1,0)\ne(0,0),$ then $(x,y)=f(1,0)\ne f(0,0)=(0,0),$ and likewise, $f(1,0)\ne(0,1).$ (Why?) It follows that $x\ne 0.$ (Why?) Observe then that for all $k\in\Bbb Z$ we have $$f(k,0)=k f(1,0)=(kx,ky)$$ and (therefore) $$f(k,1)=(kx,ky+1).$$
Now, there must be some $k\in\Bbb Z$ and some $m\in\Bbb Z_2$ such that $$f(k,m)=(1,0).$$ (Why?) It follows that $kx=1,$ so $x=1$ or $x=-1.$ (Why?) It can be shown that any choice of $y\in\Bbb Z_2$ gives us an automorphism so long as $f(0,1)=(0,1)$ and $f(1,0)=(\pm 1,y),$ so there are $4$ such automorphisms.
So, as you suspected, our standard generating set $\bigl\{(1,0),(0,1)\bigr\}$ is indeed mapped onto a generating set. In particular, $(0,1)$ must be mapped to itself, and $(1,0)$ can be sent to any of the $4$ elements $(x,y)\in\Bbb Z\oplus\Bbb Z_2$ for which $\bigl\{(x,y),(0,1)\bigr\}$ is a generating set.
I think it will be convenient to write the operator of the group in multiplication. Let $G=\langle a, b\rangle$, where $|a|=\infty$ and $|b|=2$. Clearly, $b$ is the only element of order $2$. Hence for every $\sigma \in AutG$, we have $\sigma(b)=b$. Now assume $\sigma(a)=a^ib^j$. Now $G=\langle a^ib^j, b\rangle=\langle a^i, b\rangle$. Thus, there exists $k,l$ such $a=(a^i)^kb^l$, so we get $i=1$ or $-1$. Since $j$ can be $1$ ore $0$, we get $|Aut(G)|=4$, as given by Cameron.