# How to find the number of automorphisms on $\mathbb Z_2\times \mathbb Z_4$

Well, I was trying find out the number of group homomorphisms from $U(20)$ onto $U(15)$ which I reduce to the problem of finding the number of automorphisms for $\mathbb Z_2\times \mathbb Z_4$. But don't know how shall I proceed.

Any help will be appreciated.

Let's represent $\mathbb{Z}_2 \times \mathbb{Z}_4$ by pairs $(a,b)$. Note that $(1,0)$ and $(0,2)$ are the only elements of order $2$, and furthermore $(0,2)$ is the only one which can be halved (that is, $2x=(0,2)$ for some $x$). This means that every automorphism fixes $(1,0)$ and $(0,2)$. Furthermore, it must send $(0,1)$ (a half of $(0,2)$) either to itself or to the other half $(0,3)$; and this choice determines the rest of the automorphism (since $(1,0)$ and $(0,1)$ together form a generating set). You can check that both possibilities lead to automorphisms, and so the number of automorphisms is $2$.