What is $\operatorname{Aut}(\mathbb Z_4\times\mathbb Z_2)$? I'm trying to figure out $\operatorname{Aut}(\mathbb Z_4\times\mathbb Z_2)$, but I'm not really sure how to go about this. I feel like there are some non-obvious generators.
 A: Let $A = \mathbb{Z}_4 \times \mathbb{Z}_2$, which we consider as a $\mathbb{Z}$-module. Here's one method, which is certainly not necessarily the easiest:
1) Show that any homomorphism $f:A \to A$ lifts to a homomorphism $\tilde f: \mathbb{Z}^2 \to \mathbb{Z}^2$ of the form $\tilde f = \begin{pmatrix} a & 2b \\ c & d\end{pmatrix}$ for $a, b, c, d\in \mathbb{Z}$. (That is, we have $f\pi = \pi \tilde f$, where $\pi:\mathbb{Z}^2 \to A$ is the obvious quotient map.)
2) Show that $f$ is a isomorphism iff (in the notation above) $\det \tilde f\not\equiv 0\pmod{2}$. To do so, note that $A /(2, 0) = \mathbb{Z}_2\times \mathbb{Z}_2$ with $\mathbb{Z}_2$ a field, and
$$\begin{pmatrix}a & 2b \\ c & d\end{pmatrix} \begin{pmatrix} 2x \\ y \end{pmatrix} = \begin{pmatrix} 2a & 2b \\ 2c & d\end{pmatrix}\begin{pmatrix} x \\ y\end{pmatrix}.$$
3) Given the set of $\tilde f$ for which $f$ is an isomorphism, determine the corresponding $f$. Note that $f$ depends only on $a\pmod{4}$ and on $b, c, d\pmod{2}$, and $\det \tilde f\not = 0\pmod{2}$ iff $a, d\equiv 1\pmod{2}$.
A: The following answer is an elementary approach. The idea is the find the elements of the automorphism group "by hand" and then plug them together. The answer of anomaly will work for larger groups, but I want to post this answer as it should be useful to those struggling with the concept of automorphisms and automorphism groups.
In an abelian group such as this your generating pairs need to satisfy two properties. Suppose $\phi: (1, 0)\mapsto g, (0, 1)\mapsto h$ is an automorphism of $\mathbb{Z}_4\times\mathbb{Z}_2$, then $g$ and $h$ must do the following.


*

*Preserve order. That is, then $g$ must have order $4$ while $h$ must have order $2$.

*Generate as a pair. That is $\langle g, h\rangle=\mathbb{Z}_4\times\mathbb{Z}_2$. To generate you need an element of the form $(a, x)$ where $a\neq 0$ and an element of the form $(x, 1)$.


In this specific case these are the only two things we need to worry about. In more general cases this is not true, but for abelian groups these are the conditions we are after. Note that it is not necessary that, for example, $\langle g\rangle=\langle (1, 0)\rangle$, that is, the element and its image do not need to generate the same subgroup. For example, $g=(1, 1)$ and $h=(0, 1)$ is a valid pair.
So, the group $\mathbb{Z}_4\times\mathbb{Z}_2$ has seven non-trivial elements. These, along with their orders, are as follows.


*

*$(1, 0)$, order$=4$.

*$(2, 0)$, order$=2$.

*$(3, 0)$, order$=4$.

*$(0, 1)$, order$=2$.

*$(1, 1)$, order$=4$.

*$(2, 1)$, order$=2$.

*$(3, 1)$, order$=4$.


So, if $g=(1, 0)$ then $h$ can be either $(0, 1)$ or $(2, 1)$. A more interesting example is $g=(1, 1)$, $h=(2, 1)$. I will leave you to find all of the pairs. This gives you the elements of your automorphism group. (As a check to see that you are on the right lines, you should realise that $\operatorname{Aut}(\mathbb{Z}_4\times\mathbb{Z}_2)$ has order no more than $12$. Can you see why this is?)
Then to find the isomorphism class of your group you need to multiply the elements using function composition. For example, consider the functions $\phi_1: (1, 0)\mapsto (1, 0), (0, 1)\mapsto (2, 1)$ and $\phi_2: (1, 0)\mapsto (1, 1), (0, 1)\mapsto (2, 1)$. Then $\phi=\phi_1\circ \phi_2$ is as follows. Begin by working out $\phi(1, 0)$.
$$\phi(1, 0)=\phi_2(\phi_1(1, 0))=\phi_2(1, 0)=(1, 1)$$
Next, work out $\phi(0, 1)$. This is trickier, and uses the fact that $\phi_2$ is a homomorphism.
$$\begin{align*}
\phi(0, 1)
&=\phi_2(\phi_1(0, 1))\\
&=\phi_2(2, 1)\\
&=\phi_2(2(1, 0)+(0, 1))\\
&=2\phi_2(1, 0)+\phi_2(0, 1)\\
&=2(1, 1)+(2, 1)\\
&=(4, 3)\\
&=(0, 1)\\
\end{align*}$$
Hence, $\phi_1\circ\phi_2=\phi: (1, 0)\mapsto (1, 1), (0, 1)\mapsto (0, 1)$. Similarly, you can see that $\phi_1$ has order two (because $\phi_1^2$ is the trivial automorphism) while $\phi_2$ has order four and $\phi$ has order two.
