Why are there $12$ automorphisms of $\Bbb Z\oplus \Bbb Z_{3}$? Let $A:=\Bbb Z\oplus \Bbb Z_{3}$, then what is $|\text{Aut}(A)|$?  My answer is $4$ but the correct answer  (without explanation) turns out to be $12$! How come?
Well my understanding is, it just suffices to find out all the possibilities of $f(1,\bar 1)$ where $f$ is an arbitrary automorphism, since $(1,\bar 1)$ is the generator. So I think there are altogether $4$ possibilities: $(\pm 1, \pm \bar 1),(\pm 1, \mp \bar 1)$. How could there be any more?
I'd be very grateful if anyone would solve this puzzle for me! Thanks in advance. 
 A: I think that this can be nicely understood by thinking of $\mathbb{Z} \oplus (\mathbb{Z}/3)$ as a $\mathbb{Z}$-module and then representing group homomorphisms, i.e. $\mathbb{Z}$-module homomorphisms, as matrices:
We can write every homomorphism $f \colon \mathbb{Z} \oplus (\mathbb{Z}/3) \to \mathbb{Z} \oplus (\mathbb{Z}/3)$ as matrix
$$
 f =
 \begin{pmatrix}
  f_{11} & f_{12} \\
  f_{21} & f_{22}
 \end{pmatrix}
$$
for unique homomorphisms $f_{11} \colon \mathbb{Z} \to \mathbb{Z}$, $f_{12} \colon \mathbb{Z}/3 \to \mathbb{Z}$, $f_{21} \colon \mathbb{Z} \to \mathbb{Z}/3$ and $f_{22} \colon \mathbb{Z}/3 \to \mathbb{Z}/3$, such that
$$
 f\left(
   \begin{pmatrix}
     x \\
     y
   \end{pmatrix}
  \right)
 =
 \begin{pmatrix}
  f_{11} & f_{12} \\
  f_{21} & f_{22}
 \end{pmatrix}
 \cdot
 \begin{pmatrix} x \\ y \end{pmatrix}
 =
 \begin{pmatrix}
  f_{11}(x) + f_{12}(y) \\
  f_{21}(x) + f_{22}(y)
 \end{pmatrix},
$$
where we represent the elements of $\mathbb{Z} \oplus (\mathbb{Z}/3)$ as column vectors.
Note that $f_{12} = 0$ because this is the only homomorphism $\mathbb{Z}/3 \to \mathbb{Z}$.
Thus $f$ is of the form
$$
 f =
 \begin{pmatrix}
  f_{11} & 0 \\
  f_{21} & f_{22}
 \end{pmatrix}.
$$
If $f,g \colon \mathbb{Z} \oplus (\mathbb{Z}/3) \to \mathbb{Z} \oplus (\mathbb{Z}/3)$ are two homomorphisms then we can use the usual matrix multiplication
$$
 fg
 =
 \begin{pmatrix}
  f_{11} & 0 \\
  f_{21} & f_{22}
 \end{pmatrix}
 \begin{pmatrix}
  g_{11} & 0 \\
  g_{21} & g_{22}
 \end{pmatrix}
 =
\begin{pmatrix}
  f_{11} g_{11} & 0 \\
  f_{21} g_{11} + f_{22} g_{21} & f_{22} g_{22}.
 \end{pmatrix}
$$
Because we have
$$
 \mathrm{id}_{\mathbb{Z} \oplus (\mathbb{Z}/3)} =
 \begin{pmatrix}
  \mathrm{id}_{\mathbb{Z}} & 0 \\
                         0 & \mathrm{id}_{\mathbb{Z}/3}
 \end{pmatrix}
$$
it follows that if $f$ is an isomorphism with $g = f^{-1}$, then
\begin{align*}
     f_{11} g_{11}
  &= g_{11} f_{11}
   = \mathrm{id}_{\mathbb{Z}}, \\
     f_{22} g_{22}
  &= g_{22} f_{22}
   = \mathrm{id}_{\mathbb{Z}/3},
\end{align*}
so both $f_{11}$ and $f_{22}$ must be isomorphisms with $f_{11}^{-1} = g_{11}$ and $f_{22}^{-1} = g_{22}$.
If on the other hand $f_{11}$ and $f_{22}$ are isomorphisms then
$$
\begin{pmatrix}
  f_{11} & 0 \\
  f_{21} & f_{22}
\end{pmatrix}
\begin{pmatrix}
  f_{11}^{-1}                     & 0 \\
  -f_{22}^{-1} f_{21} f_{11}^{-1} & f_{22}^{-1}
 \end{pmatrix}
 =
 \begin{pmatrix}
  \mathrm{id}_{\mathbb{Z}} & 0 \\
                         0 & \mathrm{id}_{\mathbb{Z}/3}
\end{pmatrix}
$$
as well as
$$
\begin{pmatrix}
  f_{11}^{-1}                     & 0 \\
  -f_{22}^{-1} f_{21} f_{11}^{-1} & f_{22}^{-1}
 \end{pmatrix}
 \begin{pmatrix}
  f_{11} & 0 \\
  f_{21} & f_{22}
\end{pmatrix}
 =
 \begin{pmatrix}
  \mathrm{id}_{\mathbb{Z}} & 0 \\
                         0 & \mathrm{id}_{\mathbb{Z}/3}
\end{pmatrix}.
$$
So $f$ is then already an isomorphism.
We now know that $f$ is an isomorphism if and only if both $f_{11}$ and $f_{22}$ are isomorphisms (i.e. as a lower triangular matrix, $f$ is invertible if and only if all diagonal entries are invertible). We can now simply count the number of such matrices: The two possible values of $f_{11}$ are the two automorphisms of $\mathbb{Z}$. The two possible values of $f_{22}$ are the two automorphims of $\mathbb{Z}/3$. For the entry $f_{12}$ we can pick any of the three homomorphims $\mathbb{Z} \to \mathbb{Z}/3$. Thus we have $2 \cdot 2 \cdot 3 = 12$ possible choices.
A: The problem is that $(1,\bar{1})$ does not generate $A$.  It generates the subgroup of elements $(a,b)$ where $b$ is the residue class of $a$ mod $3$.  In particular, for instance, this subgroup does not contain $(1,\bar{0})$.
To generate all of $A$, you need to take at least two elements, such as $x=(1,\bar{0})$ and $y=(0,\bar{1})$.  If $f:A\to A$ is an automorphism, $f(y)$ must be an element of order $3$, so it must be either $(0,\bar{1})$ or $(0,-\bar{1})$.  Then $f(x)$ must be an element which together with $(0,\pm\bar{1})$ generates all of $A$.  The elements with this property are $(\pm1, b)$ for any $b\in\mathbb{Z}_3$.
This gives that there are at most $12$ different automorphisms of $A$: we have $2$ choices for where to send $y$, and $6$ choices for where to send $x$.  You then have to verify that all of these choices really do define automorphisms of $A$.  Let's verify this in the case that you choose plus signs everywhere, so we want $f(y)=(0,\bar{1})$ and $f(x)=(1,b)$ for some $b\in\mathbb{Z}_3$ (the other cases are similar).  To get an automorphism with these properties, we can define $f(c,d)=(c,d+b\bar{c})$ (here $b\bar{c}$ is the product of $b$ and $c$ mod $3$).  You can then check that $f$ is a homomorphism and a bijection (to get that it is a bijection, it might be helpful to observe that its inverse is $g(c,d)=(c,d-b\bar{c})$).
A: Consider presentation of $G$: $G=\langle x,y\colon y^3=1, xy=yx\rangle$.
Let $\sigma$ be any automorphism of $G$. Then $\sigma(y)$ could be $y$ or $y^2$ only (since $\langle y\rangle$ is torsion subgroup of $G$, so it is invariant under all the automorphisms). 
What can be $\sigma(x)$? Of course, it could be $x,x^{-1}$. Anything more? Yes. $xy$, $x^{-1}y$, $xy^2$, $x^{-1}y^2$. That's all. 
Thus, $\sigma(x)$ has two choices, $\sigma(y)$ has $6$ choices; each choice of $\sigma(x),\sigma(y)$ gives similar presentation of $G$, hence defines an automorphism. There are $12$.
A: There are $6$ possible images for $(1,0)$ itself and $(1,1)$, $(1,2)$ and their negatives. Plus there are $2$ automorphisms of $\mathbb{Z}_3$. 
