Counting endomorphisms How can I find the number of endomorphisms on the direct sum of the integers mod 2 and the integers mod 4? The correct answer is supposed to be 32. I can't see why it isn't 64. And what does the fact that 2 divides 4 have to do with it?
 A: In (what I assume is) your reasoning you have your group $G$ of order $8$ with two generators (say $a$ of order $2$ and $b$ of order $4$), you say that any $(g,h)\in G^2$ will lead to an endomorphism by setting $a\mapsto g$ and $b\mapsto h$. This is false for the following reason, if $\phi\in \operatorname{End}(G)$ then $\phi(a)^2=\phi(a^2)=\phi(1_G)=1_G$.
Actually you need $g$ to be of order dividing $2$ (the order of $a$) and $h$ of order dividing $4$ (the order of $b$). 
The best way to understand this is to write your endomorphism as a matrix, if $\phi(a)=(x,y)$ and $\phi(b)=(z,t)$ then :
$$M_{\phi}:=\begin{pmatrix}x&z\\y&t\end{pmatrix} $$
One sees that $x,y\in\mathbb{Z}_2$ and $z,t\in \mathbb{Z}_4$ furthermore one needs (in order that $(x,y)$ is of order dividing $2$) that $2z=0$ so you have $2$ choices for $x$ ($0$ or $1$ mod $2$), $2$ choices for $y$ ($0$ or $2$ mod $4$), $2$ choices for $y$ ($0$ or $1$ mod $2$) and $4$ choices for $t$ ($0$, $1$, $2$ or $3$ mod $4$). On the whole there are : $8\times 4=32$ choices. 
The matrix introduced here is clearly too much for the problem, but it allows you to compute  $|\operatorname{End}(G)|$ for any finite abelian group.
