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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?

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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 $\phi\in End(G)$ 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 $4$).

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 $|End(G)|$ for any finite abelian group.

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  • $\begingroup$ Thank you. I too was using matrices but didn't see the limitation. I asked this question because I was trying to solve the question for the general finite abelian group in Jacoson' s Basic Algebra I at p. 208. $\endgroup$ – Marty Sep 10 '15 at 15:54

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