How to show two groups are isomorphic?

Question

For any abelian group $A$ and a positive integer $m$ prove that $$ Hom(\mathbb{Z}_{m},A)\cong{A[m]=\{a\in{A}|ma=0\}}. $$

I am not sure how to start to do this problem.

Thanks in advance for your help!

Answer 1

Alternatively, since it may give you a good sense of why exact sequences are so computationally powerful:

There is an exact sequence

$$0\to\mathbb{Z}\xrightarrow{\times m}\mathbb{Z}\to\mathbb{Z}/m\mathbb{Z}\to 0$$

Homing this with $A$ gives

$$0\to \text{Hom}(\mathbb{Z}/m\mathbb{Z},A)\to \text{Hom}(\mathbb{Z},A)\to\text{Hom}(\mathbb{Z},A)$$

But, this is clearly the same as

$$0\to \text{Hom}(\mathbb{Z}/m\mathbb{Z},A)\to A\xrightarrow{\times m}A$$

So

$$\text{Hom}(\mathbb{Z}/m\mathbb{Z},A)\cong \ker(A\xrightarrow{\times m}A)=A[m]$$

Answer 2

Note that a $f \in Hom(\mathbb{Z}_{m},A)$ is determined by $f(1) = a$ because $\mathbb{Z}_{m}$ is cyclic.

The order of $1 \in \mathbb{Z}_{m}$ is $m$, so $ma = 0$ is a necessary condition; but for every $a \in A[m]$ you can define $\phi \in Hom(\mathbb{Z}_{m},A)$ with $\phi(1) = a$, and so is also a sufficient condition.

Answer 3

HINT: Given $f \in \operatorname{Hom}(\Bbb{Z}_m, A)$, consider $f(1) \in A$.