Let $G$ be an abelian group and for each positive integer $n$, define
$$ G[n] = \{ g \in G: ng = 0 \}. $$
It is easy to check that if $m$ and $n$ are positive integers such that $m$ divides $n$, then $G[m]\subseteq G[n]$. Define $f: G[n] \rightarrow G $ as $f(g) = m·g$. My question is why $f$ is an homomorphism?
If $a,b \in G[n]$ then, how do I prove that $f(ab) = m·(ab) = (m·a)(m·b) = f(a)f(b)$? With that, the preservation of the identity element is straightforward, concluding the proof.
Since the group $G$ is abelian, the notation $ng$ stands for the $n$-fold sum $g + g + \ldots + g$. Now, \begin{align}f(a+b) &= m(a + b) \\ &= (a+b) + (a+b) + \ldots + (a+b)\\ &= (a + a + \ldots + a) + (b + b + \ldots + b)\\ &= ma + mb\\ &= f(a) + f(b).\end{align} You can do the same thing in multiplicative notation, but then one would write $g^n$ instead of $ng$. Note that commutativity is used essentially.