Homomorphism between Abelian groups G and H Prove the following lemma:

Let $G, H$ be Abelian groups and let $\phi : G \to H$ be a homomorphism.  Then $\phi(n g) = n \phi(g)$ for all $g \in G$, $n \in \Bbb Z$.

Could someone please lend a hand on this? I'm bad
 A: Hint: For positive integers $n$, note that
$$
ng = \overbrace{g + g + \cdots + g}^{n \text{ times}}
$$
and recall the definition of a homomorphism.  For negative integers, note that $(ng) + (-ng) = 0_G$
A: First, recall what $ng$ means. Given $n \in \mathbb Z$ and $g \in G$, we have that
$$
ng := \begin{cases}
\underbrace{g + g + \ldots + g}_{n \text{ times}} & \text{, if } n > 0, \\
0 & \text{, if } n = 0 \text{ and } \\
\underbrace{(-g) + (-g) + \ldots + (-g)}_{-n \text{ times}} & \text{, if } n < 0
\end{cases}
$$
Let me prove by induction on $n \in \mathbb Z^{+}$ (that is $n \in \mathbb Z$ and $n > 0$) that $\phi(ng) = n \phi(g)$. The case $n = 0$ is trivial and I'll leave the case $n < 0$ to you (here use an induction on $-n$ and a very similar argument).
If $n = 1$, then $ng = g$ and hence $\phi(ng) = \phi(g) = n \phi(g)$.
Thus assume that the result holds for $n$. Then
$$
\begin{align*}
\phi((n+1)g) &= \phi(ng + g) \\
&= \phi(ng) + \phi(g) \\
&\overset{\text{induction hypothesis}}{=} (n \phi(g)) + \phi(g) \\
&= (n+1) \phi(g)
\end{align*}
$$
Q.E.D.
