Homomorphism between Abelian groups G and H

Question

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

Answer 1

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.

Answer 2

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 $(ng) + (-ng) = 0_G$