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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

Stefan Mesken
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E.Kimura
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    You should be able to type that mathematics here (using mathjax) and show what you have tried and where you are stuck. http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Ethan Bolker Nov 02 '16 at 19:41
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    If I'm not mistaken, we don't actually need to assume $H$ is abelian because $< \phi(g) > = { ( \phi(g) )^k : k \in \mathbb{Z} } = \phi ( \mathbb{Z} g )$ is always abelian. – AJY Nov 02 '16 at 19:56
  • @AJY: Right, we would only change to a multiplicative notation to state the same conclusion if $G,H$ were not known to be abelian. – hardmath Nov 02 '16 at 20:01

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

Ben Grossmann
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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.

Stefan Mesken
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  • What is the inductive hypothesis here? I don't get the result for -(n+1) – E.Kimura Nov 03 '16 at 15:28
  • @E.Kimura The induction hypothesis, in my case, is that $\phi(ng) = n \phi(g)$. In the case I left to you, the induction hypothesis would be that $\phi(-ng) = -n \phi(g)$. – Stefan Mesken Nov 03 '16 at 17:35