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Problem 57 (Powers Of Products In An Abelian Group)

Suppose $G$ is an Abelian group. Prove that if $a,b\in G$, then $(ab)^2=a^2b^2$. Then use induction to prove that if $a,b\in G$, then $(ab)^n=a^nb^n$ for each $n\in \mathbb{N}$.

Solution

Let $G$ be an Abelian Group. Pick $a,b\in G$. Thus we know by definition of an Abelian Group that $ab=ba$. We compute

$$\begin{align} (ab)^2&=abab \\ &=aabb \\ &=a^2b^2. \end{align}$$

We will now prove that for all $n\in\mathbb{N}$ we have $(ab)^n=a^nb^n$. By definition of an Abelien group, we know the base case where $n=1$ is satisfied (this was shown above as $ab=ba$). Now suppose for some $k\in\mathbb{N}$ that the statement $(ab)^k=a^kb^k$ is true. We compute

$$\begin{align} (ab)^{k+1}&=(ab)^k(ab) \\ &=a^kb^kab \\ &=a^k(b^ka)b \\ &=a^kab^kb \\ &=a^{k+1}b^{k+1}. \end{align}$$

Thus, by mathematical induction, we know that $(ab)^n=a^nb^n$ for all $n\in\mathbb{N}$.

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