Prove Abelian Group: $(a \cdot b)^n = a^n \cdot b^n$

[Guest2]

Prove that if $G$ is an abelian group, then for $a, b \in G$ and all integers $n$, $(a \cdot b)^n = a^n \cdot b^n.$

Never mind. I figured it out. We proceed by induction on $n$, then use a lemma in the text.

 

[Deveno]

Just for the record, we see that:

$(ab)^1 = ab = a^1b^1$

Now if $(ab)^{n-1} = a^{n-1}b^{n-1}$, then:

$(ab)^n = ab(ab)^{n-1} = ab(a^{n-1}b^{n-1})$ (by hypothesis)

$= a(a^{n-1}b^{n-1})b$ (since $G$ is abelian)

$= a^nb^n$.

This proves it for $n \in \Bbb N^{+}$.

From there, if $n < 0$, say $n = -k$, for $k > 0$, then

$(ab)^n = (ab)^{-k} = [(ab)^k]^{-1} = (a^kb^k)^{-1}$

$= (b^k)^{-1}(a^k)^{-1} = b^{-k}a^{-k} = b^na^n = a^nb^n$.

Finally, $(ab)^0 = e = ee = a^0b^0$, by definition.