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Let $a$ and $b$ be elements of an Abelian group and let $n$ be any positive integer. Show that $(ab)^n = a^nb^n$. Is this also true for non-Abelian groups?

String
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Karina Pena
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  • You definitely need to use the fact that the group is abelian in order to successfully write the proof. Since commutativity is a necessary condition to successfully prove $(ab)^n = a^nb^n$, we cannot generalize to non-abelian groups. – amWhy Jan 18 '15 at 21:34
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    No! You cannot rewrite $(ab)^n=ababab...ab$ as $a^nb^n$ unless you can interchange $a$'s and $b$'s. – String Jan 18 '15 at 21:35

2 Answers2

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This is never true for non-abelian groups if $n=2$, but otherwise, for $n\ge 3$ this can happen for certain non-abelian groups. For more details see here. Example: The quaternion group satisfies $(ab)^n=a^nb^n$ for all $n=2^k$ with $k>1$, but is not abelian.

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Dietrich Burde
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  • Also, as a consequence of Lagrange's theorem we have the fairly trivial result that if $G$ is a finite group then $(ab)^n = a^nb^n$ whenever $|G| \mid n$ – jxnh Jan 18 '15 at 21:50
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$(ab)^n = a^nb^n$ for all $n$ if and only if $a$ and $b$ commute with each other. For example, if $a=e$, or $a=b$. Also, if $b=a^{-1}$. And for another example, if $a$ is in the center of $G$, then $a$ commutes with every element of $G$, so it commutes with $b$.

Edit: If you mean for a specific $n$ as opposed to all positive integers $n$, then see @Dietrich Burde's answer.

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