# Show that you can distribute powers to commuting elements

## Homework Statement

If $a$ and $b$ are commuting elements of $G$, prove that $(ab)^n = a^nb^n$ for all $n \in \mathbb{Z}$.

## The Attempt at a Solution

We prove two lemmas:
1) If $a$ and $b$ commute, then so do their inverses: $ab=ba \implies (ab)^{-1} = (ba)^{-1} \implies b^{-1}a^{-1} = a^{-1}b^{-1}$.

2) If $a$ and $b$ commute, then $b^n a = ab^n$: Base case is trivial. Suppose for some $k$ we have $b^ka = ab^k$. Then $b^{k+1}a = bb^ka = bab^k = abb^k = ab^{k+1}$.

Now to the actual result.

Clearly $(ab)^0 = a^0b^0$. So first we prove the result for the positive integers, by induction. The base case is trivial. Suppose for some $k \in \mathbb{Z}^+$ we have $(ab)^k = a^kb^k$. Then $(ab)^{k+1} = (ab)^k(ab) = a^kb^kab = a^kab^kb = a^{k+1}b^{k+1}$.

Now we prove the result for negative integers. $(ab)^{-n} = (b^{-1}a^{-1})^n = (a^{-1}b^{-1})^n = (a^{-1})^n(b^{-1})^n = a^{-n}b^{-n}$.

Does this argument work? Were the lemmas really necessary or could I have assumed they held since their proofs are trivial?

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mfb
Mentor
since their proofs are trivial?
So is the whole statement you have to show. Better to show it explicitly.

fresh_42
Mentor
There is another argument, why you will not need to prove the negative ones explicitly: If you prove the statement for all $n \in \mathbb{N}_0$ and for all $a,b \in G$, then you have also proven it for inverse elements. This is in words what you have written.

To do it by induction is a very formal way to prove it. In cases like the above, some dots will equally be acceptable, although from a logical point of view certainly not sufficient. But with the dots, every reader knows how the induction goes.

There is another argument, why you will not need to prove the negative ones explicitly: If you prove the statement for all $n \in \mathbb{N}_0$ and for all $a,b \in G$, then you have also proven it for inverse elements. This is in words what you have written.

To do it by induction is a very formal way to prove it. In cases like the above, some dots will equally be acceptable, although from a logical point of view certainly not sufficient. But with the dots, every reader knows how the induction goes.
But isn't it the case that I am not proving for all $a,b \in G$, rather just in the case $a,b$ commute?

fresh_42
Mentor
But isn't it the case that I am not proving for all $a,b \in G$, rather just in the case $a,b$ commute?
Yes, sure. But that doesn't change by taking the inverses: they commutate if and only if $a$ and $b$ do.

StoneTemplePython
another approach, that is of course closely related, is to show that commuting of $a$ and $b$ gives you
$ab^{-1} = b^{-1}a$ (and ditto for $ba^{-1} = a^{-1}b$),