Show that you can distribute powers to commuting elements

[Mr Davis 97]

Homework Statement

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

Homework Equations

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?

 

[mfb]

since their proofs are trivial?

So is the whole statement you have to show. Better to show it explicitly.

 

[fresh_42]

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.

 

[Mr Davis 97]

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]

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 $ba^{-1} = a^{-1}b$),
from here you can derive all the results you want by the ability to 'multiply by 1' (identity)