# Show that you can distribute powers to commuting elements

• Mr Davis 97
In summary, the conversation discusses proving the statement that if a and b are commuting elements of G, then (ab)^n = a^n b^n for all n in the integers. The conversation includes the attempt at a solution, which involves proving two lemmas and using induction to prove the result for both positive and negative integers. Other possible approaches are also mentioned, such as showing that commuting of a and b gives ab^-1 = b^-1a, which can be used to derive the desired results. The conversation also raises the question of whether or not the lemmas were necessary to prove the statement.

## 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?

Mr Davis 97 said:
since their proofs are trivial?
So is the whole statement you have to show. Better to show it explicitly.

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
fresh_42 said:
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?

Mr Davis 97 said:
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.

• Mr Davis 97
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)