Show that you can distribute powers to commuting elements

  • #1
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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?
 

Answers and Replies

  • #3
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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.
 
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  • #4
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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?
 
  • #5
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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.
 
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  • #6
StoneTemplePython
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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)
 

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