Question

the question says: prove that if a is an element of a group G, then the...

the question says:
prove that if a is an element of a group G,
then the order of a = order of its inverse.
my attempt:
Let order of a=n , so aⁿ=e , and so (a)ⁿ(a^-1)ⁿ=e=(a^-1)ⁿ , so order of a divides order of a^-1
let order of a^-1 =m. so (a^-1)^m=e if and only if a^m =e , so order of a^-1 divides order of a
so they are equal.
Q.E.D
is the proof correct?

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Homework Answers

Answer #1

You Proof is correct, what you did is that you showed order of 'a' divides order of a^-1 and order of a^-1 divides order of a. So they should be equal, I am writing down your proof in other simple way.

But this proof is valid when order of 'a' is finite. Can you think a proof when a is of infinite order?

Just proof that in later case order of a^-1 is not finite. To prove this assume order of a^-1 is finite then proceed in same manner and you will get a contradiction that order of 'a' is also finite.

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