Question

Group theory:
Let G be finite and let H be a subset if G. Let identity be in H.
Prove that if H is closed with respect to its operation, then it is closed with respect to inverses.

Answer 1

Something without induction.

Answer 2

This sounds like something you could use Lagrange's theorem on... it follows that \(H\) is a subgroup of \(H\), and by Lagrange's theorem you know that there exists some \(n\) for every \(a\in H\) s.t. \(a^n=e\in H\). It follows that \(a^{n-1}\) would be the inverse of \(a\) and must be in \(H\) since \(H\) is closed.

Answer 3

I don't know a full proof but I think that approach makes sense.

Answer 4

OOPS! \(H\) is a subgroup of \(G\).

Answer 5

I don't see the logic that (a^n)^(-1) must be in H

Answer 6

that notation only makes sense if we know the inverse is there to begin with.

Answer 7

Yeah @zzr0ck3r

Answer 8

$$a^n=e\\aa^{n-1}=e\\a^{n-1}a=e$$so \(a^{n-1}\) is the inverse of \(a\). Since \(H\) is closed under our operation we know \(a\in H\implies a^{n-1}\in H\)

Answer 9

um a*a^(n-1) = a^n

Answer 10

and again you are saying its there but not proving it

Answer 11

yes and \(a^n=e\) by Lagrange's theorem for some \(n\) that divides \(|G|\).

Answer 12

don't kill me if it's not right lol I don't have much experience with group theory or abstract algebra

Answer 13

I don't think this works, because you are just saying that e is in H and if e is in H of course its inverse is in H

Answer 14

nah nah its cool , its a good place to start.

Answer 15

its not homework, I just wanted to prove it to make my final easy next week:)

Answer 16

I have a proof with induction, It seems there should be something easier as the idea seems so obvious.