Element of a group G raise to the index of a subgroup H is in H

Let $H$ be normal in $G$ and $[G:H]=m$. Prove that $x^m$ is in $H$ for all $x$ in $G$.

Since $H$ is normal its left cosets form a group. Also since $m$ is the order of that group $(xH)^m=x^mH=H$, which implies that $x^m \in H$.

I am actually not sure what my question is, it just hard to believe that if I pick any element of a group $G$ and I raise it to $m$ the resulting element will be in $H$. Anyways, is there another way one prove this? Thanks.

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Well clearly it's true and in my opinion the most intuitive proof possible. –  anon Feb 17 '12 at 5:11
This depends on H being normal. When G is the symmetric group on 3 points and H={(),(2,3)} then [G:H]=3 but for x=(1,2), x^3=x is not contained in H; the smallest m that works for all of G is 6. When G is a p-group, G and all of its subgroups have lots of normal subgroups of index p, and so raising things to the p'th power moves elements closer and closer to the identity (sort of like a derivative lowering the degree of polynomials). More general groups are more complicated, but that just means you have to choose "m" instead of "p" more carefully. –  Jack Schmidt Feb 17 '12 at 14:09

Instead of talking (explicitly) about cosets, let's talk (equivalently, of course!) about the quotient group and the quotient map $q: G \rightarrow G/H$. For $x \in G$, since $\# (G/H) = m$, by Lagrange's Theorem, $q(x)^m = e$ (the identity) in $G/H$. But since $q$ is a homomorphism, $e = q(x)^m = q(x^m)$, so $x^m \in H$.