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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
up vote 9 down vote accepted

It's only a superficial change, but since you seem to be having issues accepting your (correct) argument, maybe you will like this better:

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 $x^m \in H$.

In any case: believe it! It's an extremely useful result: I have used it many times in my work in algebra and number theory and have several times noted that students don't seem to be as aware of it as they should be. (In fact, I like to think of this in terms of the period-index problem in Galois cohomology, but that's really beyond the scope of this answer.)

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+1 This proof is definitely more insightful and best of all for the beginner,it's not significantly more sophisticated,Pete. Moral For The Beginner: Sometimes an indirect proof yields greater insight into the conceptual meaning of a result! – Mathemagician1234 Feb 17 '12 at 7:45

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