Let $G$ be a group, with $H$ normal within it such that $|G| = r|H|$. Show that $g^r \in H$ for all $g \in G$.

Question

Let $G$ be a group, with $H$ normal within it such that $|G| = r|H|$. Show that $g^r \in H$ for all $g \in G$.

Here is my proof:

If $|G| = r|H|$, then $[G:H] = r$, which means that $G/H$ has $r$ elements in it, each one corresponding to one of the $r$ cosets of $H$. Pick a system of representatives $e, x_1, x_2 \dots, x_{r - 1}$ for these $r$ representatives, with $e = H$, the identity of $G/H$. Consider some $x_j \neq e$, and note that $|\langle x_j \rangle| = k$, the order of $\langle x_j \rangle$, must divide $|G/H| = r$. We know that $x_j^{k} = e$, so since $k|r$, $x_j^r = x_j^{r \mod{k}} = x_j^0 = e$.

Since $x_j$ is just a representative for the $j$th coset, any element within that same coset could also be a representative, and thus any element within that coset follows the rule that its $r$th power is $e$ (i.e. an element of $H$. Since the cosets form a partition of $G$, every element of $g$ is in some coset, and thus every element in $G$ follows the rule that its $r$th power is in $H$.

Is my proof correct?

Answer 1

Yes, your proof is correct.

There is a slight abuse of notation when you typed $e=H$, though, since $e$, like the $x_i$, is a representative of the coset $H$, not the coset itself.

Answer 2

Your proof is correct, but probably too long.

Since $H$ is normal in $G$, the result follows directly from Fermat–Euler's theorem for groups applied to the quotient group $G/H$:

If $G$ has order $n$ and $g \in G$, then $g^n=e$.

In its turn, this result follows directly from Lagrange's theorem applied to the subgroup $H = \langle g \rangle$, because $o(g)=|H|$, which divides $|G|$.

In your case, just note that $H=(gH)^r=g^r H$ implies $g \in H$.