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

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?

• Actually you can say something more : for any $g\in G$ we have $k_g\in \Bbb N$ such that $g^{k_g}\in H$ whenever $H$ is a subgroup (not necessarily normal) of $G$ with finite index. In case your $H$ is normal you can use Lagrange's theorem to prove that order of subgroup of the group $G/H$ divides the order of the group $G/H$ to conclude that each $g\in G$ has the property , $g^r\in H$ – 0-th User Oct 14 '18 at 5:21
• It suffices to show that for any group $G$ of order $m$, we always have $x^m=e$, for all $x \in G$. This follows directly from Lagrange's theorem – leibnewtz Oct 14 '18 at 5:25
• This is a cool question! – coreyman317 Sep 7 '19 at 13:45

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.
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 $$g \in H$$.