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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?

Shaun
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bzm3r
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  • 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 , $g^r\in H$ – Sumanta Oct 14 '18 at 05:21
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    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 – Exit path Oct 14 '18 at 05:25

2 Answers2

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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.

Shaun
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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 $g \in H$.

lhf
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  • See also my pet question: https://math.stackexchange.com/questions/28332/is-lagranges-theorem-the-most-basic-result-in-finite-group-theory. – lhf Sep 07 '19 at 14:13
  • For when $H$ is not normal, see https://math.stackexchange.com/questions/573050/if-h-is-a-subgroup-of-g-of-finite-index-n-then-under-what-condition-gn and https://math.stackexchange.com/questions/108952/if-gh-n-then-gn-in-h-for-all-g-in-g – lhf Sep 07 '19 at 14:18