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Statement: Let $G$ be a group and $H$ a subgroup with $[G:H]=8$. Assume $G/H$ is a quotient group. If $g\in G$ has odd order, then $g\in H$.

I don't quite understand this concept of $[G:H] = 8$ and how it connects to proving an element in $G$ has an odd order. Just looking for a quick clarification on this aspect of the proof. Thank you!

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Let $m$ be the order of $g$. Note that $(gH)^m = g^m H = eH = H$, so the order of $gH$ must divide the odd number $m$. But since the order of $G/H$ is 8 we also know that the order of $gH$ must divide $8$. Hence the order of $gH$ must be $1$, and so $gH=H$. Hence $g \in H$.

So, what mattered was that $[G:H]=8$ gives the size of the quotient group $G/H$, and this had no odd divisors.

(You are tacitly assuming that $H$ is normal if $G/H$ is to be a group.)

Randall
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  • So essentially we show that the order of $gH$ being 1 shows that since $G$ must have an odd order, the only odd number divisible by 8 is 1. –  Apr 24 '21 at 18:08
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    The order of $gH \in G/H$ being $1$ means that it must be the identity element, which is $H$. But $gH = H$ if and only if $g \in H$. – Randall Apr 24 '21 at 18:11