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Let $G$ be a finite group with a subgroup $H$ of index 5 and 5 is the smallest prime divisor of $|G|$.

Let $X=\{gH:g\in G\}$ be the set of left cosets of $H$ in $G$ and $H$ acts on $X$ by left multiplication.

Show that every orbit of $X$ has length 1.

I've tried by supposing that there is an orbit of length 2 but am stuck trying to show this and don't know if there is a better way?

1 Answers 1

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Notice that $\{H\}$ is one orbit.

Consider an orbit $\Delta\ne\{H\}$ with $gH\in\Delta$ for some $g\in G$.

By the Orbit-Stabiliser Theorem $|\mathrm{Stab}_H(gH)|=|H|/|\Delta|$. In particular $|\Delta|$ divides $|H|$ and therefore $|G|$, but $|X|=5$ so $|\Delta|\le 4$.

Since $5$ is the smallest prime dividing $|G|$ this leaves $|\Delta|=1$ as required.

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    I'm a bit unclear on a few things here. Firstly, how do we know ${H}$ is one orbit? Also why can't we write the orbit stabelizer theorem as $|G{_x}|=|G|/\Delta|$? and deduce this way that $|\Delta|$ divides $|G|$ rather than going through $H$? and finally why can't $|\Delta|$ be 4? – 2017-02-03
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    You wrote $H$ acts on $X$ by left multiplication. For any $h\in H$ we have $hH=H$ so the orbit containing $H$ is $\{H\}$. The action we are using is an action of $H$ on $X$, so orbit-stabilizer must be applied using $H$. If $|\Delta|=4$ then $2$ divides $|\Delta|$ and therefore $|G|$. – 2017-02-03
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    When you say for any $h\in H, hH=H$ how do we know that? Is this because the identity produces one orbit? – 2017-02-06
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    $hH\subseteq H$ by closure of $H$. $H\subseteq hH$ as for $g\in H$ we have $g=h(h^{-1}g)\in hH$ – 2017-02-06
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    How exactly does $|X|=5$ imply that $|\Delta|$ is less than or equal to 4? – 2017-02-06
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    $H\notin\Delta$ so $\Delta\subseteq X\setminus\{H\}$ and $|X\setminus\{H\}|=|X|-1=4$ – 2017-02-06