0
$\begingroup$

Let $G$ be a finite group with a subgroup $H$ of index 5. Also 5 is the smallest prime divisor of $|H|$. Let $X=\{gH:g\in G\}$ be the set of left cosets of $H$ in $G$, $K$ acts on $X$ by left multiplication.

Every orbit of X has length 1 (See here). Using this fact that every orbit has length 1, show that $hg\in gH$ for $h\in H$, and $g\in G$.

I'm not sure how I would show this, I tried by assuming it was normal and therefore $gH=Hg$ but I think this is wrong because I don't think I can assume that orbits of length 1 means it's normal?

1 Answers 1

0

Let $gH$ be one coset in $X$. Since it's orbit has only one element (i.e., one coset), you just need to figure out which coset it is. You can show fairly easily that $gH$ must be in its own orbit, so the orbit of $gH$ is just $\{gH\}$. Then you know that for any $h\in H$, $h(gH)$ must be in the orbit of $gH$, which means that $h(gH)=gH$. But $h(gH)=\{hgk|k\in H\}$, so you can use that to deduce that $hg\in gH$.

  • 0
    Thank you, do you know how I would use this fact to prove the theorem 'If 5 is the smallest prime dividing the order of a finite group G, then any subgroup of index 5 in G is normal'. My thoughts were that maybe by showing $hg\in gH$ we have shown that $H$ is normal? And because of this every H would be normal.. I'm unsure if I can just assume that. – 2017-02-08
  • 0
    Normality is equivalent $gH=Hg$ for all $g$. You've shown that $hg\in gH$, which is only a few steps away from normality. For the results of this question (and the other) you assumed that 5 was the smallest prime divisor of $\vert H\vert$. For the next question you are assuming that $H$ has index 5. Can you show that $[G:H]=5$ implies that 5 is the smallest prime dividing $\vert H\vert$? – 2017-02-10
  • 0
    Okay, so I need to show $hg$ also $\in Hg$. Using the same logic would I show this by saying: the orbit of $Hg$ would be $\{Hg\}$ and for any $h\in H$, $h(Hg)=Hg$. Then since $h(Hg)=\{hkg|k\in H\}$, $hg\in Hg$. So H is normal? Then $[G:H]=5$ implies 5 is the smallest prime diving $|H|$ since we can write $|H|=|G|/5$? But then I don't know how to tie this all together in the proof – 2017-02-10