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If $G$ is finite and $H$ is a subgroup of $G$ s.t $H\neq G$ then $G\neq\bigcup_{g\in G} H^g$, where ($H^g = gHg^{-1}$).

I managed to show a bijection $N_G(H)g \rightarrow H^g$, i.e from the right cosets of $N_G(H)$ to the set of conjugates of $H$ in $G$, therefore the number of $H$ conjugates in $G$ is $|G:N_G(H)|$

on the other hand, we know that $|G| = |G:H|\cdot|H|$ and since $H\neq G$, $|H|<|G|$

now I'm trying to combine all of those to get to my result, but no success so far.

Guidance will be appreciated.

Thanks.

Jyrki Lahtonen
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Philip L
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  • Sorry, typo. I meant $G\neq\bigcup_{g\in G} H^g$ :) – Philip L Feb 18 '19 at 15:15
  • Consider the action of $G$ on the set of cosets $G/H$. Then $H^g$ is the stabilizer of $gH$. Meaning that there are $|H|\cdot [G:H]=|G|$ pairs $(g,xH)$ such that $g\cdot xH= xH$. Meaning that, in the average, an element $g$ is a part of one just pair. The element $1\in G$ is a part of $[G:H]$ such pairs, because $1$ is in all the stabilizers. For the average to come out as exactly one, there must be some element $x\in G$ that is not involved in any such pair. Meaning that $x$ is not contained in the union of conjugates. – Jyrki Lahtonen Feb 18 '19 at 15:23
  • I'm nearly positive that we have covered this already. Did you look for a duplicate? – Jyrki Lahtonen Feb 18 '19 at 15:24
  • This is the closest I could find - https://math.stackexchange.com/questions/1172377/can-g-be-a-union-of-some-ofh-copies – Philip L Feb 18 '19 at 15:25
  • I'm not sure I understand your meaning by "average" :( – Philip L Feb 18 '19 at 15:27
  • See here for history. And here for a highly voted version. – Jyrki Lahtonen Feb 18 '19 at 15:28
  • To each element $g\in G$, let $n(g)$ be the number of cosets $xH$ such that $g(xH)=xH$. The average of $n(g)$ is $$\Bbb{E}(n(g))=\frac1{|G|}\sum_{g\in G}n(g).$$ Think of $g$ as a random variable ranging over $G$ if you like. Anyway $\sum_g n(g)$ come to $|G|$, so $\Bbb{E}(n(g))=1$. – Jyrki Lahtonen Feb 18 '19 at 15:31

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