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Since $p$ is the smallest prime dividing $|G|$, and $|G:H|=p$, we know that $H$ is the largest proper subgroup of $G$. So the normalizer $N_G(H)$ must be equal to either $G$ or $H$.

If $N_G(H)=G$, then $H$ is normal, so we are done. If $N_G(H)=H$, I have to show that it leads to a contradiction.

If we let $G$ act on conjugates of $H$ by conjugation, then the stabilizer of $H$ becomes $N_G(H)$ and it is equal to $H$ so by the orbit-stabilizer theorem we obtain the order of orbit of $H$ must be equal to $p$. This is where I have figured out so far, but don't know how to proceed from here. Any hint would be greatly appreciated.

user642721
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    https://math.stackexchange.com/questions/268236/g-is-a-finite-group-such-that-g-n-and-p-is-minimal-prime-dividing-n and https://math.stackexchange.com/questions/1825321/if-a-subgroup-has-smallest-prime-index-then-it-is-normal and https://math.stackexchange.com/questions/1560028/h-leq-g-s-t-g-h-is-the-least-prime-dividing-g-show-h-is-normal and https://math.stackexchange.com/questions/299333/let-g-be-finite-and-let-p-be-the-smallest-prime-dividing-g-let-h-le-g and many others. – Gerry Myerson Mar 23 '20 at 10:07

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Let $S = \{ H = g_1 H,g_2 H, \cdots , g_p H \}$ be the set of all left cosets of $G$. Then $G$ acts on $S$ by \begin{equation} g \cdot g_i H := (g g_i) H. \end{equation}
Thin induces a homomorphism of groups $\phi : G \rightarrow S_p$. Let $K = \text{Ker } \phi$. Observe that $K \subseteq H$ and $K$ is normal. Now $| G/K|$ divides $|G|$ and $|G/K|$ divides $|S_p|$. You can now use counting arguments to show that $K = H$.

Arpan Dutta
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