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Let $G$ be a finite group, $H \leq G$ such that $[G : H]$ is the least prime which divides $|G|$. Show that $H$ is normal in $G$.

Can someone explain what information we get from knowing that $[G : H]$ is the least prime which divides $|G|$? By Lagrange, $|H| = \frac{|G|}{[G : H]}$. Since $[G : H]$ prime, then $|G|$ should be a power of that prime. By Sylow theorem $1$, then there is a Sylow $p$-group (where $p$ is the prime) with order $p^n$. I don't know if this is the correct path and I also don't know why we care about the least prime.

If someone can explain the solution for this exercise, I would be grateful. I've been banging my head on this for forever.


marked as duplicate by Derek Holt group-theory Dec 4 '15 at 20:33

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  • 1
    $\begingroup$ This answer does a nice job of summarizing the standard proof of this result. $\endgroup$ – Bungo Dec 4 '15 at 18:55

First note that $G$ operates on $G/H$ via the operation $$ \tau * \sigma H := (\tau \sigma) H. $$

This operation is transitive, since $\tau * H = \tau H$. The fixed group of $\tau H$ $G_{\tau H}$ is given by $\{\sigma \in G | \tau^{-1} \sigma \tau \in H\}$, since $$ \sigma * \tau H = \tau H \Leftrightarrow \sigma \tau H = \tau H \Leftrightarrow \tau^{-1} \sigma \tau H = H. $$

This operation induces a homomorphism $\varphi: G \to S_{G/H}$ via the map $$ \varphi(\sigma) := \tau H \mapsto \sigma * \tau H. $$ The kernel of this homomorphism is exactly $K := \bigcap\limits_{\tau \in G} G_{\tau H}$, since $$ \varphi(\sigma) = id_{G/H} \Leftrightarrow \forall \tau \in G : \sigma \tau H = \tau H \Leftrightarrow \forall \tau \in G : \tau^{-1} \sigma \tau H = H. $$

Note that kernels of homomorphisms are always normal subgroups. There is a trick to memorise this: All normal subgroups occur as kernels, since they are the kernel of the projection homomorphism to the factor group. Hence, the kernels of homomorphisms are exactly the normal subgroups.

Hence, if you want to check that something is a normal subgroup, you may try to check that it's the kernel of something.

By the first isomorphism theorem, we have $$ G / K \cong \text{Im } \varphi. $$ Now $\text{Im } \varphi$ is a subgroup of $S_{G/H}$ and therefore its order (which by the above isomorphism is equal to the order of $G/K$) divides the order of $S_{G/H}$ by Lagranges theorem; in short, $$ |G / K| \big| \underbrace{|S_{G/H}|}_{= p!}, $$ where $p$ is the smallest prime dividing $|G|$.

But since $|G/K| \big| |G|$ as well, only the possibilities $|G/K| = p$ or $|G/K| = 1$ remain, since $p!$ has only prime divisors smaller or equal than $p$ and $|G|$ only prime divisors larger or equal than $p$, and $p^k \nmid p!$ for $k \ge 2$. But $\text{Im } \varphi$ is non-trivial since the action is transitive, and thus $|G/K| = p$.

Now we note that $K$ is a subset of $H$ since $$ K := \bigcap_{\tau \in G} G_{\tau H} \subseteq G_H = \{\sigma \in G | \sigma \in H\} = H. $$

But since by $|K| = |G|/p = |H|$ it has the same cardinality as $H$, $K = H$.

  • $\begingroup$ Holy hell that is really a lot more convoluted than I was expecting. $\endgroup$ – complexLost Dec 4 '15 at 18:56
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    $\begingroup$ I also thought it would take less. But it is an extremely beautiful one, as one has to admit (at least in my eyes). $\endgroup$ – Cloudscape Dec 4 '15 at 19:01

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