7

Suppose you have a group $G$ of order $ms$ and it is $m \leq p$ for every prime factor $p$ of $s$. We want to show that a subgroup $H$ with $(G:H)=m$ is a normal subgroup of $G$.

So let $H$ be a subgroup of $G$ with $\vert H\vert=s$ and $K$ be another subgroup of $G$ of order $s$. Then we have

$$ \vert HK\vert= \frac{\vert H \vert \vert K \vert}{\vert H \cap K \vert}=\frac{s^2}{\vert H \cap K \vert} \leq ms \Longleftrightarrow \frac{s}{\vert H \cap K\vert} \leq m.$$

Suppose now that $\frac{s}{\vert H \cap K\vert}>1$. Is it always true that in this case $m$ is the smallest prime divisor of the group order? I read this in another post, but I don't see an argument for this.

My attempt:

$\vert H \cap K \vert$ is a divisor of $s$ (how to show?). Then $m$ is the smallest prime divisor of $\frac{s^2}{\vert H \cap K\vert}$. But how do we get to the order of $G$ in this case?

Maybe there is an easier way to prove this statement...

Arnaud Mortier
  • 27,742
Jan
  • 5,622
  • 10
  • 20
  • 32
  • Yes if $|G:H|=p$ is the smallest prime dividing $|G|$, then $H$ is normal in $G$. – Wang Kah Lun Aug 05 '18 at 14:41
  • 2
    @AlanWang I know, but why is in the upper case $m$ the smallest prime divisor of $\vert G\vert=ms$? With the explanation of this the given statement follows immediately. – Jan Aug 05 '18 at 14:43
  • Are you sure you want to require $m \leq p$, not $m < p$ ? If $m < p$ is what you mean, then there is a nice proof: Consider the composition $H \to G \to S_{G/H}$, where the first arrow is the inclusion map and where the second arrow is the standard action of $G$ on the coset space $G/H$. This composition must be the trivial map (sending everything to $\operatorname{id}$), since the size $s$ of $H$ is coprime to the size $m!$ of $S_{G/H}$. Thus, $H$ acts trivially on $G/H$; but this is equivalent to $H$ being normal in $G$. – darij grinberg Aug 07 '18 at 16:05
  • Ah, I see. If some prime factor $p$ of $s$ satisfies $m \leq p$ but not $m < p$, then it must satisfy $m = p$, so that $m$ itself is prime. And then the result follows from the classical fact that @AlanWang quoted (see, e.g., https://math.stackexchange.com/questions/164244/normal-subgroup-of-prime-index for its proof). – darij grinberg Aug 07 '18 at 16:09
  • @darijgrinberg There is an easier proof if $m<p$: when one reaches $$\frac{s}{\vert H \cap K\vert} \leq m$$ the LHS is a product of some primes found in $s$, so if it is less than or equal to $m$, then it is equal to $1$, meaning $H = K$. – Arnaud Mortier Aug 07 '18 at 17:09

1 Answers1

1

You were able to prove that $$\frac{s}{\vert H \cap K\vert} \leq m.$$

Note that by Lagrange's Theorem and since $H\cap K$ is a subgroup of $H$, $\frac{s}{\vert H \cap K\vert}$ is actually an integer (which was one of your questions). Therefore if you write down the prime factorization of $s$, $\vert H \cap K\vert$ is a product of some of these primes and $\frac{s}{\vert H \cap K\vert}$ is the complementary product of the remaining primes:

If $s=p_1^{\alpha_1}\ldots p_k^{\alpha_k}$ then $\frac{s}{\vert H \cap K\vert} =p_1^{\beta_1}\ldots p_k^{\beta_k}$ for some $0\leq\beta_i\leq \alpha_i$.

So you have $$\underbrace{p_1^{\beta_1}\ldots p_k^{\beta_k}}_{\text{Thereafter called LHS}}\leq m\leq \min_{i=1,\ldots , k}\{p_i\}$$ where the second inequality is from the fact that [being less than or equal to every element in a finite list] is equivalent to [being less than or equal to the minimum element from that list].

From there, it is clear that the LHS can either be equal to $1$ or to $\min_{i=1,\ldots , k}\{p_i\}$, otherwise it would violate transitivity of $\leq$.

If you assume $\frac{s}{\vert H \cap K\vert}>1$ then the LHS is not $1$ and it must be $m$. Therefore $m=\min_{i=1,\ldots , k}\{p_i\}$.

Arnaud Mortier
  • 27,742