# double coset counting formula

the double coset counting formula is a relation inter double cosets $HaK$, where $a\in G$ and $H,K$ subgroups in $G$. This is:

$\#(HaK)=\frac{|H||K|}{|H\cap aKa^{-1}|}$

and

$\#(G/K)=\sum_a[H;H\cap aKa^{-1}]$

The proof is easy.

One is to be bounded to the study of the natural map $H\times K\stackrel{\phi_a}\to HaK$. And it uses the second abstraction lemma.

The formula allows you to see the kinds of subgroups of arbitrary $H$ versus $K$ a $p-SS$ of $G$, $p-SS$ for the set of the $p$– Sylow subgroups.

Or, you can see that through the action $H\times G/K\to G/K$ via $h\cdot aK=haK$ you can get:

• ${\rm Orb}_H(aK)=\{haK\}$ which comply the equi-partition
• $HaK=aK\sqcup haK\sqcup...\sqcup h_taK$, so $\#(HaK)=m|K|$, for some $m\in \mathbb{N}$
• ${\rm St}_H(aK)=H\cap aKa^{-1}$

then you can deduce:

$|G|=\sum_a\frac{|H||K|}{|H\cap aKa^{-1}|}$

Now, let us use those ideas to prove the next statement:

Let $G$ be a finite group, with cardinal $|G|=q_1^{n_1}q_2^{n_2}\cdots q_t^{n_t}$, where each $q_i$ are primes with $q_1 and $n_i$ positive integers.

Let $H$ be a subgroup of $|G|$ of index $[G:H]=q_1$.

Then, $H$ is normal.

Proof:

By employing $K=H$ in the double coset partition, one get the decomposition:

$G=HeH\sqcup Ha_1H\sqcup...\sqcup Ha_tH$

So by the double coset counting formula you arrive to:

$|G/H|=1+[H:H\cap a_1Ha_1^{-1}]+\cdots+[H:H\cap a_tHa_t^{-1}]$

i.e.

$q_1=1+\frac{|H|}{|H\cap a_1Ha_1^{-1}|}+\cdots+\frac{|H|}{|H\cap a_tHa_t^{-1}|}$

From this, we get $\frac{|H|}{|H\cap a_iHa_i^{-1}|}.

But $|G|=q_1|H|$ as well $|H|=|H\cap a_iHa_i^{-1}|[H:H\cap a_iHa_i^{-1}]$ so

$|G|=q_1|H\cap a_iHa_i^{-1}|[H:H\cap a_iHa_i^{-1}]$, i.e.

$[H:H\cap a_iHa_i^{-1}]$ divides $|G|$

Then $[H:H\cap a_iHa_i^{-1}]=1$. So $|H|=|H\cap a_iHa_i^{-1}|$ for each $a_i$.

This implies $H=H\cap a_iHa_i^{-1}$ and so $H=a_iHa_i^{-1}$ for all the posible $a_i$, hence, $H$ is normal.

QED.