2013/10/18 · 08:24
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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<q_2<...<q_t 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}|}<q_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.

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