If $H,K \leq G$ a finite group, then $\left\lvert HK \right\rvert = \cdots$

Question

If $H,K \leq G$ a finite group, then $$\left\lvert HK \right\rvert = \frac{\left\lvert H \right\rvert \cdot \left\lvert K \right\rvert}{\left\lvert H\cap K \right\rvert}.$$

The first part of the problem asked us to show this when $\left\lvert H\cap K \right\rvert=1$. This was doable and I used a counting argument saying how if we had some $h_ik_j=h_lk_m$ then $h_i=h_l$ and $k_j=k_m$. Then counting the size of the set $HK$ was straightforward.

My attemps for this problem have been saying something like $$\frac{\left\lvert H \right\rvert \cdot \left\lvert K \right\rvert}{\left\lvert H\cap K \right\rvert}= \frac{\left\lvert H / (H \cap K) \right\rvert \cdot \left\lvert H \cap K \right\rvert^2 \cdot \left\lvert K / (H \cap K ) \right\rvert}{\left\lvert H \cap K \right\rvert}=\left\lvert H \cap K \right\rvert \cdot \left\lvert H / (H \cap K) \right\rvert \cdot \left\lvert K / (H \cap K) \right\rvert$$ $$= \left\lvert H \right\rvert \cdot \left\lvert K / (H \cap K) \right\rvert.$$

Any ideas for ways I could solve this?

Federica Maggioni · Accepted Answer · 2013-04-30T09:01:24.947

Consider firstly the case: $H\cap K=\{e\}$. So we want to show that $$|HK|=|H|\cdot |K|$$ One should ask: how could this fail to happen?

The answer clearly must be that, if we list all the elements $hk$, with $h\in H,k\in K$, there should be some collapsing, that is some element in the list must appear at least twice. Equivalently, for some $h\neq h_1\in H$, we must have $$hk=h_1k_1$$ But then $$h_1^{-1}h=k_1k^{-1}$$ so that $$h_1^{-1}h\in H\cap K=e$$ leading to the contadiction: $h=h_1$. So no collapsing can occur.

Now the general case. We ask ourselves: how often does a given element $hk$ appear as a product of the list for $HK$. Our claim is that it must appear exactly $|H\cap K|$ times. To see this, we firstly remark that if $h_1\in H\cap K$, then $$hk=(hh_1)(h_1^{-1}k)$$ thus $hk$ is duplicated in the product at least $|H\cap K|$ times.

However, if $hk=h'k'$, then $$h^{-1}h'=kk'^{-1}=:u$$ and $u\in H\cap K$, and so $h'=hu$ and $k'=u^{-1}k$

Thus all duplications were accounted for in the previous equation: $hk=(hh_1)(h_1^{-1}k)$. Consequently, $hk$ appears in the list exactly $|H|\cdot |K|$ times. Thus the number of distinct elements in $HK$ is the total number of elements in the list of $HK$, i.e. $|H|\cdot |K|$, divided by the number of times a given elements appear, i.e. $|H\cap K|$, as we claimed.

I don't think this is quite right. $h'^{-1}h$ and $k'k^{-1}$ are certainly in $H\cap K$, but $hk$ and $h'k'$ don't have to be. Take some $h$ and $k$ such that $hk\in HK\setminus H\cap K$ and $x\in H\cap K$, then $hk=(hx)(x^{-1}k)$. — Elmar Zander, Apr 30 '13 at 08:18
@ElmarZander yes you are right, i expressed the concept really bad, i tried an edit — Federica Maggioni, Apr 30 '13 at 08:21
It still looks like you're saying that $H\cap K=\lbrace hk\rbrace$. — tomasz, Apr 30 '13 at 08:28
@tomasz yes, i was confusing, now i've tried to explain in full detail — Federica Maggioni, Apr 30 '13 at 09:02

score 4 · Answer 2 · answered Apr 30 '13 at 07:54

Consider the surjection $H\times K\to HK:(h,k)\mapsto hk$. For any $x\in H\cap K$, $(hx,x^{-1}k)\mapsto hk$, so each element $hk\in HK$ has at least $|H\cap K|$ preimages in $H\times K$.

For a fixed element of $HK$, I claim for any two preimages $(h,k)$ and $(h',k')$ in $H\times K$ we have $h'=hx$ and $k'=x^{-1}k$ for some $x\in H\cap K$. By assumption $hk=h'k'$, so $h'^{-1}h=k'k^{-1}\in H\cap K$. Let $y$ be this element. Then $$ h'=h(h^{-1}h')=hy^{-1}$$ and $$ k'=(k'k^{-1})k=yk. $$ So each element of $HK$ has $|H\cap K|$ preimages in $H\times K$, so $$ |HK||H\cap K|=|H\times K|=|H|\cdot |K| $$ and the conclusion follows.

If $H,K \leq G$ a finite group, then $\left\lvert HK \right\rvert = \cdots$

2 Answers2