# Lemma used to prove $\left|HK\right|=\frac{\left|H\right|\left|K\right|}{\left|H \cap K\right|}$

Given a group $$G$$ and $$H,K \le G$$,then :

$$\left|HK\right|=\frac{\left|H\right|\left|K\right|}{\left|H \cap K\right|}$$

Where $$HK:=\left\{hk:h \in H ,k \in K\right\}$$

Lemma:

For $$h_1,h_2 \in H$$

$$hK=h'K \iff h(H \cap K)=h'(H \cap K)$$

We have:

$$HK=\bigcup_{h \in H}hK$$

Not every such left cosets of $$K$$ in $$H$$ are distinct, on the other hand the function $$\phi:hK \to K$$ with $$hk \mapsto k$$ is a bijection, so the number of elements in $$hK$$ is the same as that $$K$$'s , here I showed that the set of left cosets (equivalently right cosets) partitions the group.

By this we see that :$$\left|HK\right|=\left|\color{blue}{\text{the set consiting of all distinct left cosets }}hK\right|\left|K\right|$$

One concludes from the lemma that the number of such distinct left cosets is the same as $$\left|H: H \cap K\right|$$ but I don't know how such a conclusion is possible, how the lemma helps us?

It looks that $$hK \ne h^{'} K$$ iff $$h(H \cap K) \ne h^{'}(H \cap K)$$ and the order of the set of all such distinct $$h(H \cap K)$$ for $$h \in H$$ is $$\left|H: H \cap K\right|$$...

Also, it would be appreciated if someone gives me an example where such left cosets $$hk$$ are identical.

• For finding example identical left cosets look at monogen groups and its sub-groups.
– EDX
Oct 26 '20 at 18:34
• The formula only makes sense when the quantities are finite. But the equality $|HK||H\cap K| = |H||K|$ holds in the sense of cardinalities in all cases, so it should be used instead. Oct 26 '20 at 18:42

Consider the map $$\varphi: H/H\cap K\longrightarrow HK/K$$ by $$h(H\cap K)\mapsto hK$$.

1. This is a well defined map by your lemma $$\impliedby$$.

2. This map is injective by your lemma $$\implies$$.

3. This map is surjective by definition of $$HK$$.

Therefore this is a natural one-to-one correspondence between these cosets, and the Product Formula follows immediately.

I happen to have written about this yesterday, so here's a link for you https://ml868.user.srcf.net/ExpositoryWritings/Groups3.pdf. There are a few typos I haven't fixed but I hope it is readable and somewhat inspiring.

You noted that in the union $$\bigcup_{h \in H} hK$$, some cosets appear more than once. If you are able to show that each distinct coset appears $$|H \cap K|$$ times in the union, then you can arrive at the desired conclusion.

The lemma implies that the only way $$hK=h'K$$ can happen (for $$h,h' \in H$$) is if $$h' = gh$$ for some $$g \in H \cap K$$. In particular, for a given coset $$hK$$, it appears in the union $$|H \cap K|$$ times as $$(gh)K$$ for each $$g \in H \cap K$$.

For simplicity: $$I=\{hK|h\in H\}$$ $$J=\{h(H\cap K)|h\in H\}$$ Notice that: $$|J|=|H:(H\cap K)|$$ And we have simply to prove that $$|I|=|J|$$ Thanks to the lemma the application: $$\omega: I \to J$$ $$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ hK \mapsto h(H\cap K)$$ Is a bijection, in fact firstly the application is well defined since: $$hK=h'K \Rightarrow^{\text{Lemma}} h(H\cap K)=h'(H\cap K)\Rightarrow \omega(hK)=\omega(h'K)$$ The application is also injective: $$\omega(hK)=\omega(h'K)\Rightarrow h(H\cap K)=h'(H\cap K)\Rightarrow^{\text{Lemma}} hK=h'K$$ And it's clearly surjective because for every $$h(H\cap K)\in J, \omega (hK)=h(H\cap K)$$ It follows $$|I|=|J|$$.

• Should not the function be defined as $\phi :I \to J$? and how do we know such a function is bijective? Oct 26 '20 at 18:41
• It's a matter of notation. Briefly this application associates to every $hK$, $h(H\cap K)$, Now I'll write the proof that the application is a bijection. Oct 26 '20 at 18:43

The equivalence relation $$(h,k)\sim (h',k')\stackrel{(def.)}{\iff} hk=h'k'$$ induces a partition of $$H\times K$$ into equivalence classes each of cardinality $$|H\cap K|$$, and the quotient set $$(H\times K)/\sim$$ has cardinality $$|HK|$$. Therefore, $$|H\times K|=|H||K|=|H\cap K| |HK|$$, whence (if $$H$$ and $$K$$ are finite, in particular if they are subgroups of a finite group) the formula in the OP. Hereafter the details.

(Note that the formula holds irrespective of $$HK$$ being a subgroup.)

Let's define in $$H\times K$$ the equivalence relation: $$(h,k)\sim (h',k')\stackrel{(def.)}{\iff} hk=h'k'$$. The equivalence class of $$(h,k)$$ is given by:

$$[(h,k)]_\sim=\{(h',k')\in H\times K\mid h'k'=hk\} \tag 1$$

Now define the following map from any equivalence class:

\begin{alignat*}{1} f_{(h,k)}:[(h,k)]_\sim &\longrightarrow& H\cap K \\ (h',k')&\longmapsto& f_{(h,k)}((h',k')):=k'k^{-1} \\ \tag 2 \end{alignat*}

Note that $$k'k^{-1}\in K$$ by closure of $$K$$, and $$k'k^{-1}\in H$$ because $$k'k^{-1}=h'^{-1}h$$ (being $$(h',k')\in [(h,k)]_\sim$$) and by closure of $$H$$. Therefore, indeed $$k'k^{-1}\in H\cap K$$.

Lemma 1. $$f_{(h,k)}$$ is bijective.

Proof.

\begin{alignat}{2} f_{(h,k)}((h',k'))=f_{(h,k)}((h'',k'')) &\space\space\space\Longrightarrow &&k'k^{-1}=k''k^{-1} \\ &\space\space\space\Longrightarrow &&k'=k'' \\ &\stackrel{h'k'=h''k''}{\Longrightarrow} &&h'=h'' \\ &\space\space\space\Longrightarrow &&(h',k')=(h'',k'') \\ \end{alignat}

and the map is injective. Then, for every $$a\in H\cap K$$, we get $$ak\in K$$ and $$a=f_{(h,k)}((h',ak))$$, and the map is surjective. $$\space\space\Box$$

Now define the following map from the quotient set:

\begin{alignat}{1} f:(H\times K)/\sim &\longrightarrow& HK \\ [(h,k)]_\sim &\longmapsto& f([(h,k)]_\sim):=hk \\ \tag 3 \end{alignat}

Lemma 2. $$f$$ is well-defined and bijective.

Proof.

• Good definition: $$(h',k')\in [(h,k)]_\sim \Rightarrow f([(h',k')]_\sim)=h'k'=hk=f([(h,k)]_\sim)$$;
• Injectivity: $$f([(h',k')]_\sim)=f([(h,k)]_\sim) \Rightarrow h'k'=hk \Rightarrow (h',k')\in [(h,k)]_\sim \Rightarrow [(h',k')]_\sim=[(h,k)]_\sim$$;
• Surjectivity: for every $$ab\in HK$$ , we get $$ab=f([(a,b)]_\sim)$$. $$\space\space\Box$$

Finally, the formula holds irrespective of $$HK$$ being a subgroup, which was never used in the proof.