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


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.

  • $\begingroup$ For finding example identical left cosets look at monogen groups and its sub-groups. $\endgroup$
    – EDX
    Oct 26 '20 at 18:34
  • $\begingroup$ 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. $\endgroup$ 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 $|I|=|J|$.

  • $\begingroup$ Should not the function be defined as $\phi :I \to J$? and how do we know such a function is bijective? $\endgroup$
    – 45465
    Oct 26 '20 at 18:41
  • $\begingroup$ 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. $\endgroup$
    – Eureka
    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.


\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.


  • 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.