# Proof that $|HK|=|H||K|/|H \cap K|$ for $H,K,HK$ subgroups of $G$

I've found this problem in a book and devised my own proof. (took me like 5 hours and it seems trivial - just build a bijection). I am not sure that I haven't made any errors though. Perhaps some other proof would be simpler.

Statement:

$$G$$ is a group and $$H,K,HK \subseteq G$$

Prove that $$|HK|=\frac{|H||K|}{|H \cap K|}$$

Proof:

The statement above is equivalent to: $$\frac{|HK|}{|K|}=\frac{|H|}{|H \cap K|}$$

So now we look at the cosets of $$K$$ in $$HK$$ - ie. the elements of $$HK/K$$. They are exactly $$\frac{|HK|}{|K|}$$ because of Lagrange's theorem.
Then we look at the cosets of $$H \cap K$$ in $$H$$ - ie. the elements of $$H/H \cap K$$. They are exactly $$\frac{|H|}{|H \cap K|}$$ because of Lagrange's theorem.

So if we could find a bijection from $$HK/K$$ to $$H/H \cap K$$, we are done.

So let's look at the elements of $$HK/K$$, they are cosets of the form: $$h_1k_1K$$, but $$k_1K=K$$, so we get $$h_1K$$.

Then lets look at the elements of $$H/H \cap K$$, they are cosets of the form: $$h_1H \cap K$$.

So let's define $$f: HK/K \to H/H \cap K$$, $$f(hK)=hH \cap K$$.

To see that it's a function we need to show that it's well defined.

Let $$h_1K=h_2K$$, ie. $$h_2^{-1}h_1 \in K$$ which also implies $$h_2^{-1}h_1 \in H \cap K$$

Then we need to show that $$f(h_1K)=f(h_2K)$$.

So $$f(h_1K)=h_1H \cap K$$ and $$f(h_2K)=h_2H \cap K$$

To prove that $$h_1H \cap K = h_2H \cap K$$ we need $$h_2^{-1}h_1 \in H \cap K$$.

Hence f is well defined.

Now we need to show that it's injective.

Suppose $$f(h_1K)=f(h_2K)$$, but $$h1K \neq h2K$$.

Ie. $$h_1H \cap K = h_2H \cap K$$ but $$h_1K \neq h_2K$$.

$$h_1K \neq h_2K$$ implies $$h_2^{-1}h_1 \notin K$$ which implies $$h_2^{-1}h_1 \notin H \cap K$$.

Hence $$h_1H \cap K \neq h_2H \cap K$$.

So we know f is injective.

Now to check for surjectivity:

Since $$H/H \cap K$$ has elements of the form $$h_1H \cap K$$, for each of them we have, $$f(h_1K)=h_1H \cap K$$.

QED

• Use $h_2$ for $h_2$. Sep 23, 2020 at 0:15
• You can't quite use Lagrange's theorem on $HK/K$, because $HK$ isn't necessarily a group. (You can repeat the argument in the proof of Lagrange's theorem and check that it still works here - essentially, showing that $HK$ is a union of $|HK|/|K|$ cosets of $K$ in $G$ - but that's an extra step.) Sep 23, 2020 at 0:23
• I wrote in the title that $HK$ is a subgroup of $G$. Yes, I know that generally $HK$ is not a subgroup of $G$. So you're saying this statement is true, even if $HK$ is not a subgroup of $G$? Sep 23, 2020 at 0:26
• The statement is still true if $HK$ is not a subgroup. If $H$ or $K$ is infinite, then $HK$ is also infinite (because it contains both $H$ and $K$), and then you have $\infty = \infty$, so there's nothing to prove; if $H$ and $K$ are both finite, then $HK$ is also finite. Sep 23, 2020 at 0:33
• JCAA's proof works whether or not it's a subgroup (so does yours, except for the Lagrange's theorem statement). Sep 23, 2020 at 1:32

The best and standard way to prove it is to consider the map from $$H\times K$$ to $$HK$$ which sends $$(h,k)$$ to $$hk$$ and look at the equivalence classes of pairs mapped to the same element of $$HK$$. That is for every $$(h,k)$$ count pairs $$(h',k')$$ such that $$hk=h'k'$$.

Edit To make it complete, $$hk=h'k'$$ is equivalent to $$h^{-1}h'=k'k^{-1}$$. The LHS is in $$H$$, the RHS is in $$K$$, so both are in $$K\cap H$$. So the number of pairs $$(h',k')$$ such that $$hk=h'k'$$ is the same as the number of elements in $$h(H\cap K)$$ which is $$|H\cap K|$$.

• I don't know why, but this proof seems more complex to me. Ie. the proof I wrote seems trivial to me, like I could follow it and understand it without thinking. But maybe that's just cause I wrote it and everyone understands his proofs easily. Sep 23, 2020 at 0:08
• That is not "my" proof. It is in every abstract algebra book. But yes, usually proofs made by yourself seem easier. For me it is the easiest probably because I came up with it many years ago. Sep 23, 2020 at 0:15
• I see. Then what about my proof? Is it a standard proof too? Or perhaps not, because it's too long or I made some mistake? Also are there other interesting proofs? Sep 23, 2020 at 0:21
• Your proof does not seem to be correct because some phrases are hard to understand. Say, "we look at the cosets of $K$ as a subgroup of $HK$". Second, the statement you are proving is not general enough: the set $HK$ does not need to be assumed a subgroup. So this proof cannot be "standard". But other than that it is or is very close to a correct proof. Sep 23, 2020 at 0:42
• $HK$ is already partitioned into cosets $hK$ each of which has $|K|$ elements. You need to show that the number of cosets is $|H|/|(H\cap K)|$. I think it can be done as in your proof. But it is longer than "my" proof. Sep 23, 2020 at 1:01

There is a quite easy way to prove this problem. If you understand chinese, there is a classic proof on Yang Zixu's 'Abstract Algebra'. Here is the proof:

Since $$H\cap K\le H$$, let $$|H|/|H\cap K|=m$$ and $$H = h_1(H\cap K)\cup h_2(H\cap K)\cup \cdots \cup h_m(H\cap K),$$ here $$h_i\in H, h_i^{-1} h_j \notin K,i\neq j.$$ Clearly, $$HK=h_1K\cup h_2K\cup\cdots\cup h_m K,$$ while $$h_iK\cap h_jK = \varnothing,i\neq j,$$ thus $$|HK|=m|K|,$$ which means $$|HK|=|H||K|/|H\cap K|.$$ QED

It is an application of coset decomposition theory. There is no need of considering the bijection or maps etc.

• Why $HK= h_1K\cup...$ etc.?
– user810157
Sep 23, 2020 at 5:39
• Since $H$ is some union of $h_i (H\cap K)$,$HK$ is the sets of all $h_i \in H$ left times $k_j\in K$, thus it is the union of $h_i K$ Sep 23, 2020 at 5:54
• This proof only satisfies the case that $H,K$ are finite groups. Sep 23, 2020 at 6:09
• So you "mimic Lagrange's" even in the case $HK$ not a subgroup of $G$, nice. +1
– user810157
Sep 23, 2020 at 8:16
• Interesting. Also it seems I have to mimic Lagrange too in my proof, if HK is not a subgroup of G as another poster has noted in his comments. Sep 23, 2020 at 20:48

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, if $$H$$ and $$K$$ are finite (in particular if they are subgroups of a finite group), we get: $$|H\times K|=|H||K|=|H\cap K| |HK|$$, whence the formula in the OP. Hereafter the details.

(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.

In the case that$$H\triangleleft G$$, this follows immediately from the second isomorphism theorem.

But, actually your result is well known, and called the product formula. Neither $$H$$ nor $$K$$ is required to be normal. See "Product of group subsets - Wikipedia" https://en.m.wikipedia.org/wiki/Product_of_group_subsets

• Isn't it needed for H to be normal? Hm, I know that $HK=KH$ is equiv to $HK$ being a group. Btw a comment above said that $HK$ doesn't have to be a group for this statement to be true. Sep 23, 2020 at 0:33
• But do we need $HK$ to be a group for the statement to be true? A comment above is saying that we don't. Sep 23, 2020 at 0:39
• That I would have to think about.
– user403337
Sep 23, 2020 at 0:40
• Neither $H$ nor $K$ needs to be normal for $HK$ to be a subgroup. For example in a solvable group products of Sylow subgroups can be a subgroup (a Hall subgroup) even if the Sylow subgroups are not normal. This can be seen in $S_4$. It has a Sylow 2-subgroup $H$ of order $8$ (three of them, none of them normal) and $4$ subgroups $K$ of order $3$ (none of them normal), but $HK$ contains $8\cdot 3=24$ elements so $HK$ is a subgroup, in fact equal to $S_4$. Sep 23, 2020 at 1:18
• @JCAA you're correct. So this is more general than the Second Isomorphism Theorem, meaning you need other things to happen for it to be not just a bijection but a morphism. Sep 23, 2020 at 20:53