Question

If H and K are any subgroups of G, prove that $\mathbf{\left|}\mathbf{HK}\mathbf{\right|}\mathbf{=}\frac{\mathbf{\left|}\mathbf{H}\mathbf{\right|}\mathbf{·}\mathbf{\left|}\mathbf{K}\mathbf{\right|}}{\mathbf{|}\mathbf{H}\mathbf{\cap }\mathbf{K}\mathbf{|}}$

Short answer

It is proved that,$\left|HK\right|=\frac{\left|H\right|·\left|K\right|}{|H\cap K|}$.

Step-by-step solution

Referring to part(a)                             

Considering HK denotes the set$\mathbf{\left\{}\mathbf{h}\mathbf{k}\mathbf{\in }\mathbf{G}\mathbf{|}\mathbf{h}\mathbf{\in }\mathbf{H}\mathbf{,}\mathbf{k}\mathbf{\in }\mathbf{K}\mathbf{\right\}}$.

Given that H and K are any subgroups of G.

Proving that |HK|  =  |H|·|K||H∩K|

Let $\varphi :H\times K\to HK$ given by $\varphi \left(hk\right)=hk$ is a surjective mapping.

Assume that $h\in H$and $k\in K$, role="math" localid="1654611381709" ${\varphi }^{-1}\left(hk\right)=\left\{(h{a}^{-1},ak)|a\in K\cap H\right\}$.

Then, if $a\in K\cap H$ and $h\text{'}=h{a}^{-1},k\text{'}=ak$, we get $h\text{'}\in H$and $k\text{'}\in K$and $hk=h\text{'}k\text{'}$.

Or, we can also say that, if $h\text{'}\in H$ and $k\text{'}\in K$and $hk=h\text{'}k\text{'}$then, $k\text{'}{k}^{-1}=h{\text{'}}^{-1}h\in k\cap H$.

We get,$h\text{'}=h{\left(h{\text{'}}^{-1}h\right)}^{-1}$ and $k\text{'}=\left(h{\text{'}}^{-1}h\right)k$.

Hence, our assumption is correct.

Therefore, for each $hk\in HK,\left|{\varphi }^{-1}\left(hk\right)\right|=|H\cap K|$ and simplify as:

$\begin{array}{c}\left|HK\right||H\cap K|=|H\times K|\\ \left|HK\right||H\cap K|=\left|H\right|\cdot \left|K\right|\\ \left|HK\right||H\cap K|=\left|H\right|\cdot \left|K\right|\\ \left|HK\right|=\frac{\left|H\right|\cdot \left|K\right|}{|H\cap K|}\end{array}$

Thus, it is proved that$\left|HK\right|=\frac{\left|H\right|\cdot \left|K\right|}{|H\cap K|}$.