Question

If H and K are subgroups of G, then HK denotes the set

$\mathbf{\left\{}\mathbf{h}\mathbf{k}\mathbf{\in }\mathbf{G}\mathbf{|}\mathbf{h}\mathbf{\in }\mathbf{H}\mathbf{,}\mathbf{k}\mathbf{\in }\mathbf{K}\mathbf{\right\}}$If $\mathit{H}\mathbf{\cap }\mathit{K}\mathbf{=}\mathbf{⟨}\mathbf{e}\mathbf{⟩}$, prove that$\mathbf{\left|}\mathbf{H}\mathbf{K}\mathbf{\right|}\mathbf{=}\mathbf{\left|}\mathbf{H}\mathbf{\right|}\mathbf{·}\mathbf{\left|}\mathbf{K}\mathbf{\right|}$ .

Short answer

It is proved that$\left|HK\right|=\left|H\right|\cdot \left|K\right|$

Step-by-step solution

Step by step solution Step 1: Given that

Given that H and K are subgroups of G and HK is given by set$\left\{hk\in G|h\in H,k\in K\right\}$ .

Proving that |HK| = |H|⋅|K|

From the definition of group, we know that $\left|HK\right|\le \left|H\right|\cdot \left|K\right|$

The equality holds when $hk=h_1{k}_1$

Since both H and K are normal subgroups, they must be closed under the inverse operation.

Therefore,${h_1}^{-1}h\in H$ and $k{k_1}^{-1}\in k$.

Hence,${h_1}^{-1}h=k{k_1}^{-1}\in H\cap K$ .

Since $H\cap K=⟨e⟩$,role="math" localid="1652863439353" ${h_1}^{-1}h=k{k_1}^{-1}=e$

Simplify the equation as:

$h=h_1$ and $k=k_1$

$\Rightarrow hk=h_1{k}_1$

Therefore, equality holds.

Thus,$\left|HK\right|=\left|H\right|\cdot \left|K\right|$