Order of the product of two groups

Question

Suppose we have a group $G$ and $H,N \leq G$ such that $HN \leq G$. Then how do I justify the formula: $$ |HN| = \frac{|H||N|}{|H\cap N|} $$

Also, are there any other assumptions neccesary for the formula to hold? Thanks

Answer 1

Hint:

If one of the subgroups is normal in $G$, $HN$ is a subgroup of $G$ and you have an exact sequence of group homomorphisms $$\{1\}\longrightarrow H\cap N\longrightarrow H\times N \longrightarrow HN \longrightarrow \{1\},$$ where the map $H\times N\longrightarrow HN$ is defined by $\;(h,n)\longmapsto hn^{-1}$.

Answer 2

Suppose $H$ and $N$ are finite subgroups

We have that $H \cap N \leq H$ and $H \cap N \leq N$

And $|N| = |H \cap N| \cdot [N:H\cap N] \implies [N: H\cap N] = \frac{|N|}{|N\cap H|} \in \mathbb{N}$

So we have that $N = \cup_{i = 1}^{n} H \cap N \cdot c_{i}$ a disjoint union with $c_{i}\in N$ the coclass representative

And we have $H(H\cap N) = H \implies HN = \cup_{i = 1}^{n} H(H\cap N)c_{i} = \cup_{i = 1}^{n} Hc_{i}$ this is a disjoint union too because if $h \in Hc_{i} \cap Hc_{j}$ with $i \not=j$ then $h_{1}c_{i} =h = h_{2}c_{j}\implies h_{2}^{-1}h_{1} =c_{j}c_{i}^{-1} $ as $h_{2}^{-1}h_{1} \in H$ and $c_{j}c_{i}^{-1} \in N$ then $h_{2}^{-1}h_{1}=c_{j}c_{i}^{-1} \in H\cap N \implies h_{2}^{-1}h_{1}c_{i} = c_{j} \implies c_{j} \in ((H\cap N)c_{j})\cap ((H\cap N)c_{i}) = \emptyset $ this is a contradiction.

Then we have that $|HN| = |H|n = \frac{|H||N|}{|HN|}$