Let $N$ be a normal subgroup of $G$ and $H$ be a subgroup of $G$ such that $ H N =G$. Prove that $H$ is a system of representatives of $N$ in $G$.

Question

Let $G$ be a group and let $N \subseteq G$ be a normal subgroup.
Suppose there is a subgroup $H \subseteq G $ such that :

$HN=\{ hn : h \in H , n \in N\} = G$

$ H \cap N = \{e \} $

Prove that $H$ is a system of representatives for the costs of $N$ in $G$

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Now I see that for all elements $g$ in G there is an element $h \in H $ and an element $n \in N$ such that $g=hn$ but how do I show that $ \#H = [G:N] $?

Answer 1

The first condition $HN=G$ says that, for every coset $gN$ of $N$ in $G$ there exists an element $h\in H$ such that $gN=hN$. Just write $g=hn$ for some $h\in H$ and $n\in N$. Another way to say this is to observe that $G$ may be decomposed as follows $$G=\bigcup_{h\in H}hN$$ Another way could be to consider the map $H\to \frac{G}{N}$, defined by $h\longmapsto hN$, and observe that it is surjective.

The second condition $H\cap N=\{e\}$ says that each left coset $gN$ of $N$ in $G$ can be written uniquely in the form $gN=hN$, because if you have $hN=h'N$ for some $h,h'\in H$ then $h'^{-1}h\in H\cap N=\{e\}$ and so $h=h'$. This is equivalent to say that the union $$G=\bigcup _{h\in H}hN$$ is indeed a disjoint union, or that $\{hN:h\in H\}$ is a partition of $G$, or equivalently that the map $h\longmapsto hN$ is also injective, hence is a bijection. (It is something more, namely a group isomorphism).

Answer 2

Consider images $[h_1]$, $[h_2]$ for two elements $h_1$, $h_2$ of H in the quotient group K=G/N. if $[h_1]^{-1}[h_2]=e$, this means $h_1^{-1}h_2 \in N$, but as $H\cap N=\{e\}$ this entails $h_1=h_2$. Therefore for two different elements in H, $h_1N \neq h_2N$, so each coset of N has only one h in it.