Proof of existence of simple group of Order 168 in Dummit and Foote

Question

So I want to show that GL(V) is a simple group, where V is a three-dimensional vector space over the field of 2 elements. I am following Dummit and Foote (last paragraph on p.211) but there is one part I don't understand and it's driving me crazy. GL(V) is acting on V. Assume there exists a nontrivial, proper normal subgroup H of GL(V). Let x be a point of V (i.e. a 1 dimensional subspace) and let N be the stabilizer of x in GL(V). So I understand the N has index 7 in GL(V) and that H is not a subgroup of N, but how do these two things imply that HN = GL(V)? The point is to show that 7 divides the order of H. The only other thing I have seen on the internet is the claim that the latter follows from H being transitive. I would be very grateful if someone could explain this to me.

Answer 1

Because $H$ is normal, $HN$ is a subgroup containing $N$.

Since $N$ has prime index, there are only two possibilities: $HN=N$, or $HN=GL(V)$.

But if $HN=N$, then $H\leq HN \leq N$, so $H\leq N$.