Let p be a prime number and $N_1 , N_2 \triangleleft G $ with $N_1 \cap N_2 ={e} $, where $N_1$ and $N_2$ are non trivial subgroups of G, $[G:N_1]=[G: N_2]=p$.
Show that $G=N_1 N_2$
I'm stuck on how to do this, I'm thinking from $[G: N_1]=[G: N_2]=p$, we have the quotient groups of each of these have order p, but i'm not really sure if this shows anything. If we can show that the order of $N_1 N_2$ is equal to G will this suffice? has anyone got any idea how to progress with this problem?
I would first like to point out that you mean $[G:N_1]$ and $[G:N_2]$ ($[G,N_1]$ does mean something else, so be careful).
So you don't actually need $[G:N_2]=p$. My strong hint is to show $N_1<N_1N_2\le G$ and compute $[G:N_1N_2]$.