Let $G$ be a group and let$H,K$ be normal subgroups of $G$. Let $\pi_H,\pi_K$ be the projections on $H$ and $K$ respectively. Show that the map $$f:G/(H \cap K) \to G/H \times G/K$$
defined as $f(\overline{x})=(\pi_H(x),\pi_K(x))$ is a monomorphism.
First of all I am not so sure what the projection means in this case, for example, take $\overline{x}=x(H \cap K)$, is the projection $\pi_H(x)=Hx$?
Another doubt that I have is: if $H$ and $K$ are normal subgroups, does this imply $H \cap K$ is normal? I am pretty sure that condition is necessary in this problem, but I don't know how to show $H \cap K$ is normal.
Assuming $H \cap K$ and if I've understood the function correctly, I would like to know if this proof is correct:
- Injectivity: suppose $f(x(H \cap K))=f(y(H \cap K))$,
then $(\pi_H(x),\pi_K(x))=(\pi_H(y),\pi_K(y))$, which means $(Hx,Kx)=(Hy,Ky)$, so $$Hx=Hy,$$ $$Kx=Ky.$$
From the fact that $H,K$ are normal subgroups, it follows $xy^{-1} \in H \cap K$, which implies $x (H \cap K)=y (H \cap K)$.
- Homomorphism: $f((H \cap K)x)((H \cap K)y))=f((H \cap K)(xy))$ $$=(\pi_H(xy),\pi_K(xy))=(H(xy),K(xy))$$ $$=((Hx)H(y),(Kx)(Ky))=(\pi_H(x)\pi_H(y),\pi_K(x)\pi_K(y))$$$$=f(x)f(y).$$
I would appreciate if someone could tell me how can I show $H \cap K$ is normal and if my solution is correct (btw, anyone can write a better well-written/different/shorter solution as well). Thanks in advance.
another way of looking at the normality of $H \cap K$ is that if $\sigma$ is any inner automorphism of $G$ then $$ (H \cap K)^{\sigma} = H^{\sigma} \cap K^{\sigma} = H \cap K $$ in your proof the normality of $H$ and $K$ is used (a) to guarantee that $G/H \times G/K$ is a group, and (b) (via normality of $H \cap K$) in your demonstration that $f$ is a homomorphism - $$ f((H \cap K)x)((H \cap K)y))=f((H \cap K)(xy)) $$ but it is not required where you invoke it ("From the fact that H,K are normal subgroups, it follows...")
otherwise your proof seems fine