Prove that $f∶N \times K \rightarrow G, f(a, b) = ab$ is an injective homomorphism where $N$ and $K$ are disjoint normal subgroups of $G$

Question

Let $G$ be a group and let $N$ and $K$ be normal subgroups of $G$. Suppose $N \bigcap K = \{e_G\}$. Prove that $f∶N \times K \rightarrow G, f(a, b) = ab$ is an injective homomorphism.

Is the homomorphism as simple as: $f((a,b),(c,d))=abcd=f(a,b)f(c,d).$ I am thinking that I am missing something in that. For injective I know I want to use the fact that $f(a)=e_G$ implies $a=e_H$. I never know how to exactly write it out.

Answer 1

That's right, you want to show the kernel is trivial.

So say $f(a,b)=e$. Then $ab=e$. So $b=a^{-1}\in N\cap K=e$. So $a=b=e$.

For the first part (homomorphism): $f((a,b)\cdot (c,d))=f(ac,bd)=acbd=abcd=f(a,b)f(c,d)$, where you need the result from your previous question, that the elements of $N$ and $K$ commute.

Answer 2

Given that N and K are normal sub-groups to of G such that $N\cap K =\{e_G\} \implies hk=kh \forall h\in N,k\in K$

Let $ x=(a,b)\in N×K,y=(c,d)\in N×K$ Then$f(xy)=f(ac,bd)=acbd=abcd=f(ab,cd)=f(a,b)f(c,d)=f(x)f(y)$ Therefore $f$is a homomorphism. $ker f=\{(a,b)\in N×K :f(a,b)= e_G\}$ $ f(a,b)= e\implies ab=e_G\implies a=b^{-1}$ Since $N\cap K=\{e_ G\}$ only possible case is $a=b=e_G$ Thus $ker f=\{(e_G,e_G)\}$ $f$ is injective