# Surjectivity of product map

Let $H$ be a subgroup of a group $G$, let $\phi: G \rightarrow H$ be a homomorphism with kernel $N$, and suppose that the restriction of $\phi$ to $H$ is the identity.

I am trying to determine whether the product map $f : H \times N \rightarrow G$ defined by $f(h,n) = hn$ is surjective. I've managed to show that it is injective and that it need not be a homomorphism, but I don't know how to prove it is surjective, or come up with a counterexample if it is not. Any help is appreciated!

-
@FlybyNight: The restriction of $\phi$ to $H$ is the identity, so it is surjective. The definition of $f$ does not use $\phi$, but for $f$ to be a homomorphism, the multiplication on $H\times N$ needs to be changed using $\phi$. –  Jack Schmidt Sep 29 '12 at 22:00

Fix $g\in G$, and let $h=\varphi(g)$. Then $\varphi(g)=\varphi(h)$, so $gN=hN$, so ... ?

-
So $g = hn$ for some $n \in N$. Thank you! –  Jonas Sep 29 '12 at 22:19
@Jonas: You’re welcome! –  Brian M. Scott Sep 29 '12 at 22:21

First part of the hypothesis says that $G$ has a short exact sequence

$1\rightarrow N\rightarrow G\xrightarrow{f} H\rightarrow 1$.

As $H\leq G$, let $i\colon H\rightarrow G$ be the identity map (homomorphism).

The second part of the hypothesis implies that $f\circ i=I_{H}$.

This means that the above short exact sequence is split; equivalently, $G$ is semidirect product of $N$ by $H$
(see exercise in Alperin-Bell's Groups and Representations, in the section on semidirect products).

In particular, $G=NH(=HN)$, which is your conclusion.

-

For any $g \in G$, you know that $g^{-1}\phi(g) \in \ker\phi = N$, so $g^{-1}\phi(g) = n$ for some $n \in N$, meaning $g = \phi(g)n^{-1}$.

-