Retraction onto subgroup?

Question

Let $N\subset G$ be a nontrivial subgroup of finite group $G$ , is there a "retraction" onto $N$ , i.e. a homomorphism $\varphi : G \rightarrow N$ s.t. $ Im(\varphi) = N $ , $\varphi |_N = id_N $ . Its obviously yes when $N = G$ or $ N = \{e\}$ , if no other cases possible (I could not find them) how to proof ? What about infinite groups ? Is statement true for abelian groups ?

Answer 1

Isn't this equivalent to $N$ having a normal complement $K$ in $G$, which would be the kernel of $\phi$?

The case in which $N$ is a Sylow subgroup of a finite group $G$ is studied in transfer theory. For example, Burnside's Transfer Theorem states that if $N \in {\rm Syl}_p(G)$ and $N \le Z(N_G(N))$, then $N$ has a normal complement. This occurs for example in dihedral groups of order $2r$ with $r$ odd and $p=2$, or in $A_4$ with $p=3$.

Answer 2

Your question is related to "splitting exact sequence".

There are many examples: let $G = N_1 \times N_2 \times \cdots \times N_k$, $k > 1$ and $N_i \not= \{e\}$. Then there are natural injections $N_i \longrightarrow G$ and natural projections $p_i: G \longrightarrow N_i$.

The following proposition solves part of the problem and is easy to prove.

Proposition. Let $G$ be an abelian group and $N$ is a nontrivial subgroup of $G$. There is a retraction onto $N$ $\Leftrightarrow$ $G\cong N \times H$ for some subgroup $H$.

Answer 3

Consider the statement:

There is a retraction $\varphi:G\to N$.

If $N$ is proper nontrivial, then $\ker(\varphi)$ is a proper, nontrivial and normal. In particular pick any simple group $G$ of nonprime order, and any of its proper nontrivial subgroups (which exist by the fact that $G$ has composite order). It will be a counterexample.

This won't even be true even for abelian groups. Take $G=\mathbb{Z}_{p^n}$ for prime $p$. It is not decomposable as direct product of nontrivial groups. And as we will see soon those conditions are equivalent for abelian groups.

More generally if there is a retraction $\varphi:G\to N$ then we have a short exact sequence

$$1\to \ker(\varphi)\to G\to N\to 1$$

which is right split, and so by splitting lemma for groups our $G$ is a semidirect product of $\ker(\varphi)$ and $N$. Conversely any semidirect product will induce such retraction.

In particular if $N$ is normal (e.g. when $G$ is abelian) then there is a retraction $G\to N$ if and only if $N$ is a direct summand of $G$. Which also means that any group of the form $G=N\times K$ will be a valid example.