Let $G,G'$ be groups, Let $\phi:G \longrightarrow G' $ an surjective homomorphism. So we know that exists $f : G'\longrightarrow G$ and that $ \phi \ \circ \ f = id_{G'} $. Is $f$ a homorphism?
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If $\Phi\circ f=id_{G'}$ then $\Phi$ has to be surjective hence is bijective; then $f$ must be the unique inverse $\Phi^{-1}$ and the answer ist "yes". I'm not sure if this is what you want to ask. – user 59363 Feb 14 '15 at 22:41
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Sorry I meant surjective ^^ – Sewer Keeper Feb 14 '15 at 22:49
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Since there are in general several mappings $f$ satisfying $\phi\circ f=id_{G'}$, the correct question would be: does among these exist one which is a homomorphism? (The answer is still no, se below.) – user 59363 Feb 14 '15 at 22:57
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In general no.
For example the homomorphism $\phi\colon\mathbb Z/4 \to \mathbb Z/2$ given by reducing modulo $2$ is surjective, but the only homomorphisms $f\colon\mathbb Z/2 \to \mathbb Z/4$ are the trivial homomorphism and the map $0 \mapsto 0, 1 \mapsto 2$. You can check that for either of these choices $\phi\circ f$ is the trivial zero homomorphism.
An $f$ which is also a homomorphism exists when $G$ is a semidirect product of $G'$ and $\ker\phi$.
Jim
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The answer is no. Take the morphism $C_{p^2}\to C_p$ ($C_n$ denoting the cyclic group $\left< a\right>$ of order $n$ generated by $a$) determined by $a\mapsto a$. Then there is no (splitting) morphism $C_p\to C_{p^2}$ ($p$ any prime).
user 59363
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2Let $a$ be a generating element of $C_{p^2}$ and $b$ be some (generating element) of $C_p$. Let $\phi: C_{p^2}\to C_p$ be determined by $\phi(a)=b$; then $\phi(a^p)=1$ and $\mathrm{ker}\phi={1,a^p,a^{2p},\dots, a^{(p-1)p}}$. The image of a morphism $f: C_p\to C_{p^2}$ is either trivial or a subgroup of $C_{p^2}$ of order $p$. The only subgroup of $C_{p^2}$ of order $p$ consists of the elements $1,a^p,a^{2p},\dots, a^{(p-1)p}$. In any case this gives that $\phi\circ f$ is the trivial (constant) morphism. – user 59363 Feb 15 '15 at 21:43