On epimorphisms of groups.

Question

I want to prove by absurdity that every epimorphism of groups is a surjective morphism. So let $f\colon A\to B$ be an epimorphism of groups. Let by absurd assume that $f$ is non surjective, so that the image $f(A)$ of $f$ is properly contained in $B$ and let us also assume that $f(A)$ is not a normal subgroup of $B$, because in that case we can form the quotient group of classes of elements of $B$ modulo $f(A)$ and use the same argument that is used for abelian groups. So, the index of $f(A)$ in $B$ must be at least $3$, otherwise $f(A)$ would be normal in $B$. Let $f(A)$, $f(A)u$ and $f(A)v$ be three distinct cosets of $f(A)$ in $B$. Then, we consider the permutation $\sigma$ on $B$, defined by: $\sigma(xu)=xv$ for all $x\in f(A)$; $\sigma(xv)=xu$ for all $x\in f(A)$ and $\sigma(b)=b$ on the remaining part of $B$. Then, for all $x,b\in B$, let us define $\psi_b(x)=bx$ and $\overline{\psi}_b=\sigma^{-1}\circ\psi_b\circ\sigma$. Let us finally define $\psi,\overline{\psi}\colon B\to\operatorname{Sym}_B$ by $\psi(b)=\psi_b$ and $\overline{\psi}(b)=\overline{\psi}_b$. Since $\psi\circ f=\overline{\psi}\circ f$, then $\psi=\overline{\psi}$, because $f$ is assumed to be an epimorphism. So I need to prove that $\psi\neq\overline{\psi}$ to get a contradiction, but I can't. Can you please help me to show that actually $\psi\neq\overline{\psi}$.

Answer 1

You probably meant to define $\sigma(xu) =xv$ for $x\in f(A)$, not $x\in B$ (similarly for $\sigma(xv)$)

If $\psi = \overline{\psi}$ this means that for all $b$, $\psi_b$ and $\sigma$ commute.

Take $b= xu, x\in f(A)$. $\psi_b\circ \sigma(e) = \psi_b(e) = b$ and $\sigma\circ\psi_b(e) = \sigma(b) = \sigma(xu) = xv \neq b$.