TITLE: Showing epis in $\mathbf{Grp}$ are surjective
QUESTION [5 upvotes]: I'm working through Maclane's Categories, and I got to exercise 5 of Section 5.  In this question, the reader is asked to prove that all epis in $\mathbf{Grp}$ are surjective, and gives a sketch of the proof in the hint.
His hint goes like this:

If $\varphi: G \to H$ has image $M$ not $H$, use the factor group $H/M$ if $M$ has index $2$.  Otherwise, let $\operatorname{Perm} H$ be the group of all permutations of the set $H$, choose three different cosets $M$, $Mu$ and $Mv$ of $M$, define $\sigma\in \operatorname{Perm} H$ by $\sigma(xu)=xv$, $\sigma(xv)=xu$ for $x\in M$, and $\sigma$ otherwise the identity.  Let $\psi: H \to \operatorname{Perm} H$ send each $h$ to left multiplication $\psi_h$ by $h$, while $\psi_h'=\sigma^{-1}\psi_h \sigma$.  Then $\psi \varphi = \psi'\varphi$, but $\psi\neq \psi'$

The hint makes it seem like this is a very non-trivial result, so I'm wondering why my attempt at a solution below is incorrect:

My Solution:
Suppose $\varphi: G \to H$ is an epimorphism.  Then define $\pi : H\to H/\varphi[G]$ to be the quotient map, and $\rho: H \to H/\varphi[G]$ to be the zero map (sending all of $H$ to the identity of $H/\varphi[G]$).  We have $\pi \circ \varphi = \rho\circ \varphi$, and so because $\varphi$ is epi, we should have that $\pi = \rho$.  $\pi$ is surjective, so $\rho$ must also be, meaning that $H/\varphi[G]=\{1\}$, and hence that $H=\varphi[G]$.  In other words, $\varphi$ is surjective.

What am I missing here?  And if this is correct, what is Maclane's intention in proving the result in a less trivial manner?  Does his proof give a stronger result?

REPLY [8 votes]: Your proof depends on $\varphi[G]$ being a normal subgroup (otherwise the quotient map you need will not exist), but a priori you have no reason to expect that.