How to prove that this construction is a group homomorphism? Let $\phi:G \rightarrow H$ a group homomorphism such that $M=\phi(G) \neq H$ and $M$ having at least 3 different cosets in $H$. Take $K$ as being the group of all permutations of $H$. Choose 3 different cosets
$M$, $Mh'$, $Mh''$ of $M$ in $H$ and define $\sigma$ in $K$ by $\sigma(xh'')=xh'$, $\sigma(xh')=xh''$ for $x \in M$, while otherwise $\sigma(h)=h$. Define $\psi,\psi':H \rightarrow K$ by $\psi(h) = \text{left multiplication by}\,\,\,h$, and $\psi'(h) = \sigma^{-1}\psi(h)\sigma$.
I'm having a hard time proving that $\psi'$ is a group homomorphism, the reason is that the only way I see on how to solve the problem is testing different cases for the argument, and there exists a lot of that cases. There is a easier way to prove that $\psi'$ is a homomorphism?
 A: If you remove all of the extraneous details from the question, it will become a lot simpler; the main content of the question is not related to $\phi$ or $M$ or the choice of $\sigma$. Stripping out these details, we have
Let $H$ be a group and $K$ the symmetric group on $H$. Fix any permutation $\sigma \in K$. Let $\psi: H \rightarrow K$ be the map sending $h$ to the permutation corresponding to left multiplication by $h$. Prove that the map $\psi'$ defined by
$$\psi'(h) = \sigma^{-1}\psi(h)\sigma$$
is a homomorphism.
One can prove this in two ways. First, a direct proof:

*

*$\psi'(e) = \sigma^{-1}\psi(e)\sigma = \sigma^{-1}\sigma = e$

*$\psi'(ab) = \sigma^{-1}\psi(ab)\sigma = \sigma^{-1}\psi(a)\psi(b)\sigma = \sigma^{-1}\psi(a)\sigma\sigma^{-1}\psi(b)\sigma = \psi'(a)\psi'(b)$
Second: observe that conjugation by $\sigma$ induces a homomorphism $K\rightarrow K$, and that $\psi'$ is just the composition of that homomorphism with $\psi$. The composition of homomorphisms is a homomorphism, so you're done (this is implicit in the above).
