How do i show that $\psi : D_n \rightarrow D_m$ defined by $\psi(\mu^i \rho^j) = \delta^i \gamma^j$ is a homormophism from $D_n$ to $D_m$ where n =qm

Question

Let $n = dm$
Consider these groups:
$D_n = \langle \mu, \rho \mid \mu^2 = \rho^n = \epsilon, \mu\rho = \rho^{-1}\mu \rangle$

$D_m = \langle \delta, \gamma \mid \delta^2 = \gamma^m = \epsilon, \delta\gamma = \gamma^{-1}\delta \rangle$

$\psi : D_n \rightarrow D_m$ defined by $\psi(\mu^i \rho^j) = \delta^i \gamma^j$
show that $\psi$ is a homomorphism
This is what i tried:

$\psi((\mu^i \rho^j)(\mu^x \rho^y))$ = $\psi(\mu^i \mu^x\rho^{-j} \rho^y))$ = $\psi(\mu^{i+x} \rho^{-j+y})$ = $\delta^{i+x} \gamma^{-j+y}$ = $\delta^{i}\delta^{x} \gamma^{-j}\gamma^{y}$

However, from here I have no idea what to do.

Answer 1

You are essentially done, since by the same manipulations you also have $$ \delta ^i \delta ^x \gamma ^{-j}\gamma ^y=\delta ^i \gamma ^j \delta ^x \gamma^y=\psi (\mu ^i \rho ^j) \psi(\mu ^x \rho^y) $$ Then since $n=dm$ you also have $\psi(\epsilon)=\psi (\mu ^0\rho^0) =\epsilon$. Finally you have to prove that $\psi(g^{-1})=\psi(g)^{-1}$. But this simply follows from the two facts above and $$ \psi(g)\psi(g^{-1})=\psi(gg^{-1})=\psi(\epsilon)=\epsilon $$