Suppose that $n = dm$ where $d$ and $m$ are positive integers with $m\ge 3$. Consider the dihedral group $D_n = \langle \{\mu, \rho\}\rangle,$ where $|\mu| = 2$, $|\rho| = n$ and $\rho\mu = \mu\rho^{−1}$, and the dihedral group $D_m = \langle \{s, r\}\rangle,$ where $|s| = 2$, $|r| = m$ and $rs = sr^{−1}$.
Define $\psi : D_n \to D_m$ by $\psi(\mu^a\rho^b)=s^ar^b$, for any integers $a,b$
Find kernel and image of $\psi$.
I think :
$\ker(\psi)=\{\rho^{2x}\mu^{my}\}$ for $x,y$ integers so that $a=2x,b=my$
${\rm im}(\psi)=\{D_m\}$.
Did I do this correctly?
So, by first isomorphism theorem, have that $D_n/\ker(\psi) \cong {D_m} $. Is there anything else the theorem says here, or is this it?
Your answers make no sense. The following equations at least make sense:
$\operatorname{im}(\psi)=D_m$ and
$\operatorname{ker}(\psi)=\{\rho^{mk}\mid k=0,1,\ldots,d-1\}$.
Indeed, $$ \mu^a\rho^b\in\operatorname{ker}(\psi) \Leftrightarrow\psi(\mu^a\rho^b)=e \Leftrightarrow s^at^b=e \Leftrightarrow a\equiv0\,(\operatorname{mod}2),b\equiv0\,(\operatorname{mod}m). $$ Hence $\mu^a\rho^b=\mu^{2l}\rho^{mk}=\rho^{mk}$, where $a=2l$, $b=mk$.
Also, write $D_n=\langle\mu,\rho\rangle$ but not $D_n=\langle\{\mu,\rho\}\rangle$.