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I have been asked to find all groups H (up to isomorphism) st. there is a surjective homomorphism from $D_{2p}$ ($p$ prime) onto $H$.

The normal subgroups of $D_{2p}$ are: <$e$>,

$<\rho>$ (cyclic subgroup of $D_{2p}$) consisting of the rotations of the Dihedral group of size $2p$ and

$D_{2p}$

Now if I define $H=\frac{G}{N}$, then $\phi:G\rightarrow \frac{G}{N}$ ($N$ a normal subgroup)

we have that $\phi$ is a homomorphism and $ker\phi=N$, if $\phi(x)=Nx \quad \forall x \in G$

Now $\frac{D_{2p}}{<\rho>} \cong C_2$, $\frac{D_{2p}}{<e>} \cong D_{2p}$, $\frac{D_{2p}}{D_{2p}} \cong C_1$

The first isomorphism theorem states:

$\frac{D_{2p}}{ker\phi} \cong Im \phi \implies \frac{D_{2p}}{N} \cong Im \phi$

I understand for a surjective homomorphism $Im\phi=H$, so for $\phi:D_{2p}\rightarrow \frac{D_{2p}}{N}$ does the fact that $\frac{D_{2p}}{N} \cong Im \phi$ mean that because $\phi$ maps to $\frac{D_{2p}}{N}$; $\frac{D_{2p}}{N}=Im \phi$ and hence the map is surjective?

Also how do I know that I do indeed have all of the $H$ st. there exists a surjective homomorphism.

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You already solved the question. If you have a normal subgroup $N\leq G$ then the canonical homomorphism $G\rightarrow G/N$ is always surjective. Conversely, if there is a surjective homomorphism $\varphi:G\rightarrow H$, then $Ker(\varphi)$ is a normal subgroup of $G$ and $G/(Ker(\varphi))$ is isomorphic to $H$,(by the first isomorphism theorem). So up tp isomorphism of groups, if $G$ as your case has only 3 normal subgroups, there are only 3 groups $H$ such that there exists a surjective homomorphism $G\rightarrow H$.

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