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


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.


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 $G\rightarrow H$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.