# Finding groups $H$ for which there exists surjective homomorphisms $f:D_4 \rightarrow H$?

How can I find out for which groups $H$ there exists surjective homomorphisms $f: D_4 \rightarrow H$?

$D_4$ is the dihedral group of the square.

I have a theorem that says that there exists such surjective homomorphism, where $N$, which is a normal subgroup of $D_4$, is its kernel.

Can I use this?

• The image in $H$ under any homomorphism of $D_4$ must isomorphic to $D_4/N$ for some normal subgroup $N$ of $D_4$. (This is essentially the content of the theorem you have, correct?) This leads quickly the list of possible $H$ such that $f$ is surjective. – Justin Benfield Jun 12 '16 at 10:25
• @JustinBenfield Yes. – mavavilj Jun 12 '16 at 10:32

Of course you can use that, together with the fact that it must be $\;|H|=1,2,4,8\;$ and the knowledge of all the normal subgroups of $\;D_4\;$ you get all possible candidates.

For example, you have one unique normal subgroup of order $\;2\;$ , which is also the group's center (= the commutator subgroup, in this particular case), but three normal subgroups of order four...

Yes, your approach is correct. By the fundamental isomorphism theorem, if $f$ is such a morphism, then $\dfrac {D_4} {\ker f} \simeq \text{range } f$, and since $f$ is surjective, this means $\dfrac {D_4} {\ker f} \simeq H$.

Conversely, if $N$ is a normal subgroup of $D_4$, then the natural canonical projection $\pi : D_4 \to \dfrac {D_4} N, \ \pi (x) = \hat x$ is surjective.

Therefore, listing all groups to which you can find surjective morphisms from $D_4$ is synonymous to listing all normal subgroups of $D_4$. Assuming $D_4 = \langle R, F \mid R^4 = F^2 = (RF)^2 = 1 \rangle$, these are:

• the trivial subgroups $1$ and $D_4$
• $\{1, R^2\}$
• $\langle R^2, F \rangle$
• $\langle R^2, RF \rangle$

Factoring $D_4$ through the subgroups listed above will yield all the possible groups $H$ that you are looking for.

• A more detailed description of the normal subgroups of dihedral groups: math.stackexchange.com/questions/595884/…. – Alex M. Jun 12 '16 at 10:38
• What is $\hat x$ in $\pi (x) = \hat x$? My notes claim that $\pi(x)=Nx$ could be a surjective homomorphism, where $N$ is a normal subgroup. – mavavilj Jun 12 '16 at 11:02
• I also don't understand the "surjective morphisms from $D_4$ is synonymous to listing all normal subgroups of $D_4$.". How are the normal subgroups equal to $\frac{D_4}{N}$? – mavavilj Jun 12 '16 at 11:05
• @mavavilj: $\hat x$ is another notation for $Nx$. And no, the normal subgroups are not equal to $\frac {D_4} N$, nobody said that! What I said is that listing all the possible values of $\frac {D_4} N$ is synonymous to listing all the possible values of $N$ - which is what I have done. – Alex M. Jun 12 '16 at 11:18
• I don't understand how's that synonymous. Rather, that one must first find the normal subgroups, then $\frac{D_4}{N}$ using them. – mavavilj Jun 12 '16 at 11:19