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

Question

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.

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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?

Answer 1

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...

Answer 2

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.