# Find all the normal subgroups of the dihedral group $D_4$ (of order $8$) and find all the factor groups up to isomorphism.

I'm immensely confused on how to do this. I can create many subgroups of $$D_4$$ but I don't know how to know when I have found them all.

I don't know how to check if a subgroup is normal, although I understand the definition. I'm very confused on how to find the factor groups.

Can some one please explain a solution?

• There are only $8$ elements, with two generators $s,r$. Even finding all the subgroup and verifying are they normal isn't too much work. To check are they normal, you simply check are they closed under conjugation by $s,r$. – David Cheng Oct 17 at 19:12
• Can you give an example of a normal subgroup (and checking that it is one?) – MathEntrepeneur100 Oct 17 at 19:15
• If you are afraid that you missed some subgroups, you can easily check the answer since the properties of these groups are well documented. – David Cheng Oct 17 at 19:15
• Thanks. How do you find the factor groups up to isomorphism? – MathEntrepeneur100 Oct 17 at 19:17
• You know the order of the factor group, and it's only $4$ elements. You can check the operations of the elements by using the ones of original group. – David Cheng Oct 17 at 19:18

By Langrange's Theorem, the order of a subgroup of $$D_{4}$$ is $$1,2,4$$ or $$8$$. So, you must begin checking the order of all the elements. In this ways you will find all cyclic subgroups of $$D_4$$. After this, you must find the 2-generated subgroups of $$D_4$$. Clearly you have to consider only two elements wich are not in the same cyclic subgroup. In this way, you will find all subgroups of order $$4$$. Now you have found all subgroups of $$D_4$$ because the others are just $$1$$ and $$D_4$$.
Now, let us consider the normality of the subgroups. Every subgroup of order $$4$$ is normal because its index is $$2$$. It is easy to check that $$Z(D_4)=\langle a^2\rangle$$ where $$D_4=\langle a,b~|a^4=b^2=1,~b^{-1}ab=a^{3}\rangle$$. Thus, if $$H$$ is a normal subgroup of order $$2$$, it must be equal $$Z(D_{4})$$ (you can check easily that the other subgroups of order 2 are not normal). Therefore, the normal subgroups of $$D_4$$ are $$1,\langle a^{2}\rangle, \langle a\rangle, \langle a^{2},b\rangle,\langle a^{2},ab\rangle, D_4$$.
Finally, we will consider, up to isomorphism, the factor groups of $$D_4$$. If $$H$$ is a normal subgroup of $$D_4$$ and the order of $$H$$ is $$1,4$$ or $$8$$, it is easy to see that $$D_4/H$$ is isomorphic $$1, C_{2}$$ or $$D_{4}$$. Finally, if $$H$$ has order $$2$$, then $$H=Z(G)=\langle a^{2}\rangle$$. Now, $$D_4/H=\{\langle a^2\rangle,a\langle a^2\rangle, b\langle a^2\rangle, ab\langle a^2\rangle\}$$ and all of its elements have order $$2$$. It shows that $$D_4/H$$ is isomorphic to the Klein group $$V_4$$. [
Comment: For checking a subgroup $$H$$ of a subgroup $$G$$ is normal in $$G$$, you 'just' have to check that $$g^{-1}hg\in H$$ for all $$h\in H$$ and $$g\in G$$.