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?

  • $\begingroup$ 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 $s,r$. $\endgroup$ – David Cheng Oct 17 at 19:12
  • $\begingroup$ Can you give an example of a normal subgroup (and checking that it is one?) $\endgroup$ – MathEntrepeneur100 Oct 17 at 19:15
  • $\begingroup$ If you are afraid that you missed some subgroups, you can easily check the answer since the properties of these groups are well documented. $\endgroup$ – David Cheng Oct 17 at 19:15
  • $\begingroup$ Thanks. How do you find the factor groups up to isomorphism? $\endgroup$ – MathEntrepeneur100 Oct 17 at 19:17
  • $\begingroup$ 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. $\endgroup$ – 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 $g\in G$.

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