Abstract Algebra – Drawing Subgroup Diagram of Dihedral Group D4

abstract-algebragroup-theory

I am reading Fraleigh p. 80 in A First
Course in Abstract Algebra, and in the book i see the elements and subgroup diagram of the dihedral group $D_4$. Here are they:

enter image description here

Here is how i try to draw the subgroup diagram: p₀ must be included in every subgroup since it is identity element. Then i look at p₁, and try to find the subgroups including p₁, since p₁ is included, the inverse of it must be included also, and p₁op₁ must be included also, and so on . I need to check whether the result of these computations make it closed. But this lookslike a very long process. Is there an easy way to do that? For example, by looking at u₁ can we immediately say that it is included or not in a subgrouo without checking all compositions of u₁ with u₁? Or by looking at p₁ can we find < p₁ > easily? Thanks

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Personally, I am familiar to the following presentation of $D_4$ (or $D_8$). I added it here maybe you find it easier. In fact, You can feel the elements in this presentation easier. enter image description here

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[Math] Commutator subgroup of a dihedral group.

When D&F say "The image of $g$" in such contexts, they mean its image under the canonical homomorphism that takes $g$ to the coset $gH$. (Here the subgroup $H$ is $\left\langle r^{2}\right\rangle $.)

The next sentence is about the orders of the cosets $r\left\langle r^{2}\right\rangle ,s\left\langle r^{2}\right\rangle $ in the quotient group $D_{2n}/\left\langle r^{2}\right\rangle$, which are indeed $\leq 2$. (Not the orders of $r,s$ in the dihedral group.)

Now, by the important defining relation of the dihedral group $srs=r^{-1}$, one can directly verify that these generators commute. Since they are also of order $\leq 2$ , this implies that $D_{2n}/\left\langle r^{2}\right\rangle$ is abelian. Now, using Proposition 7(4) on page 169, ('the commutator subgroup is the largest abelian quotient'), we conclude $D_{2n}' \leq \left\langle r^{2}\right\rangle$. Therefore, $\left\langle r^{2}\right\rangle =D_{2n}'$ .

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