# Drawing subgroup diagram of Dihedral group $D4$

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:

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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This is because either $u_1p_1$ or $p_1u_1$ is not in it, assuming $D_4=u_1,u_1^2,u_1^3.u_1^4,p_1,p_1u_1,p_1u_1^2,p_1u_1^3$ – NECing Mar 16 '13 at 14:13
yes, since it is not in the subgroup diagram i know it is not in it, but why? – bigO Mar 16 '13 at 14:14
group has to be closed under operation – NECing Mar 16 '13 at 14:18
Thanks, but checking all possibilities looks like takes very long time. Is there any short way to understand whether a composition is an element of D4? – bigO Mar 16 '13 at 14:26
There isn't many combinations. For instance, if $u_1$ is in it, then $\langle u_1\rangle$ is in it. – NECing Mar 16 '13 at 14:29

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.

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Nice presentation – Adi Dani Mar 16 '13 at 22:18
Great diagram! ;-) – amWhy Mar 17 '13 at 0:45

Geometric interpretation is convenient for D_4. It is the invariance group of the square.

D_4 has three pi/2 rotations, making up the subgroup R_0, R_1, R_2, R_4 (replacing greek rho of the textbook of the original query with upper case R). It is also clear that R_0 and R_2 ( pi rotation) make up a subgroup.

M_1 and M_2 can be taken to be reflections in lines joining the opposite sides of the square and d_1 and d_2 are reflections in the two diagonals. It is obvious that each of these four elements squared is the identity, making four more proper subgroups of order 2.

A little visualisation shows that R_2 has the property of exchanging M_1 and M_2 and separately d_1 and d_2. This explains the other order 4 subgroups in the figure from the textbook.

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Please use MathJax to improve your answer. – Michael Albanese Nov 17 '15 at 15:10