Lesson 29: Group Constructions: Subgroups Finale

In lesson 27 we explored subgroups of integers and integers with modular arithmetic defined on them, and in lesson 28 we explored subgroups of permutation groups.  In this lesson, we’ll explore subgroups of our final cool example of groups, which are dihedral groups.  Rest assured, there are many more (and wildly different) groups than integers (with or without modular arithmetic), permutation groups, and dihedral groups.  It just happens to be the case that these are the three classes of groups that we’ve discussed so far, so we might as well consider their subgroups in turn.  If dihedral groups (groups of rotations and reflections of regular polygons in the plane) are unfamiliar, go ahead and review them in lesson 26.

In order to see how this all goes, it will suffice to do our work in a particular dihedral group, and to make things easier for us let’s just take the dihedral group that we explicitly explored in lesson 26, which was the dihedral group of a regular four-gon (i.e., a square).  Let’s denote this group by D_4, and we recall that it has 8 elements which we write as D_4=\{e, \rho_1, \rho_2, \rho_3, r_1, r_2, r_3, r_4\} where the \rho_i‘s are rotations and the r_i‘s are reflections.  Namely, \rho_1 is the rotation counterclockwise by 90 degrees, so that if the square starts off in the position shown in figure 1



then the corner currently at position 1 goes to position 2, while the corner currently at position 2 goes to position 3, position 3 to 4, and 4 to 1, so that we end up with the square in the configuration shown in figure 2.  Similarly, \rho_2(=\rho_1\cdot \rho_1) rotates the square counter clockwise by 180 degrees, and \rho_3(=\rho_2\cdot \rho_1) rotates the square counter clockwise by 270 degrees (where we recall also that group multiplication now is composition of rotations/reflections, where we first do the rotation/reflection on the right and then move to the left).

Additionally, r_1 is the reflection about the diagonal line connecting corner 2 with corner 4, so that if the square starts off in the configuration shown in figure 1, then r_1 represents the act of moving the square to the position in figure 3, which is just exchanging the corners 1 and 3.  In a similar way, r_2 is the reflection about the diagonal connecting 1 and 3, and therefore exchanges the corners 2 and 4, r_3 is the reflection about the line connecting the midpoints between corners 1 and 2, and 3 and 4, so it exchanges corners 1 and 2, as well as the corners 3 and 4, and finally r_4 is the reflections about the line connecting the midpoints between corners 1 and 4, and 2 and 3, so it therefore exchanges corners 1 and 4, and 2 and 3.



As in the case of the group of permutations, we have some obvious subgroups lying around.  Namely, we know that once we have an element that is its own inverse, we can form a subgroup by simply taking the subset that contains only the identity element and this self-inverting element.  Thus, we immediately see that \{e, r_i\} is a subgroup of D_4 for any choice of i\in \{1, 2, 3, 4\}, since each reflection is its own inverse (I won’t go through all of the details of this reasoning, as this was done in the previous lesson).

We also have a couple other obvious subgroups, which are subgroups of rotations.  Namely, compositions of rotations are always rotations, where as the same is not true for reflections (i.e., r_1\cdot r_2 is not another reflection, whereas \rho_i\cdot \rho_j can always be expressed as a rotation, i.e. as \rho_k for some appropriate k, where we note that \rho_4=e, \rho_5=\rho_1, etc.).  Thus, it is clear that the subset \{e, \rho_1, \rho_2, \rho_3\} is a subgroup, since it has the identity and clearly a) is closed under multiplication (i.e., composition) and b) has all of its inverses (since, for example, the inverse of \rho_1 is \rho_3).  Similarly, \{e, \rho_2\} is a subgroup, which can also be seen be noting that \rho_2 is its own inverse.



Let’s now take a moment to step back and think like mathematicians.  We’ve come a long way so far, and we’ve seen how these sorts of lines of questioning go.  Namely, we’ve constructed a world of mathematics from things as basic as sets, and elements, and we’ve defined ways to form more interesting objects called groups.  We’ve explored dihedral groups and now we’ve done a decent amount of exploration into the types of subgroups that dihedral groups have.  But we notice something.  We notice that each subgroup of this dihedral group that we’ve seen so far is such that it either only consists of rotations, or it only consists of reflections.  A natural question to ask—and the type of question that mathematician should now ask—is whether or not there are subgroups of this dihedral group that contain both rotations and reflections.

Before actually studying this question, let me just wax philosophic for a moment about why we want to ask this question.  We’ve made our definitions and we’ve studied the consequences of those definitions.  We saw that once we defined what a group is abstractly (i.e., a set with special properties), we can then construct explicit examples of these abstract objects, and one set of such examples are the dihedral groups.  We then saw a pattern in the examples of groups that we were constructing, namely that a lot of them had certain special subsets, and this motivated us to make another abstract definition—i.e., that of a subgroup.

We’ve now spent a couple of lessons exploring explicit examples of subgroups, and in this lesson we’ve found another pattern—namely, that the most obvious subgroups of dihedral groups have either only rotations or only reflections.  Now, the notion of “most obvious” is a very human one, and what is most obvious to us might be the least obvious to some alien life form (and levels of obviousness differ even from person to person).  Since we’ve found this possible pattern in the subgroups of dihedral groups, we should try to find out how general the statement “all subgroups of dihedral groups have either only rotations or only reflections” is.  This process of i) making an abstract definition, ii) exploring this abstract definition (usually through explicit examples), iii) trying to draw general conclusions about this abstract definition (which are usually motivated by our explicit examples), iv) using these general conclusions to motivate new abstract definitions, and v) repeating at step i), is a lot of what math is about, and in my personal opinion is one of the most beautiful uses of human logic that exists.  The reason I believe this is that by going through the above process (which is in no way as linear as I may have made it sound), we often find out a lot more about our initial definitions than we had ever bargained for.  Namely, we may make our definitions for one reason, and through lots of exploration and creativity find that these definitions (or the general results we derive from them) are actually telling us something about some other ideas that we did not dream of making contact with.  In this way, one can’t help but feel like the above mode of reasoning exposes us—ever so slowly—to some kind of truth with a capital T.  I’ll end my rambling here and get back to actual math now.

Since it is the case that any group is a subgroup of itself, we technically already know that there is a subgroup of this dihedral group with both rotations and reflections—the group itself.  But this is obviously not that interesting of an answer, so let us rephrase our question.  The question now is: Are there any subgroups of D_4 other than D_4 itself that contains both a rotation and a reflection?  The answer is yes, but before we see an example of such a subgroup let’s first attack this problem assuming we don’t know what the answer is.  This will help us to see why it is that this question has a positive answer, and will also help us generalize our results to other dihedral groups.

Since we’re after a subgroup with at least one rotation and at least one reflection, let’s see what happens when we put in the rotation \rho_1 by 90 degrees (and worry about the reflection later).  Since our subgroup must be closed under group multiplication, we know that we need to put all products of \rho_1 with itself in our group.  But, this rotation “generates” all the other rotations, in the sense that each rotation can be obtained by simply doing the rotation \rho_1 over and over again (i.e., \rho_2=\rho_1\cdot \rho_1, \rho_3=\rho_1\cdot \rho_1\cdot \rho_1, etc.).  Thus, we immediately see that if we put \rho_1 in our subset (which we’re trying to make a subgroup), then we must also put in \rho_2 and \rho_3 (note that \rho_4=\rho_1\cdot \rho_1\cdot \rho_1\cdot \rho_1=e since four successive 90 degree rotations is just a 360 degree rotation, i.e., the identity).  So if we start with the assumption that we want \rho_1 in our subgroup, we know that our subgroup will look something like \{e, \rho_1, \rho_2, \rho_3, ...\}, and we know we’re not done because we still need to put in a reflection (because of the specific question that we’re trying to answer).

But here’s the kicker.  If we try to add any reflection into this subset, we’ll automatically have to include the whole group.  The reason for this is that if the enemy hands us one reflection and lets us generate all possible products of this reflection with any of the rotations, then we can arrive at any other reflection.  Intuitively, the reason for this is as follows.  Recall the “napkin” analogy that we used when introducing dihedral groups, where rotations simply rotate the napkin (as the name implies) and reflections “flip” the napkin over.  The idea is that once the napkin is flipped (it doesn’t matter how it’s flipped), we can then rotate it to any other flipped position we want by simply using the rotations that we have at our disposal.  Thus, using one reflection and all of the rotations, we can generate all the other reflections.

A more mathematical manifestation of this fact is as follows.  Let’s say the reflection r_1, which swaps corners 1 and 3 (and leaves the others alone), is the reflection that we want to try to add into our subgroup.  Now let’s ask if we can generate the reflection r_2, which swaps corners 2 and 4 (and leaves the others alone), by using only r_1 and the rotations.  Let’s say we first flip the napkin using r_1, so that so far we have 1\leftrightarrow 3.  We know that the reflection we’re after (r_2) leaves these corners alone, and that we can now only use rotations.  Thus, we need a rotation that swaps corners 1 and 3 right back.  But this is precisely \rho_2, which acts as 1\leftrightarrow 3, 2\leftrightarrow 4.  Thus, by doing the action \rho_2\cdot r_1, we first swap 1 and 3 (using r_1 because we do the action that is right-most first), and then we swap 1 and 3 and swap 2 and 4 (using \rho_2), and the net effect moves 1 and 3 back to their original places (since swapping twice is like doing nothing at all) but swaps 2 and 4.  And this is precisely the action of r_2!  We therefore have that r_2=\rho_2\cdot r_1, and it is left as an exercise to show that r_3 and r_4 can also be expressed as a product of r_1 with some rotation.

So what have we found so far?  We’ve found that if we have all the rotations to play with as well as a single reflection to play with, then we can generate the whole dihedral group D_4.  Let’s first comment on the generality of this statement, namely that this will clearly apply to all dihedral groups (i.e., D_n for any n).  Intuitively this is because our ability to get to any reflection by using only a single reflection and any rotation is independent of whether or not our napkin is square.  Namely, if our napkin were an octagon (or a regular 1000-gon), and we had all possible rotations to play with, then once the napkin is flipped we can always rotate it to any other flipped position, regardless of the shape of the napkin.  So our exploration into this example has actually established a result much more general than we bargained for: any subgroup of any dihedral group that contains a) all rotations and b) at least one reflection is indeed the entire dihedral group.

This is indeed a negative result if we want to find a subgroup of D_4 that has at least one rotation and one reflection, and which isn’t the whole group itself.  This is because in the above scenario, when we have all of the rotations to play with, the act of adding in a single reflection automatically gives back the whole group.  Thus, we see that the only way out of this is to construct a subgroup that has a rotation, but not all of the rotations (for otherwise adding in a reflection ruins the whole construction, as we’ve just seen).  Is this possible?  Well, we know that we can’t include the rotation \rho_3, because again this rotation generates all the other rotations in an analogous way as \rho_1.  But luckily, for D_4, we can include \rho_2, since \rho_2\cdot \rho_2=e, and we therefore stand a chance of constructing a subgroup that has one rotation but not all rotations.

So let’s try putting in \rho_2 as our “at least one rotation” and let’s try r_1 as our “at least one reflection”.  Both of these elements are their own inverses, so let’s consider what their product is.  We calculated above that \rho_2\cdot r_1=r_2, which is a simple swap of corners 2 and 4.  Similarly, we also find that r_1\cdot \rho_2=r_2, so that it doesn’t matter in which order we apply these actions.  Also, r_2 is its own inverse, as all reflections are.  Therefore so far we have the subgroup \{e, \rho_2, r_1, r_2, ...\}.  Moreover, the only remaining product that we need to check is r_1\cdot r_2 and r_2\cdot r_1.  But r_1 simply swaps corners 1 and 3, and r_2 simply swaps corners 2 and 4.  Thus, doing them together (in either order) gives us precisely \rho_2, and our subset is indeed already closed under multiplication.  It also clearly has the identity as well as the inverses of each element, since each element is its own inverse.  Thus we’re done, and we’ve successfully found a positive answer to our above question.

One last remark, before going on to the exercises and then on to completely new material in the next lesson, is that there is yet another completely general statement that we can make lurking in the work we’ve done so far.  Namely, we’ve seen that if we ever have a rotation that “generates” all the other rotations in the aforementioned sense, then we can’t construct a subgroup that a) contains this rotation, b) also contains a reflection, and c) is not the whole group itself.  In the special case of D_4, we were able to find a rotation that didn’t generate all of the rotations, since there are some rotations of the square that are never “hit” by products of \rho_2.  This is clearly a result of the fact that 2 divides 4, since \rho_4=e and so products of \rho_2 get back to e before ever hitting \rho_3, for example.  But in a dihedral group D_n where n is prime, every rotation will generate all the other rotations, since each rotation will have to “hit” every other rotation before finally getting back to the identity (an example of this is given in the exercises).  Since in these cases any rotation we choose will force us to include all rotations, and since the subsequent inclusion of any reflection will then force us to include the whole group, we get the following general result:

Theorem: If D_n is the dihedral group of a regular n-gon, and if n is prime, then there is no subgroup of D_n that contains at least one rotation and at least one reflection, except for the entire group itself.

The proof of this theorem is this entire lesson.


1) Show that r_3 and r_4 can both be expressed as a product of r_1 with some rotation \rho_i.  Namely, find such a product for each reflection.

2) What is the subgroup containing at least one rotation as well as r_3, and which isn’t the whole group D_4?

3) Show that any rotation \rho_i in the dihedral group D_7 can be expressed as a product of \rho_4.  I.e., show that by multiplying \rho_4 to itself over and over again, we cycle through all possible rotations before coming back to the identity.  This is an example of the special nature of dihedral groups of prime-gons discussed in the final paragraph.

Solutions to Exercises

On to Lesson 30

Back to Lesson 28

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