# Lesson 28: Group Constructions: Subgroups Part Deuce

In the previous lesson we defined subgroups and took a look at some examples of subgroups of integers and of integers modulo 8.  In this and the next lesson we’ll take a look at subgroups of some more exotic groups, like groups of permutations and dihedral groups.  If either of these sets of groups is unfamiliar, feel free to go brush up on groups of permutations in lesson 24 and lesson 25, and dihedral groups in lesson 26.

Let us begin with asking about subgroups of permutation groups.  For concreteness, let’s pick the particular permutation group $S_4$, which is defined to be the group of permutations of four objects.  This group has $4!=4\times 3\times 2\times 1=24$ elements, and we’ll list them all here in their full and tedious glory:

$S_4=\{e, (1,2), (1,3), (1, 4), (2, 3), (2, 4), (3,4),$

$(1, 2, 3), (1, 3, 2), (1, 2, 4), (1, 4,2),(1, 3,4), (1, 4, 3), (2, 3,4), (2, 4, 3),$

$(1, 2)(3,4), (1, 3)(2,4), (1,4)(2,3), (1,2, 3, 4), (1, 2, 4, 3), (1, 3, 2, 4), (1, 3, 4, 2), (1, 4, 3, 2), (1, 4, 2, 3)\}$

An easy exercise would be to ensure that there are indeed 24 elements written above, and a perhaps harder exercise would be to look away from this page for 5 minutes and then try to write all 24 of the permutations of 4 objects down on your own.  The easiest exercise of all would be to trust my judgement and believe that these are all 24 of them (though math isn’t faith-based, so I encourage the reader to check my work (especially since you’re a reader who has made it to lesson 28)).

I wrote all 24 of these elements out because it will now be much easier to study the subgroups of $S_4$.  Let’s begin by simply writing down a couple subgroups and proving that they’re subgroups.  Then, we’ll talk about a way to generate subgroups from specific elements.  Now, we know that subgroups are nothing but subsets that “remember” the structure of its parent group in a particular way—namely, in a way that makes the subset a group in its own right, with the group structure being inherited from its parent group.  Thus, let’s consider some subsets of $S_4$ and ask whether or not they’re also subgroups.

Let’s first consider the subset $\{e, (1,2)\}$.  Is this a subgroup?  The answer is yes, now let’s see why.  First of all, it has the identity of $S_4$, which we know any subgroup must have.  So that’s good.  Second, it is closed under group multiplication, meaning that if we take any two elements in this subset and group multiply them together, their product will also be in the subset.  This is true for our subset because the only non-trivial product we need to consider is $(1,2)\cdot (1,2)$ (where we recall that the product in the group of permutations is to simply compose permutations “one after the other” (again, feel free to review the above lessons if this is unfamiliar (this lesson will still be here when you get back (I promise)))).  But it is easy to see that $(1,2)\cdot (1,2)=e$, since $(1,2)$ swaps, and then $(1,2)$ swaps right back.  Moreover, I say that this is the only non-trivial product that we need to consider because every other product involves $e$, and since $e$ leaves everything fixed (by definition of the group identity), we know that we don’t need to verify that the product of anything in the subset with $e$ is still in the subset.  Since the non-trivial product $(1,2)\cdot (1,2)=e$, and since $e$ is in our subset, we have proved that our subset is closed under the group multiplication adopted from the parent group $S_4$.  Finally, we need to make sure that for every element in our subset, the inverse of that element is also in our subset.  But this is also easy, because the inverse of $e$ is $e$, and the inverse of $(1,2)$ is $(1,2)$ (which the above calculation shows).  Thus, the inverses of the elements in our subset are also in our subset, and our proof that the subset $\{e, (1,2)\}$ is also a subgroup is complete.

When we note that the above analysis relied only on the facts that a) our subset had $e$ in it and that b) our subset only had one other element in it, and that this element was its own inverse, we can immediately see that the subsets $\{e, (1,3)\},$ $\{e, (1,4)\},$ $\{e, (2,3)\},$ $\{e, (2,4)\},$ and $\{e, (3,4)\}$ are all also subgroups of $S_4$.

Now, before going on to find more exotic subgroups of $S_4$, let’s first look at subsets which are not subgroups.  This is a standard tactic for understanding things in math—finding examples and non-examples.  Sometimes it’s the non-examples that are most enlightening, as they often highlight certain things about our mathematical structures that motivate why we make the definitions we make.  Let’s then consider the subset $\{e, (1, 2), (1,3)\}$ and ask if it’s a subgroup of $S_4$.  Is it?  Go ahead and give it a good think over for a second.  Now, we know that the subsets $\{e, (1, 2)\}$ and $\{e, (1, 3)\}$ are both subgroups, so what about $\{e, (1, 2), (1, 3)\}$?

In fact, the answer is no, this subset is not a subgroup.  This may seem surprising (and it may not).  This subset does have the identity, which is good, and does have the inverse of every element it contains (since each element in our subset is its own inverse), which is also good.  What it does not have, however, are the products $(1, 2)\cdot (1,3)=(1, 3, 2)$ and $(1, 3)\cdot (1,2)=(1, 2, 3)$.  I.e., our subset is not closed under multiplication.  We now see again why we need to ask for closure under group multiplication, for if we want our subset to be a group in its own right, then we need to be able to define a group multiplication on it for every pair of its elements.  However, we’re also forced to use the same multiplication rule that its parent group has, which means that we can’t define this multiplication if there are a pair of elements whose product is not in our subset.

So let’s just jam these two missing elements into our subset so that we have $\{e, (1, 2), (1, 3), (1, 2, 3), (1, 3, 2)\}$.  Do we finally have a subgroup?  We still have the identity, and in fact we still have inverses (because the inverse of $(1, 2, 3)$ is $(1, 3, 2)$ and vice versa).  However, we’re still not quite there, because we don’t have the product $(1, 2)\cdot (1, 2, 3)=(2, 3)$ (this is also the product $(1, 3)\cdot (1, 3, 2)$).  So let’s also jam $(2, 3)$ into our subset and see what happens.  Well, we won’t really see what happens, and I’ll leave it for the reader to check that $\{e, (1,2), (1, 3), (2, 3), (1, 2, 3), (1, 3, 2)\}$ is indeed a subgroup.  Namely, by checking the products of all possible pairs, one will find that the subset is indeed closed under group multiplication.  We also know that this subset contains inverses of all of its elements, and it clearly contains the identity, so we’ve finally found another more interesting subgroup.

The procedure that we used to create this subgroup was somewhat ad hoc, in the sense that we just kind of looked around for products that our subset didn’t have, then added them in, and then looked again until we got to a point where we had all of our products.  We were also lucky that at every step of the game it was easy to see that our subset always had inverses of its elements.  At this stage, there are a couple somewhat natural questions to as.  Is there a more mechanical way of creating subgroups of $S_4$?  If the enemy hands me an element of $S_4$, can I construct a subgroup that contains this element?  If I can, will I be able to find the smallest such subgroup, i.e., the smallest subgroup containing that element?  It turns out that all of these questions have a positive answer, and we can answer them all in one go.

Suppose the enemy hands me an element of $S_4$, and for concreteness let’s actually choose one: $(1, 3, 4)$ (my reason for choosing this element is non-existent).  What is the smallest subgroup I can make that contains this element?  Well, I know that $e$ needs to be in my subset, and the statement of the problem requires me to put $(1, 3, 4)$ in there as well.  Thus, so far we have $\{e, (1, 3,4), ...\}$ where the “…” is there to remind us that we may need to add more elements to our set.  Now, we’ve already satisfied the criterion for a subgroup that the identity is a member of it, so let’s now focus on the criterion of being closed under group multiplication.  The only non-trivial product we need to consider is $(1, 3, 4)\cdot (1, 3, 4)$, which gives $(1, 4, 3)$.  For obvious reasons, we can denote this product as $(1, 3, 4)^2$, so that $(1, 3, 4)^2=(1, 3, 4)\cdot (1, 3, 4)=(1, 4, 3)$.  Now, $(1, 4, 3)$ is not in our subset, so let’s throw it in there.  We therefore have $\{e, (1, 3, 4), (1, 4, 3),...\}$, where again the “…” reminds us that we might not be done yet.

Let’s now see if we have $(1, 3, 4)^3=(1, 3, 4)\cdot (1, 3, 4)\cdot (1, 3, 4)=(1, 3, 4)^2\cdot (1, 3, 4)$ in our subset.  We have $(1, 3, 4)^3=(1, 3, 4)^2\cdot (1, 3, 4)=(1, 4, 3)\cdot (1, 3, 4)=e$, which is indeed in our set.  Now, it’s easy to see that our set is closed under multiplication, because the only remaining non-trivial product to examine is $(1, 4, 3)\cdot (1, 4, 3)$, but this is $(1, 3, 4)^2\cdot (1, 3, 4)^2=(1, 3, 4)^3\cdot (1, 3, 4)=e\cdot (1, 3, 4)=(1, 3, 4)$, which is in our set.  It is also easy to check that our subset contains all of its inverses as well, so that $\{e, (1, 3, 4), (1, 4,3)\}$ is indeed a subgroup.  Moreover, it is the smallest subgroup containing $(1, 3,4)$, because at every step of our discussion we did the bare minimum necessary for ensuring closure under multiplication, which is a prerequisite for a subset to be a subgroup.

It is now clear how to always generate a subgroup of $S_4$ containing a particular element.  Namely, we take this element and include all the powers of it (i.e., itself composed with itself over and over again) until we get back to the identity.  At each step, if we form a product that isn’t already in our subset, we place it in there.  Once we cycle back to the identity all that will happen by considering higher powers of that element will be that we’ll cycle through our subset again.  This is an example of what’s called a “cyclic group”, of which we’ll have more to say later.  By doing this, we’re doing the bare minimum necessary for ensuring closure under multiplication, and in the process we get inverses for free.  Namely, any element in the subset that we construct in this manner will be written as some power of the element with which we started.  Since the identity itself can be written as some power of this element (in our example above, $e=(1, 3, 4)^3$), the inverse of any element is just that element with the correct exponent.  For example, in our example above, the inverse of $(1, 3, 4)^2=(1, 4, 3)$ is $(1, 3, 4)^1=(1, 3, 4)$.  This procedure of generating “smallest subgroups containing a certain element” naturally generalizes to any permutation group, and not just the group of permutations of 4 objects.

This will do it for now, and I’ll provide a couple explicit calculations in the exercises.  We’ll come back to this when we explore cyclic groups more thoroughly in the future, but for now lets do the exercises and then go on to learn about subgroups of dihedral groups (which are similar to those found here).

Exercises

1) What is the smallest subgroup in $S_4$ containing the element $(1, 2, 3, 4)$?

2) What is the smallest subgroup in $S_5$ (the group of permutations of 5 objects) containing the element $(1, 4)(2,5)$?

Solutions to Exercises

Back to Lesson 27

On to Lesson 29