# Lesson 27: Group Constructions: Subgroups

Now that we’ve seen several different specific examples of groups, it is time that we ask a few more questions about their more general, abstract existence.  And now that we’re becoming more mature and sophisticated mathematicians, we know relatively well how this should go.  Namely, we’ve just defined an abstract structure (a group), and it is almost always the case that the next logical question to ask is what sort of substructure this structure has.  Let us therefore proceed along these lines.

In the case of sets, we were motivated to define a subset as the proper notion of substructure for a set.  Well, seeing as a group is a set (with some added structure), we can indeed ask about subsets of groups.  However we quickly run into a very obvious problem, which is that some subsets are obviously “more special” than other subsets, though we don’t yet have any machinery to encapsulate this.

For example, let’s suppose we were talking about the familiar group of integers $\mathbb{Z}$ under addition.  There are many subsets of this group—indeed, infinitely many.  But it is somehow clear to us as reasonable humans that the subset $\{15, -24, 1,633,131, 417\}$ is fundamentally different from the subset $\{..., -8, -6, -4, -2, 0, 2, 4, 6, 8, ...\}$ of even numbers.

How might one describe the way in which these two subsets are different?  Well, one could say that one subset has finitely many elements, and the other has infinitely many.  That’d be correct, but it doesn’t really get to the heart of how these subsets are different.  For I could define a subset like $\{18, 23, 45, -17, 457, 1,673,900, -29, ...\}$ where the “…” stands for some infinitely long list of completely random integers (unfortunately I can’t explicitly write out such a list on this page, for reasons which I hope are clear and forgivable to the reader).  I could define such a list and we could try to make sense of it by saying it’s “first the number of eggs in one of those extra large cartons, then the number that Michael Jordan is remembered for, then the number that Michael Jordan is not remembered for (45, for the non-basketball fans), then negative the age at which Kobe was drafted, then s0me number which is much bigger than the previous ones, then some gigantic number, then oh jeez I give up!”.  The point being that there really is no way to make sense of this set, whereas the other set that we wrote down can simply be specified by “all the even integers”.

So it isn’t the infinite-ness of one of the subsets that separates it from the other, but rather some sort of “simplicity of specification”.  But this still isn’t quite right, because I can very easily specify the subset $\{63\}$ as “the number 63”, but this fact doesn’t make this subset any more interesting.  There’s nothing cool about the subset $\{63\}$ whereas there seems to be something interesting about the set $\{...,-6, -4, -2, 0, 2, 4, 6, ...\}$.

And whatever it is that is interesting about the subset $\{..., -6, -4, -2, 0, 2, 4, 6, ...\}$, the same thing will be interesting about the subset $\{..., -12, -9, -6, -3, 0, 3, 6, 9, 12, ...\}$, as well as the subset $\{..., -30, -20, -10, 0, 10, 20, 30, ...\}$.

As a last piece of motivation for our next definition, let’s take a look at another group that we know and love—modular arithmetic.  Let’s consider the set $\{0, 1, 2, 3, 4, 5, 6, 7\}$ with addition modulo 8, meaning that we continually “cycle back through” once we hit 8, in precisely the same way we do with the number 12 when we tell time (or the number 24 for the non-12-hour-clock readers).  We note that for this group, the subset $\{0, 2, 4, 6\}$ is somehow very different from the subset $\{1, 3, 4, 7\}$.  Perhaps the reasoning for this is less clear, but it will become clear when we define the proper notion of substructure for a group.  And by “proper notion”, I simply mean that which properly encapsulates how some of these subsets of these various groups are “more interesting” than other subsets.  Let us therefore just make the definition and see how it goes.

Definition: Let $G$ be a group and let $H$ be a subset of $G$.  We call $H$subgroup of $G$ if $H$ is itself a group, with its group structure adopted from $G$.

A few words need to be said about this.  First, I need to clarify what I mean by “with its group structure adopted from $G$”.  By this, I mean that the rule of abstract group multiplication in $H$ is the same as that in $G$, which is well-defined because any element in $H$ is also an element in $G$ by virtue of its being a subset.  This also means that the identity in $H$ is the same as the identity in $G$.  The fact that the group multiplication for $H$ is associative follows from the fact that $G$ is a group, so its multiplication is associative, and $H$ simply adopts this multiplication and is therefore associative.

The second and perhaps most important thing that should be noted here is that $H$ must be what’s called “closed” under the group multiplication that it adopts from $G$.  This means that if two elements $g$ and $h$ are in $H$, then their product $g\cdot h$ is also in $H$.  Indeed, this is required by the statement “$H$ is itself a group”, since the multiplication on a group is a function from “itself cross itself” to “itself”, and therefore two elements in $H$ can’t jump out of $H$ if $H$ is to be a group in its own right.

Now it turns out (as it should) that all of the example subsets that motivated this definition are indeed subgroups.  For example, the set $\{..., -6, -4, -2, 0, 2, 4, 6, ...\}$ of all even integers is a subgroup of the integers, since it has the identity element of the integers, it is closed under addition (i.e., an even number plus an even number is again an even number), and it has inverses (i.e., the negative of an even number is again an even number).  Similarly, the set $\{..., -30, -20, -10, 0, 10, 20, 30, ...\}$ of all multiples of 10 is a subgroup of the integers, since again it has the identity element, it is closed under addition (the sum of multiples of 10 is again a multiple of 10), and it has inverses (the negative of a multiple of ten is again a multiple of 10).

However, the subsets $\{15, -24, 1,633,131, 417\}$, and $\{18, 23, 45, -17, 457, 1,673,900, -29, ...\}$ are not subgroups, which can be easily seen by noting that they don’t have the identity element of $\mathbb{Z}$ (they also aren’t closed under addition nor do they contain the inverses of all of their elements, so these subsets are very not-subgroup-ey).

What about the set $\{..., -5, -3, -1, 1, 3, 5, 7, ...\}$ of all odd integers?  This also is not a subgroup, for two reasons.  Although this subset does contain the inverses of all of its elements, it neither has the identity element nor is it closed under addition (an odd number plus an odd number is an even number, which isn’t in this subset).

Let’s look at one more extremely important example of what is not a subgroup.  Namely, let’s consider $\{0, 1, 2, 3, 4, 5, 6, 7\}$.  Is this subset of $\mathbb{Z}$ also a subgroup?  The answer is no, which might have been obvious, or might be surprising.  The reason it might be surprising is that we know that we can turn this set into a group by using addition modulo 8.  However, we must recall that the definition of a subgroup involves having the subset adopt the same group structure from its parent group.  Therefore, this subset is not a subgroup of $\mathbb{Z}$ because it requires a different form of group multiplication to be turned into a group.  Once we recall this fact, it is easy to see that this subset is not a subgroup.  It does indeed contain the identity element, but it isn’t closed under addition (for example, $1+7$ is not in this subset because under the regular addition of integers $1+7=8$) and it doesn’t contain the inverses of its elements (under the regular addition of integers).

If we do consider addition modulo 8, though, then the subsets $\{0, 2, 4, 6\}$ and $\{0, 4\}$ are both subgroups.  For example, if we focus on the second subset, we have that $0+0=0,\ 0+4=4+0=4,\ 4+4=0$, so it is closed under addition (modulo 8).  Moreover, it contains the identity element, and it contains inverses because the inverse of $0$ is $0$, and the inverse of $4$ is $4$.  In a completely similar fashion, we can see that the subset $\{0, 2, 4, 6\}$ is also a subgroup, since it is closed under addition (which the reader should check/convince herself of), contains the identity, and has inverses (for example, the inverse of 6 is 2, which is in the set).

To end this lesson, let me note that any group comes along with 2 subgroups for free.  Namely, for any group $G$, $G$ is a subgroup of itself.  This is painfully obvious and trivial, yet also extremely important (for reasons we won’t see for a LONG time).  We can see that this is true by noting that any set is a subset of itself, so $G$ satisfies the requirement of being a subset of itself.  And $G$ is obviously a group when we consider it adopting the group structure of its parent group, because its parent group is itself, which is a group!

The second subgroup that we always get for free is the subgroup formed by the 1-element subset $\{e\}$ of only the identity of the group.  This is indeed closed under group multiplication (the identity fixes everything, including itself, so multiplication in this group boringly multiplies the identity to itself over and over again, always coming right back to the identity), it contains the identity (obviously), and it contains inverses of every element, since the inverse of the identity is the identity.  Since every group by definition has an identity, we always get this subgroup for free.

These two “free” subgroups are also very boring (the phrase “there’s no free lunch” applies to math as well—anything that is free is always boring (well, sometimes)).  We therefore make the following definition.

Definition: Let $G$ be a group and let $H$ be a subgroup of $G$.  We say $H$ is a trivial subgroup of $G$ if $H=\{e\}$ or if $H=G$, otherwise we say $H$ is a non-trivial subgroup.

The reason for using this terminology is hopefully clear.  I’ll finish with an exercise, and then we’ll see some more examples of subgroups in the next lesson, where we examine subgroups of groups that are non-commuting (and therefore more complex).

Exercise:

1) Find 3 non-trivial subgroups of the group $\{0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$ with addition modulo 12.

Solution to Exercise

On to Lesson 28

Back to Lesson 26