* Definition.* Let G be a group. A subset H of G
is a * subgroup* of G if:

(a) (Closure) H is closed under the group operation: If , then .

(b) (Identity) .

(c) (Inverses) If , then .

The notation means that H is a subgroup of G.

Notice that associativity is *not* part of the definition of a
subgroup. Since associativity holds in the group, it holds
automatically in any subset.

Look carefully at the identity and inverse axioms for a subgroup; do you see how they differ from the corresponding axioms for a group?

In verifying the identity axiom for a subgroup, the issue is not the
*existence* of an identity; the *group* must have an
identity, since that's part of the definition of a group. *The
question is whether the identity for the group is actually contained
in the subgroup*.

Likewise, for subgroups the issue of inverses is *not* whether
inverses *exist*; every element of a group has an inverse. The
issue is whether the inverse of an element in the subgroup is
actually contained in the subgroup.

* Lemma.* Let G be a group. Then
and G are subgroups of G.

is called the * trivial
subgroup*.

* Proof.* The proofs are almost too easy!
Consider . The only possible multiplication is , which shows is closed.

obviously contains the identity 1.

is closed under taking inverses, since .

The proof that G is a subgroup is equally easy; I'll let you do it.

* Example.* (* Subgroups of the
integers*) Let . Let

Show that is a subgroup of , the group of integers under addition.

consists of all multiples of n.

First, I'll show that is closed under addition. If , then

Therefore, is closed under addition.

Next, the identity element of is 0. Now , so .

Finally, suppose . The additive inverse of in is , and . This is n times something, so it's in . Thus, is closed under taking inverses.

Therefore, is a subgroup of .

I'll show later that every subgroup of the integers has the form for some .

Notice that is not a subgroup of . I have and , so 2 and 3 are elements of the union . But their sum is not an element of , because 5 is neither a multiple of 2 nor a multiple of 3.

This example shows that the union of subgroups need not be a subgroup.

* Example.* (* A subset that
isn't closed under inverses*) is a group under
addition. Consider , the set of nonnegative
integers. Check each axiom for a subgroup. If the axiom holds, prove
it. If the axiom doesn't hold, give a specific counterexample.

If , then and , so . Therefore, , and the set is closed under addition.

0 is a nonnegative integer, so .

However, , but the inverse -3 is not an element of . Therefore, is not closed under taking inverses, so it's not a subgroup of .

* Example.* (* The integers as a
subgroup of the rationals*) Show that the set of integers is a subgroup of , the group of
rational numbers under addition.

If you add two integers, you get an integer: is closed under addition.

The identity element of is 0, and .

Finally, if , its additive inverse in is . But is also an integer, so is closed under taking inverses.

Therefore, is a subgroup of .

* Example.* (* A subgroup under
multiplication*) Let be the group
of nonzero integers under multiplication. Consider the set

Is H a subgroup of ?

Let , where . Then

Thus, H is closed under multiplication.

The identity of is 1, and .

Finally, let . Then , and . Therefore, H is closed under taking inverses.

Therefore, H is a subgroup of .

* Example.* denotes the set of pairs of integers:

It is a group under "vector addition"; that is,

The identity is and the inverse of is .

Taking this for granted, consider the set

Check each axiom for a subgroup. If the axiom holds, prove it. If the axiom doesn't hold, give a specific counterexample.

In words, an element is in H if the sum of its components is nonnegative.

Suppose . This means

Then

Therefore,

Thus, H is closed under addition.

Since , I have .

, because . But , because

Thus, the inverse axiom fails (so H is not a subgroup).

* Definition.* If G is a group and g is an
element of G, the * subgroup generated by* g (or
the * cyclic subgroup generated by* g) is

In other words, consists of all (positive or
negative) *powers* of g.

This definition assumes *multiplicative* notation; if the
operation is addition, the definition reads

In this case, you'd say that consists
of all (positive or negative) *multiples* of g.

Be sure you understand that the difference between the two forms is simply notational: It's the same concept.

Since I'm calling a *subgroup*, I'd
better verify that it satisfies the subgroup axioms.

* Lemma.* If G is a group and , then is a subgroup of G.

* Proof.* For closure, note that if , then

. Finally, if , its inverse is , which is also in .

Therefore, is a subgroup of G.

In fact, is the *smallest* subgroup
of G which contains g.

* Example.* (* Subgroups of a
finite cyclic group*) List the elements of the subgroups
generated by elements of .

The way the subgroups are contained in one another can be pictured in
a * subgroup lattice diagram*:

The following result is easy, so I'll leave the proof to you. It says that the subgroup relationship is transitive.

* Lemma.*(* Subgroup
transitivity*) If and , then : A subgroup of a subgroup is a subgroup of the (big)
group.

If you want to show that a subset H of a group G is a subgroup of G, you can check the three properties in the definition. But here is a little shortcut.

* Lemma.* Let G be a group, and let H be a
nonempty subset of G. if and only if implies .

* Proof.* ( ) Suppose , and let . Then (since H is closed under inverses), hence (since H is closed under products).

( ) Suppose that implies . Since , take . Then .

If , then (since I already know ). This shows H is closed under taking inverses.

Finally, suppose . Then , so . Therefore, .

Note: In order to use this criterion, *you have to show that the
set in question is nonempty* before doing the " " check. Usually you show the set is
nonempty by showing that it contains the identity element. So you
really have to do two checks, not just one.

* Example.* (* A subgroup of a
matrix group*) Let be the set of invertible
matrices with real entries.

(a) Show that is a group under matrix multiplication.

(b) Show that the following set is a subgroup of :

(a) If , then and . Hence,

Therefore, is invertible, so matrix multiplication is a binary
operation on . (The point is that the set is
*closed* under the operation.)

From linear algebra, I know that matrix multiplication is associative.

The identity matrix is invertible, so it's in . It's the identity for under matrix multiplication.

Finally, if , then exists. It's also an element of , since its inverse is A.

This proves that is a group under matrix multiplication.

(b) First,

Therefore, D is nonempty.

Next, suppose , where and . Note that

Then

Therefore, D is a subgroup of .

* Definition.* Let G be a group. * commute* if .

The * center* of G is the set of
elements which * commute* with everything in G:

* Lemma.* .

* Proof.* Suppose . I'll show
. To do this, I must show that
commutes with everything in G.

Let . Then

Therefore,

Next, for all , so .

Finally, let . I need to show that . Let . I need to show that .

I have

Therefore, .

Hence, is a subgroup of G.

The union of subgroups is not necessarily a subgroup, but the intersection of subgroups is always a subgroup. Before I prove this, a word about notation.

In this result, I want to talk about a bunch of subgroups of a group G. How should I denote these subgroups? I don't want to write , , ..., , because I may want an infinite number of subgroups. Well, how about , , ... (where I think of the sequence as continuing forever)?

The problem in the second case is that I might not be able to list
the subgroups in a sequence. You may know that there are different
kinds of "infinity" and some a bigger than others.
Specifically, if the number of subgroups under consideration is not
* countable*, I can't list them as "
, , ...".

I'll use notation like in
situations like these. Each is a subgroup, and A is an * index set*. In other words, A is an unspecified set
whose elements I use to subscript the H's. Since A could be
arbitrarily big, this gets around the problems I had with the other
notations.

Rather than get into technicalities, I will leave things at that and illustrate by example how you work with infinite index sets. If the next proof confuses you, try writing out the proof for two subgroups: That is, if H and K are subgroups of a group G, then is a subgroup of G.

* Lemma.* The intersection of a family of
subgroups is a subgroup.

* Proof.* Let G be a group, and let be a family of subgroups of G. Let . I claim that H is a subgroup of G.

First, for all , because each is a subgroup. Hence, , and the intersection is nonempty.

Next, let . I want to show that . Since , I know for all a. Then for all a, since each is a subgroup. This implies that , so .

Here is how I can use the preceding construction. Suppose G is a group, and S is a collection of elements of G. S might not be a subgroup of G --- it might not contain 1, or it might be missing the inverses of some of its elements --- but intuitively I ought to be able to add the "missing elements" and enlarge S to a subgroup.

If you try to say precisely what you need to add to S, and how you will add it, you will quickly find yourself tied in knots. Do you add elements one at a time? If you throw in an element, you have to throw in the products of that element with everything else that is there (to ensure closure). If you do this sequentially, how do you know the process actually terminates?

Instead of building up the subgroup from S, I'll get at it "from above". Consider the collection of all subgroups such that . The collection is nonempty, because G is a subgroup of G and .

Let . H is a subgroup of G, and . H is * the subgroup generated
by* S. It is clearly * the smallest subgroup
of* G * containing* S, in the following
sense: If K is a subgroup of G and , then .

It's common to write for the subgroup generated by S. So in case (a finite set), write for the subgroup generated by the x's.In the case of a single element , the subgroup generated by x is the cyclic subgroup generated by x that I discussed earlier.

* Example.* (* Subgroups
generated by elements*) Let , the cyclic group of
order 6. Show

The first statement is easy: , .

What about the second? By definition, is the smallest subgroup which contains 2 and 3. Since subgroups are closed under addition, must be in as well. But I can make any element of by adding 1 to itself enough times, so must contain everything in --- that is, .

* Example.* is a group under
vector addition. Give an example of two subgroups whose union is not a subgroup.

consists of the points in the x-y-plane, or equivalently 2-dimensional vectors with real components.

Two elements of are added as 2-dimensional vectors:

The following sets are subgroups of :

A is the x-axis, and B is the y-axis.

For example, I'll verify that A is a subgroup of . It's closed under addition: If , then

The identity for is , which is contained in A.

If , then

Try writing out the proof for B yourself.

However, the union is *not* a subgroup of . is the union of the x-axis and the
y-axis. This set is not a subgroup because it's not closed under
addition. For example, and , but

This example shows that the union of subgroups need not be a subgroup.

Copyright 2018 by Bruce Ikenaga