Introduction to Some of Group Theory

Steven Miller^{∗}

Abstract

We will review some group theory.

1 Lagrange’s Theorem

1.1 Basic group theory

Group G is a set of elements g_{i} satisfying the four conditions below, relative to some binary operation. We often use multiplicative notation (g_{1}g_{2}) or additive notation (g_{1}+g_{2}) to represent the binary operation. For definiteness, we use multiplicative notation below; however, one could replace xy with b(x, y) below.

If the elements of G satisfy the following four properties, then G is a group.

1. ∃e ∈ G s.t. ∀g ∈ G : eg = ge = g. (Identity.) We often write e = 1 for multiplicative groups, and e = 0 for additive groups.

2. ∀x, y, z ∈ G : (xy)z = x(yz). (Associativity.)

3. ∀x ∈ G, ∃y ∈ G s.t. xy = yx = e. (Inverse.) We write y = x^{−1} for multiplication, y = −x for addition.

4. ∀x, y ∈ G : xy ∈ G. (Closure.)

∗E-mail: millerj@cims.nyu.edu or sjmiller@math.princeton.edu

If commutation holds (∀x, y ∈ G, xy = yx), we say the group is Abelian. Non-abelian groups exist and are important. For example, consider the group of N ×N matrices with real entries and non-zero determinant. Prove this is a group under matrix multiplication, and show this group is not commutative.

H is a subgroup of G if it is a group and its elements form a subset of those of G. The identity of H is the same as the identity of G. Once you’ve shown the elements of H are closed (ie, under the binary operation, b(x, y) ∈ H if x, y ∈ H), then associativity in H follows from closure in H and associativity in G.

For the application to Fermat’s Little Theorem you will need to know that the set {1, x, x^{2}, ⋯ x^{n−1}} where n is the lowest positive integer s.t. x^{n} = 1, called the cyclic group, is indeed a subgroup of any group G containing x, as well as n divides the order of G.

For a nice introduction to group theory see: M. Tinkham, Group Theory and Quantum Mechanics, (McGraw-Hill, 1964) or S. Lang, Undergraduate Algebra.

1.2 Lagrange’s Theorem

The theorem states that if H is a subgroup of G then |H| divides |G|.

First show that the set hH, i.e. all the elements of H premultiplied by one element, is just H rearranged (Cayley’s theorem). By closure hH falls within H. We only need to show that hh_{i} can never equal hh_{j} for two different elements i ≠ j. If it were true, since a unique h^{−1} exists we could premultiply the equation hh_{i} = hh_{j} by h^{−1} to give h_{i} = h_{j}, which is false. Therefore hh_{i} ≠ hh_{j}, and we have guaranteed a 1-to-1 mapping from H to hH, so hH = H.

Next we show that the sets g_{i}H and g_{j}H must either be completely disjoint, or identical. Assume there is some element in both. Then g_{i}h_{1} = g_{j}h_{2}. Multiplying on the right by h_{i}^{−1} ∈ H (since H is a subgroup) gives g_{i} = g_{j}h_{2}h_{1}^{−1}. As H is a subgroup, ∃h_{3} ∈ H such that h = h_{2}h_{1}^{−1}. Thus g_{i} = g_{j}h_{3}. Therefore, as h_{3}H = H, g_{i}H = g_{j}h_{3}H = g_{j}H, and we see if the two sets have one element in common, they are identical. We call a set gH a coset (actually, a left coset) of H.

Clearly

G = ⋃_{g∈G} gH  (1)

Why do we have an equality? As g ∈ G and H ⊂ G, every set on the right is contained in G. Further, as e ∈ H, given g ∈ G, g ∈ gH. Thus, G is a subset of the right side, proving equality.

There are only finitely many elements in G. As we go through all g in G, we see if the set gH equals one of the sets already in our list (recall we’ve shown two cosets are either identical or disjoint). If the set equals something already on our list, we do not include it; if it is new, we do. Continuing this process, we obtain

G = ⋃_{i=1}^{k} g_{i}H  (2)

for some finite k. If H = {e}, k is the number of elements of G; in general, however, k will be smaller.

Each set g_{i}H has |H| elements. Thus, |G| = k|H|, proving |H| divides |G|.

2 Quotient groups

Say we have a finite Abelian group G (this means for all x, y ∈ G, xy = yx) of order m which has a subgroup H of order r. We will use multiplication as our group operation. Recall the coset of an element g ∈ G is defined as the set of elements gH = g{h_{1}, h_{2}, ⋯, h_{r}}. Since G is Abelian (commutative) then gH = Hg and we will make no distinction between left and right cosets here.

The quotient group (or factor group), symbolized by G/H, is the group formed from the cosets of all elements g ∈ G. We treat each coset g_{i}H as an element, and define the multiplication operation as usual as g_{i}Hg_{j}H. Why do we need G to be Abelian? The reason is we can then analyze g_{i}Hg_{j}H, seeing that it equals g_{i}g_{j}HH. We will analyze this further when we prove that the set of cosets is a group.

There are several important facts to note. First, if G is not Abelian, then the set of cosets might not be a group. Second, recall we proved the coset decomposition rule: given a finite group G (with n elements) and a subgroup H (with r elements) then there exist elements g_{1} through g_{k} such that

G = ⋃_{i=1}^{k} g_{i}H.  (3)

The choices for the g_{i}’s is clearly not unique. If g_{1} through g_{k} work, so do g_{1}h_{1} through g_{k}h_{k}, where h_{i} is any element of H. Recall this was proved by showing any two cosets are either distinct or identical.

We will show below that, for G Abelian, the set of cosets is a group. Note, however, that while it might at first appear that there are many different ways to write the coset group, they really are the same. For example, the cosets gH and gh_{1}h_{2}^{4}h_{3}H are equal. This is similar to looking at integers mod n; mod 12, the integers 5, −7 and 19 are all equal, even though they look different.

We now prove that the set of cosets is a group (for G Abelian).

Closure. By commutivity g_{i}Hg_{j}H = g_{i}g_{j}HH. What is “HH”? Just the set of all r^{2} possible combinations of elements of H. By closure, and the existence of the identity, this just gives H again (recall no element in a group can appear more than once—duplicates are removed). Therefore g_{i}Hg_{j}H = g_{i}g_{j}H. Now, as G is a group and is closed, g_{i}g_{j} ∈ G. Thus, there is a α such that g_{i}g_{j} ∈ g_{α}H (as G = ⋃_{β=1}^{k} g_{β}H. Therefore, there is an h ∈ H such that g_{i}g_{j} = g_{α}h, which implies g_{i}g_{j}H = g_{α}hH = g_{α}H. Thus, the set of cosets is closed under coset multiplication. Note, however, that while the coset g_{i}g_{j}H is in our set of cosets, it may be written differently.

Identity. If e is identity of G, then eHg_{i}H = g_{i}H and g_{i}HeH = g_{i}H, so eH is the identity of this quotient group.

Associativity. Since as you may have noticed, the quotient group elements behave just like those of G, associativity follows from that of G.

Inverse. It is easy to guess g^{−1}H is the inverse of gH. Check it: g^{−1}HgH = g^{−1}gH = eH = identity, also true the other way round of course by commutativity. Unfortunately, g^{−1}H might not be listed as one of our cosets! Thus, we must be a little more careful. Fortunately, as g^{−1} ∈ G = ⋃_{β=1}^{k} g_{β}H, there is an α such that g^{−1} ∈ g_{α}H. Then, there is an h ∈ H with g^{−1} = g_{α}h. Thus, g^{−1} = g_{α}hH = g_{α}H, and direct calculation will show that the coset g_{α}H is the inverse (under coset multiplication) of gH.