These four conditions must be satisfied for a set S and an operation * to be considered a group:

Groups must contain a set of elements, an operation (*), and satisfy four conditions.

Closure (If x is in set S and y is in the set S, then x*y must be in set S)
* must be an associative operation. Hence (x*y)*z = x*(y*z) assuming that x,y, and z are all in set S.
Set S must contain an identity element (e) such that x*e=e*x=x for all x in set S.
Every element in set S must have an inverse (If x is in set S, then there must be a y in set S such that x*y = y*x = e where e is the identity element).

For example, consider the set S = {-1,1}. Even though this set is a rather boring set to consider it is clearly a group with respect to multiplication. This can be verified by checking all of the conditions that need to be satisfied for a set and an operation to be considered a group.

Closure: 1 * 1 = 1; -1 * -1 = 1; 1 * -1 = -1; -1 * 1 = -1

As you can see, no new elements are produced that are not already in the set S satisfying the group property of closure.

Real number multiplication is associative. Example: (1*1)*-1 = -1 = 1*(1*-1)

The identity element in set S with respect to multiplication is 1:

1*1 = 1; 1*-1=-1; -1*1=-1

Each element in set has an inverse with respect to multiplication. The inverse of the element 1 is itself (1*1=1) and the inverse of the element -1 is also itself (-1*-1=1). Remember that an element operated with its inverse yields the identity element.

It is easy to illustrate all of the group properties with the set {-1,1} and multiplication. If any of the group properties are not satisfied the set and the operation cannot be considered a group.

Application Problem (Problem taken from the Modern Algebra text)

Determine if the set {i,-i,1,-1} is a group with respect to multiplication. If you determine that this set is not a group with respect to multiplication, state all of the group conditions that do not hold true. On the other hand, if this set is a group, verify each of the four group conditions.

Solution (by Ben Mueller)

Closure:

This can be verified by completing a multiplication table with each element in the set.

As you can see, no new elements appear on our table. Thus, this set is closed with respect to multiplication and the group condition of closure is satisfied.

Associative:

Complex number multiplication is associative. Example: (i*i)*-1 = i*(i*-1)

(-1)*-1 = i*(-i)

1 = 1

Identity element:

The identity element for this set with respect to multiplication is 1.

1*i=i=i*1

1*-i=-i=-i*1

1*1=1

1*-1=-1=-1*1

Inverse:

Every element in this set has an inverse with respect to multiplication.

Inverse of 1 = 1 (1*1=1)

Inverse of -1 = -1 (-1*-1=1)

Inverse of i = -i (i*-i=1)

Inverse of –i = i (-i*i=1)

All four conditions for a group hold true. Thus the set {1,-1,i,-i} is indeed a group with respect to multiplication.

**Note: If we changed our operation to addition (instead of multiplication) with the same set, we would no longer have a group. In fact, the only condition that would hold true then would be the associative property. The group condition of closure would not be satisfied because i is in set S and i+i=2i (2i is not in the set). The set S also would not contain an identity element since 0 is the identity element with operation addition and 0 is not in the set. Since there is no identity element in the set, we cannot talk about inverses.**

Now, there are also some other properties that hold true for groups. For example, the identity element of a group must be unique.

Note: If G is a group, the identity element of G is unique.

Proof by contradiction:

Assume that the identity element in G is not unique. Then there is an e and e’ in G such that e*e’=e’ (e is identity) and also, e*e’=e (e’ is identity).

Now we have that e*e’=e’=e and thus, e=e’. This contradicts our original statement that the identity element in G is not unique and completes the proof./(Proof complete)

**Recall that the concept of abelian means that the elements within the group are commutative with respect to the groups operation. For example, if x and y are both in the group G, we can be assured that xy=yx.**

Prove that H = {h element of G such that h^(-1) = h} is a subgroup of the group G if G is abelian. (Note h^(-1) denotes the inverse of h)

Proof (written by Ben Mueller):

To prove that a group H is a subgroup of a group G, we must first show that H is non-empty. Now, since G is a group we can be assured that there is an identity element (e) in G. Since e = e^(-1), we know that e is also in H and thus, H is non-empty.

To prove that H is a subgroup of the group G, we must also prove that the set H is closed under its respective operation. We know that if x and y are in H, then x = x^(-1) and similarly, y = y^(-1). From this, xy = x^(-1)y^(-1) and by the reverse order law,

x^(-1)y^(-1) = (yx)^-1. Now since G is an abelian group and x,y are both in G we know that xy=yx. Because of this, (yx)^-1 = (xy)^-1. This is exactly what we were trying to prove as this shows that xy = (xy)^-1 which implies that xy is in the set H.

Lastly we want to show that if x is in H, then the inverse of x is in H. This is exactly the condition that must hold true for an element to be in H and thus, this part of the proof is easily proven./(Proof complete)

**The order of an element in a group is the least amount of times that element must be operated on to produce the identity element.**

Prove that if a is an element of order m in a group G and a^k = e, m divides k.

Proof (Written by Ben Mueller):

Since the element a has order m, a^m = e. We also have that a^k = e and we know that

m<k or m=k since m is the order of a. Now, a^m=a^k=e and that means that m is congruent to k modulo m (m = k (mod m)). From this, we get that (m-k)/m = z where z is an integer. The equation from the last statement can be rewritten as m/m – k/m = z or 1 – k/m = z. By subtracting 1 from both sides of this equation, we see that –k/m = z-1 and since z is an integer, z-1 must also be an integer. This implies that –k/m produces an integer which means that m must divide k./(Proof complete) A more direct proof to this is shown below:

a has order mèa^m=e=a^k è m<k or m=k (a has order m which means m is the least integer such that a^m=e) è m = k (mod m) è (m-k)/m = z where z is an integerè

m/m – k/m = zè1 – k/m = zè-k/m=z-1èm divides k (z-1 is an integer)./(Proof complete)