Subgroup

We will now look at groups within groups which we will call a subgroup. 

Definition

Let G be a group with operator * and H a subset of G.  Then H is called a subgroup of G if  H is a group under the same operator *.

Notice that to prove that H is a subgroup of G, we do not need to show that the associative property holds for H since it the operator is the same as G.

Theorem

If G is a group and H is a subset of G, then H is a subset if and only if the following are true:

1. If a and b are in H, then a*b is in H.
2. If a is in H, then a^-1 is also in H.

Proof

If H is a subgroup of G, then 1. and 2. hold by the definition of a subgroup.  We need to show that if 1. and 2. hold then H is a subgroup of G.  All that is needed is to show that the identity element e is in H.  By 2. a^-1 is in H, so by 1. a*a^-1 = e is in H.

Example

If G is a group, then the subset of G containing only e is a subgroup.  Also G is a subgroup of itself.  These two subgroups are called the trivial subgroups of G. 

Example

If a is any element of G, then consider the set (a) = {aⁿ | n is an integer}.  Then this is a subgroup of G.  If we let n = -1, then we get the inverse.  Also a^x * a^y = a^x+y gives 1.  This subgroup is called the cyclic subgroup generated by a.  If G contains an a with (a) = G then G is called a cyclic group.

Theorem

If G is a finite group, then (a) = {a^k | k is a natural number)

Proof

Consider the elements

{a, a², a³, a⁴, ...}

Since G is a finite group, these cannot all be distinct.  So let a^x =  a^y with x < y.  Then  (a^x)^-1 (a^x) = (a^x)^-1 (a^y) so that

e = a^y-x = a^1+y-x-1 = a*a^y-x+1

This shows that

a^-1 = a^y-x+1

is in (a).

Example

Let Z₁₂ be the group of clock arithmetic numbers.  Then Z₁₂ is a cyclic group generated by (1).  Notice that (3) is a cyclic group of order 4.

Example

Instead of the clock having 12 hours, we can pick any number of hours to start over again.  We let Z_n be the set of integers with * defined by addition with the extra property that 1ⁿ = 0.  Notice that 1ⁿ means 1 + 1 + ... + 1 where the 1 appears n times.  Every cyclic group of order n can be thought as being equivalent to Z_n . 

Example

Let {a₁, a₂, ..., a_k} be elements of a group G.  Let H be the subgroup that consists of products of the form a₁^m1* a₂^m2 *...* a_k^mk where all of the powers are integers.  We say that H is the subgroup generated by a₁, a₂, ..., a_k. 

Definition

Let H be a subgroup of G and let a and b be elements of G.  Then we say that a is congruent to b mod H if a*b^-1 is in H.

We leave it as an exercise to show that congruency is an equivalency relationship.  That is the following three properties hold:

1. a is congruent to a
2. If a is congruent to b then b is congruent to a
3. If a is congruent to b and b is congruent to c then a is congruent to c.

Example

Let G be the integers consider the subgroup Z₄.  Then 7 is congruent to 3 mod Z₄ since 7 *(-3) = 7 - 3  = 4 is a multiple of 4.

Definition

If H is a subgroup of G and a is an element of G then Ha = {h*a | h is in H} is called a right coset of H in G.

Theorem

Ha = {x in G | a is congruent to x mod H}

Proof

Let h*a be in Ha.  Then

a*(h*a)^-1 = a*(a^-1*h^-1) = (a*a^-1)*h^-1 = e*h^-1 = h^-1

Since h is in H, so is h^-1.

Similarly if a is congruent to x mod H, then x is congruent to a.  That is

x*a^-1 = h

for some h in H.  Now just multiply both sides by a to get

(x*a^-1)*a = ha

or

x*(a^-1*a) = ha

x*e = ha

x = ha

This tells us that any two right cosets are either the same or disjoint (have no elements in common).  This is true since either two elements are equivalent or not.

Theorem

There is a one-to-one correspondence between any two right cosets of H in G.

Proof

What we need to prove is that there is a one-to-one function from Ha to Hb.  The function that works is

f(h*a) = h*b

Notice that if

h₁*b = h₂*b

then

h₁ = h₂

and then

h₁*a = h₂*a

Corollary

If G is a finite group, then any two cosets of H have the same number of elements.

We define the order of H (written o(H)) to be the number of elements of H.

Theorem (Lagrange)

If G is a finite group and H is a subgroup of G, then the o(H) is a divisor of G.

Exercises

1. Let G be the group of integers with * defined by addition and let H be the set of multiples of 10.  Prove that H is a subgroup of G.
2. Let G be the group of nonzero real numbers with * defined by multiplication and let H be the nonzero rational numbers.  Prove that H is a subgroup of G.
3. Let G be the group of all nonzero complex numbers a + bi with a and b real numbers not both 0.  Prove that
   H = {a + bi | a² + b² = 1}
   is a subgroup of G.
4. If H and K are both subgroups of G, prove that the intersection of H and K is also a subgroup of G.
5. Prove that if H is a subgroup of G, then aHa^-1 = { a*h*a^-1 | and h is in H} is a subgroup of G.
6. Let G be a group and a be in G.  Prove that N(a) = {n | n*a = a*n} is a subgroup of G.
7. Prove that congruency is an equivalency relationship.
8. If G is a finite group prove that if a is in G, then a^o(G) = e.
9. Prove that any subgroup of a cyclic group is itself a cyclic group.