Excerpted from Beachy/Blair, Abstract Algebra, 2nd Ed. © 1996

## § 3.2 Subgroups

Definition 3.2.1. Let G be a group, and let H be a subset of G. Then H is called a subgroup of G if H is itself a group, under the operation induced by G.

Example 3.2.1. Q× and R× are subgroups of C×, the multiplicative group of complex numbers.

Example 3.2.2. SLn(R), the set of all n × n matrices over R with determinant 1, is a subgroup of GLn(R).

Proposition 3.2.2. Let G be a group with identity element e, and let H be a subset of G. Then H is a subgroup of G if and only if the following conditions hold:

(i) for all a,b in H, the product ab is in H;

(ii) e belongs to H;

(iii) for all a in H, the inverse a-1 is in H.

Corollary 3.2.3. Let G be a group and let H be a subset of G. Then H is a subgroup of G if and only if H is nonempty and ab-1 is in H for all a,b in H.

Corollary 3.2.4. Let G be a group, and let H be a finite, nonempty subset of G. Then H is a subgroup of G if and only if ab is in H for all a,b in H.

Definition 3.2.5. Let G be a group, and let a be any element of G. The set

<a> = { x in G | x = an   for some   n in Z }

is called the cyclic subgroup generated by a.
The group G is called a cyclic group if there exists an element a in G such that G=<a>. In this case a is called a generator of G.

Proposition 3.2.6. Let G be a group, and let a be an element of G.

(a) The set <a> is a subgroup of G.

(b) If K is any subgroup of G such that a is an element of K, then <a> K.

Examples 3.2.7 - 3.2.9. Z and Zn are cyclic groups, but Zn× may not be cyclic. (See Corollary 7.5.11.)

Definition 3.2.7. Let a be an element of the group G. If there exists a positive integer n such that an = e, then a is said to have finite order, and the smallest such positive integer is called the order of a, denoted by o(a).
If there does not exist a positive integer n such that an = e, then a is said to have infinite order.

Proposition 3.2.8. Let a be an element of the group G.

(a) If a has infinite order, and ak = am for integers k,m, them k=m.

(b) If a has finite order and k is any integer, then ak = e if and only if o(a) | k.

(c) If a has finite order o(a)=n, then for all integers k, m, we have

ak = am   if and only if   k m (mod n).

Furthermore, | <a> | = o(a).

Lemma 3.2.9. Let H be a subgroup of the group G. For a,b in G define a ~ b if ab-1 is in H. Then ~ is an equivalence relation.

Theorem 3.2.10 (Lagrange). If H is a subgroup of the finite group G, then the order of H is a divisor of the order of G.

Corollary 3.2.11. Let G be a finite group of order n.

(a) For any a in G, o(a) is a divisor of n.

(b) For any a in G, an = e.

Example 3.2.12. (Euler's theorem) Let G be the multiplicative group of congruence classes modulo n. The order of G is given by (n), and so by Corollary 3.2.11, raising any congruence class to the power (n) must give the identity element.

Corollary 3.2.12. Any group of prime order is cyclic.

## § 3.2 Subgroups: Solved problems

If the idea of a subgroup reminds you of studying subspaces in your linear algebra course, you are right. If you only look at the operation of addition in a vector space, it forms an abelian group, and any subspace is automatically a subgroup. Now might be a good time to pick up your linear algebra text and review vector spaces and subspaces.

Lagrange's theorem is very important. It tells us that in a finite group the number of elements in any subgroup must be a divisor of the total number of elements in the group. This is a useful fact to know when you are looking for subgroups in a given group.

It is also important to remember that every element a in a group G defines a subgroup < a > consisting of all powers (positive and negative) of the element. This subgroup has o(a) elements, where o(a) is the order of a. If the group is finite, then you only need to look at positive powers, since in that case the inverse a-1 of any element can be expressed in the form an, for some n > 0.

For a group G, the centralizer C(a) of an element a in G is defined in Exercise 3.2.14 of the text as

C(a) = { x in G | xa = ax } .

Exercise 3.2.14 shows that C(a) is a subgroup of G that contains <a>.

The center of the group G, denoted by Z(G), is defined in Exercise 3.2.16 as

Z(G) = { x in G | xg = gx for all g in G } .

Exercise 3.2.16 shows that Z(G) is a subgroup of G.

23. Find all cyclic subgroups of Z24×.     Solution

24. In Z20×, find two subgroups of order 4, one that is cyclic and one that is not cyclic.     Solution

25. (a) Find the cyclic subgroup of S7 generated by the element (1,2,3)(5,7).
(b) Find a subgroup of S7 that contains 12 elements. You do not have to list all of the elements if you can explain why there must be 12, and why they must form a subgroup.
Solution

26. In G = Z21×, show that

H = { [x]21 | x 1 (mod 3) }   and   K = { [x]21 | x 1 (mod 7) }

are subgroups of G.     Solution

27. Let G be an abelian group, and let n be a fixed positive integer. Show that

N = { g in G | g = an   for some   a in G }

is a subgroup of G.     Solution

28. Suppose that p is a prime number of the form p = 2n + 1.
(a) Show that in Zp× the order of [2]p is 2n.
(b) Use part (a) to prove that n must be a power of 2.
Solution

29. In the multiplicative group C× of complex numbers, find the order of the elements

=   and   = .

Solution

30. Let K be the following subset of GL2 (R).

K =

Show that K is a subgroup of GL2 (R).     Solution

31. Compute the centralizer in GL2 ( R) of the matrix .     Solution

32. Let G be the subgroup of GL2 (R) defined by G = .
Let A = and B = .
Find the centralizers C(A) and C(B), and show that C(A) C(B) = Z(G), where Z(G) is the center of G.     Solution

## § 3.2 Lab questions

To answer these questions experimentally, you can use the Groups15 Java applet written by John Wavrik of UCSD.

Lab 1. For each group G of order 8 on the list of groups given by Groups15, find the elements of order 2. Do these elements (together with the identity element) form a subgroup of G?

Lab 2. For each group G of order n < 16, find the largest possible order of an element a in G.

Lab 3. Find all groups G of order n < 16 for which there is a divisor m of n, but no corresponding subgroup of G of order m. (These are the groups of order 15 or less for which the converse of Lagrange's theorem fails.)

Lab 4. If |G| = n, where n < 16, and p is a prime divisor of n, can you always find a subgroup H of G with |H| = p? What can you say if you ask the question for prime power? (You only need to worry about n = 8 and n = 12.)

Lab 5. List the groups G in Groups15 for which the center

Z(G) = { x in G | xg = gx for all g in G }

contains only the identity element.

Lab 6. For a group G, the centralizer of an element a in G is defined as

C(a) = { x in G | xa = ax } .

In the group of order 12 called A4 in Groups15, find the centralizers C(B) and C(D) of the elements listed as B and D.

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