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

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§ 3.3 Constructing Examples

 
Definition 3.3.1. Let G be a group, and let S and T be subsets of G. Then

ST = { x in G | x = st   for some   s in S, t in T } .

 
Proposition 3.3.2. Let G be a group, and let H and K be subgroups of G.
If h-1kh is in K for all h in H and k in K, then HK is a subgroup of G.

 
Definition 3.3.3. Let G1 and G2 be groups. The set of all ordered pairs (x1,x2) such that x1 is in G1 and x2 is in G2 is called the direct product of G1 and G2, denoted by

G1 × G2.

 
Proposition 3.3.4. Let G1 and G2 be groups.

(a) The direct product G1 × G2 is a group under the multiplication defined for all
(a1,a2), (b1,b2) in G1 × G2 by

(a1,a2) (b1,b2) = (a1b1,a2b2).

(b) If the elements a1 in G1 and a2 in G2 have orders n and m, respectively, then in
G1 × G2 the element (a1,a2) has order lcm[n,m].

 
Example 3.3.3. The group Z2 × Z2 is called the Klein four-group.

 
Definition 3.3.5. Let F be a set with two binary operations + and · with respective identity elements 0 and 1, where 1 is distinct from 0. Then F is called a field if

(i) the set of all elements of F is an abelian group under +;

(ii) the set of all nonzero elements of F is an abelian group under ·;

(iii) a · (b+c) = a · b + a · c for all a,b,c in F.

 
Definition 3.3.6. Let F be a field. The set of all invertible n × n matrices with entries in F is called the general linear group of degree n over F, and is denoted by GLn(F).

 
Proposition 3.3.7. Let F be a field. Then GLn(F) is a group under matrix multiplication.

 
Example. 3.3.7. (The Quaternion group Q)
Consider the following set of invertible 2 × 2 matrices with entries in the field of complex numbers.

± ,   ± ,   ± ,   ± .

If we let

1 = ,   i = ,   j = ,   k =

then we have the identities

i2 = j2 = k2 = -1;

ij = k,   jk = i,   ki = j;

ji = -k,   kj = -i,   ik = -j.

These elements form a nonabelian group Q of order 8 called the quaternion group, or group of quaternion units.



§ 3.3 Constructing Examples: Solved problems

The most important result in this section is Proposition 3.3.7, which shows that the set of all invertible n × n matrices forms a group, in which we can allow the entries in the matrix to come from any field. This includes matrices with entries in the field Zp, for any prime number p, and this allows us to construct very interesting finite groups as subgroups of GLn (Zp).

The second construction in this section is the direct product, which takes two known groups and constructs a new one, using ordered pairs. This can be extended to n-tuples, where the entry in the ith component comes from a group Gi, and n-tuples are multiplied component-by-component. This generalizes the construction of n-dimensional vector spaces (that case is much simpler since every entry comes from the same set).


16. Show that Z5×Z3 is a cyclic group, and list all of the generators for the group.     Solution

17. Find the order of the element ([9]12, [15]18) in the group Z12×Z18.     Solution

18. Find two groups G1 and G2 whose direct product G1×G2 has a subgroup that is not of the form H1×H2, for nontrivial subgroups H1 G1 and H2 G2.     Solution

19. In the group G = Z36×, let

H = { [x] | x 1 (mod 4) }   and   K = { [y] | y 1 (mod 9) }.

Show that H and K are subgroups of G, and find the subgroup HK.     Solution

20. Show that if p is a prime number, then the order of the general linear group GLn(Zp) is

(pn -1)(pn - p) · · · (pn - pn-1).

Solution

21. Find the order of the element A = in the group GL3 (C).     Solution

22. Compute the centralizer in GL2 ( Z3 ) of the matrix .     Solution

23. Compute the centralizer in GL2 ( Z3 ) of the matrix .     Solution

24. Let H be the following subset of the group G = GL2 (Z5).

H = .

    (a) Show that H is a subgroup of G with 10 elements.
    (b) Show that if we let A = and B = , then BA = A-1B.
    (c) Show that every element of H can be written uniquely in the form Ai Bj, where i = 0,1,2,3 or 4 and j = 0 or 1.
    Solution


§ 3.3 Lab questions

To answer the experimental questions, use the Groups15 applet written by John Wavrik of UCSD.

The goal of this lab is to find some conditions under which we can see that a given group is actually put together from smaller subgroups. Before stating the first problem, we need to give a definition. (It appears later in the text, as Definition 3.7.5.)

Definition. Let G be a group, and let H be a subgroup of G. The subgroup H is called a normal subgroup of G if for any element x in H and any element g in G, the product gxg-1 belongs to H.

Lab 1. Show that every subgroup of an abelian group is a normal subgroup.     Solution

Lab 2. Let G1 and G2 be groups, and let G be the direct product G1 × G2.
Let H = { (x1,x2) in G1 × G2 | x2 = e }
and K = { (x1,x2) in G1 × G2 | x1 = e }.
Prove that H and K are normal subgroups of G.     Solution

Let G1 and G2 be groups, and let G be the direct product G1 × G2.
The result above (Lab 2) shows that if H = { (x1,e) in G1 × G2 } and K = { (e,x2) in G1 × G2 }, then H and K are normal subgroups. Exercise 3.3.9 in the text shows that H and K are subgroups for which HK = G and HK = {e}.

Sometimes G = G1 × G2 is called the external direct product of G1 and G2, to distinguish it from the construction given in the next definition.

Definition. The group G is called the internal direct product of a normal subgroup H and and a normal subgroup K if its has normal subgroups H and K for which HK = G and HK = {e}.

This raises the question of finding an easy way to calculate the product HK of two subgroups H and K. The multiplication table for the symmetric group

S3 = { e, a, a2, b, ab, a2b }

is given in Table 3.3.3 of the text (see page 104 of the text or the online table).
If we let H = { e, a, a2 } and K = { e, b }, then to compute HK we can compute the products
    He = { ee, ae, a2e } = H,
    Hb = { eb, ab, a2b } = { b, ab, a2b }.
Thus we can see that HK = He Hb = G.
We could also find HK by computing
    eK = { ee, eb } = { e, b },
    aK = { ae, ab } = { a, ab },
    a2K = { a2e, a2b } = { a2, a2b }.
This again shows that HK = G.

The next definition appears later in the text, as Definition 3.8.2.

Definition. Let H be a subgroup of the group G, and let a be an element of G. The set

aH = { x in G | x = ah for some h in H }

is called the left coset of H in G determined by a. Similarly, the right coset of H in G determined by a is the set

Ha = { x in G | x = ha for some h in H } .

When you use Groups15, if you ask to list the subgroups of a group, you can highlight a subgroup H and ask to compute the cosets of H. You will notice that unless H is normal (shown by an N in front of the subgroup) the left cosets of H will differ from the right cosets. The set aH is the left coset that contains a, while the set Ha is the right coset that contains a. Remember that the point of this discussion of left and right cosets is to help you to use Groups15 to compute the product of two subgroups.

Lab 3. Using the notation of Groups15, show that Z10 is the internal direct product of the subgroups H = {A,F} and K = {A,C,E,G,I}.

Lab 4. Using the notation of Groups15, show that the group D6 (of order 12) is the internal direct product of the normal subgroups {A,D} and {A,C,E,H,J,L}. On the other hand, explain why the groups D5 and D7 cannot be written as the internal direct product of proper normal subgroups.

Lab 5. Using the notation of Groups15, explain why none of these groups can be written as the internal direct product of proper normal subgroups:
    Q (of order 8),
    A4 (of order 12), and
    Z3 Z4 (of order 12).

Sometimes, even though a group cannot be expressed as an internal direct product of proper normal subgroups, it is almost possible to do this, in the sense of the following definition.

Definition. The group G is called the internal semidirect product of a normal subgroup H and and a subgroup K if its has a normal subgroup H and a subgroup K for which HK = G and HK = {e}.

Lab 6. Using the notation of Groups15, show that the group D4 (of order 8) is the internal semidirect product of the normal subgroup {A,B,C,D} and the subgroup {A,E}. Show that D4 is also the internal semidirect product of the normal subgroup {A,C,E,G} and the subgroup {A,F}. List all of the other ways in which D4 can be expressed as an internal semidirect product.

Lab 7. Using the notation of Groups15, show that the group D5 (of order 10) is the internal semidirect product of the normal subgroup {A,B,C,D,E} and the subgroup {A,F}. List all of the other ways in which D5 can be expressed as an internal semidirect product.

Lab 8. Using the notation of Groups15, show that the group D6 (of order 12) is the internal semidirect product of the normal subgroup {A,B,C,D,E,F} and the subgroup {A,G}. Show that D6 is also the internal semidirect product of the normal subgroup {A,C,E} and the subgroup {A,D,G,J}. List all of the other ways in which D6 can be expressed as an internal semidirect product.

Lab 9. Using the notation of Groups15, show that the group A4 (of order 12) is the internal semidirect product of the normal subgroup {A,D,I,K} and the subgroup {A,B,C}. List all of the other ways in which A4 can be expressed as an internal semidirect product.

Lab 10. Using the notation of Groups15, show that the group Z3 Z4 (of order 12) is the internal semidirect product of the normal subgroup {A,C,E} and the subgroup {A,D,G,J}. List all of the other ways in which Z3 Z4 can be expressed as an internal semidirect product.

 


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