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**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^{-1}kh is in K for all h in H and k in K,
then HK is a subgroup of G.

**Definition 3.3.3.**
Let G_{1} and G_{2} be groups.
The set of all ordered pairs
(x_{1},x_{2})
such that x_{1} is in G_{1}
and x_{2} is in G_{2} is called the
**direct product**
of G_{1} and G_{2}, denoted by

G_{1} × G_{2}.

**Proposition 3.3.4.**
Let G_{1} and G_{2} be groups.

**(a)**The direct product G_{1}× G_{2}is a group under the multiplication defined for all

(a_{1},a_{2}), (b_{1},b_{2}) in G_{1}× G_{2}by(a

_{1},a_{2}) (b_{1},b_{2}) = (a_{1}b_{1},a_{2}b_{2}).**(b)**If the elements a_{1}in G_{1}and a_{2}in G_{2}have orders n and m, respectively, then in

G_{1}× G_{2}the element (a_{1},a_{2}) has order lcm[n,m].

**Example** 3.3.3.
The group **Z**_{2} × **Z**_{2}
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 GL_{n}(F).

**Proposition 3.3.7.**
Let F be a field.
Then GL_{n}(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** =

**i**^{2} =
**j**^{2} =
**k**^{2} = -**1**;

**i****j** = **k**,
**j****k** = **i**,
**k****i** = **j**;

**j****i** = -**k**,
**k****j** = -**i**,
**i****k** = -**j**.

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 *i*th component comes from a group G_{i},
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 **Z**_{5}×**Z**_{3} 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 **Z**_{12}×**Z**_{18}.
*Solution*

**18.**
Find two groups G_{1} and G_{2}
whose direct product G_{1}×G_{2}
has a subgroup that is not of the form H_{1}×H_{2},
for nontrivial subgroups
H_{1} G_{1} and
H_{2} G_{2}.
*Solution*

**19.**
In the group G = **Z**_{36}^{×},
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.
**20.**
Show that if p is a prime number,
then the order of the general linear group
GL_{n}(**Z**_{p}) is

(p^{n} -1)(p^{n} - p) · · ·
(p^{n} - p^{n-1}).

**21.**
Find the order of the element
A =
in the group GL_{3} (**C**).
*Solution*

**22.**
Compute the centralizer
in GL_{2} ( **Z**_{3} ) of the matrix
.
*Solution*

**23.**
Compute the centralizer
in GL_{2} ( **Z**_{3} ) of the matrix
.
*Solution*

**24.**
Let H be the following subset of the group
G = GL_{2} (**Z**_{5}).

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

(c) Show that every element of H can be written uniquely in the form A

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 G_{1} and G_{2} be groups,
and let G be the direct product G_{1} × G_{2}.

Let H = { (x_{1},x_{2}) in G_{1} × G_{2}
| x_{2} = e }

and
K = { (x_{1},x_{2}) in G_{1} × G_{2}
| x_{1} = e }.

Prove that H and K are normal subgroups of G.
*Solution*

Let G_{1} and G_{2} be groups,
and let G be the direct product G_{1} × G_{2}.

The result above (Lab 2) shows that if
H = { (x_{1},e) in G_{1} × G_{2} } and
K = { (e,x_{2}) in G_{1} × G_{2} },
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 = G_{1} × G_{2}
is called the **external** direct product of
G_{1} and G_{2},
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

S_{3} = { e, a, a^{2}, b, ab, a^{2}b }

If we let H = { e, a, a

He = { ee, ae, a

Hb = { eb, ab, a

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 },

a

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

When you use
**Lab 3.**
Using the notation of
** Groups15**,
show that Z

**Lab 4.**
Using the notation of
** Groups15**,
show that the group D

**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),

A

Z

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 D

**Lab 7.**
Using the notation of
** Groups15**,
show that the group D

**Lab 8.**
Using the notation of
** Groups15**,
show that the group D

**Lab 9.**
Using the notation of
** Groups15**,
show that the group A

**Lab 10.**
Using the notation of
** Groups15**,
show that the group
Z

Forward to §3.4 | Back to §3.2 | Up | Table of Contents

Forward to §3.4 | Back to §3.2 | Up | Table of Contents