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Chapter 3
Finite Groups and
Subgroups
In this chapter we focus on …nite groups, that is groups with a …nite number of
elements. We will derive some very important properties they have. But …rst,
we introduce some needed terminology. Let us also remind the reader that all
the de…nitions and results are given using the multiplicative notation and are to
be adapted to whatever operation the group has. For example, when we write
an , it means a a::: a with n factors if the operation is multiplication but it
means a + a + ::: + a with n terms if the operation is addition. The inverse of
an element a will be denoted a 1 , the identity element will be denoted e.
3.1
Order of a Group, Order of an Element
De…nition 151 (Order of a Group) Let G be a group. The order of G,
denoted jGj, is de…ned to be the number of elements G has. Note that this
de…nition also applies to groups which are not …nite.
Example 152 Z under addition has in…nite order.
Example 153 Zn under addition mod n has order n since Zn = f0; 1; 2; :::; n
1g.
Example 154 U (10) under multiplication mod 10 has order 4 since U (10) =
f1; 3; 7; 9g.
Example 155 If n is prime, then U (n) under multiplication mod n has order
n.
De…nition 156 (Order of an Element) Let G be a group (…nite or in…nite)
and g 2 G. The order of g, denoted jgj, is the smallest positive integer n such
that g n = e. If no such integer exists, we say that g has in…nite order.
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CHAPTER 3. FINITE GROUPS AND SUBGROUPS
Remark 157 Remember that if the operation is addition, g n = e becomes ng =
0.
To compute jgj, we compute the sequence of products g, g 2 , g 3 , ... until we
reach the identity for the …rst time.
Example 158 Consider U (10) = f1; 3; 7; 9g.with multiplication mod 10 Compute j3j. 32 mod 10 = 9. 33 mod 10 = 7, 34 mod 10 = 1 hence j3j = 4.
Example 159 Consider Z3 with addition mod 3. Then j1j = 3 since 1 6= 0,
12 mod 3 = (1 + 1) mod 3 = 2 mod 3 = 2, 13 mod 3 = (1 + 1 + 1) mod 3 = 0.
Example 160 Consider Z with addition. Then j1j = 1 since 1 6= 0, 1 + 1 6= 0,
1 + 1 + 1 6= 0, etc.
1
0
Example 161 Consider GL (2; R) then the order of
1
0
3.2
0
1
6=
1
0
0
1
and
1
0
0
1
2
=
1
0
0
1
0
1
is 2 since
.
Subgroups
Looking back at some examples we studied, some of the groups were subsets of
other groups. For example, GL (2; R) is a group under matrix multiplication.
SL (2; R) is also a group under matrix multiplication. SL (2; R) is also a subset
of GL (2; R) since it contains the elements of GL (2; R) with a determinant equal
to 1. As such, one might think that it might inherit certain of the properties of
the larger set. Indeed it does. It is what we study in this section.
De…nition 162 (Subgroup) Let G be a group and H a subset of G. If H is
a group under the operation of G then we say that H is a subgroup of G. We
will write H G to mean that H is a subgroup of G. We will write H < G to
mean that H is a proper subgroup of G.
Remark 163 It is important to note that the two sets must share the same
operation. Zn is a group with addition mod n. Z is a group with addition.
Zn Z. Yet, (Zn ; +) is not a subgroup of (Z; +) since the operation is not the
same. The …rst is addition mod n, the second is addition.
Example 164 If G is a group, then feg is a subgroup of G. It is called the
trivial subgroup of G.
Before looking at examples, we give a test to see if a subset is also a subgroup.
As noted earlier, if G is a group and H G, we might expect H to inherit some
of the properties of G. Indeed it does. Associativity will be inherited. Also,
since every element of H is also in G, it will have an inverse. The problem
is that the inverse might not be in H. The identity of g might not be in H.
3.2. SUBGROUPS
49
H might not be closed under the operation of G:So, being a subset of a group
is clearly not a guarantee to be a group even if some properties of G will be
inherited. The following theorem is a test to determine if a subset of a group G
is itself a group, that is a subgroup of G.
Theorem 165 (Subgroup Test) Let G be a group and H a nonempty subset
of G. H is a subgroup of G if and only if the following two conditions are
satis…ed:
1. For all a, b in H, ab is also in H that is H is closed under the operation
of G.
2. For all a in H, a
1
is also in H that is H is closed under inverses.
Proof. We prove both directions.
=) Suppose that H is a subgroup of G. Then the …rst condition is
satis…ed, it is part of the de…nition of a group. H is also closed
under inverses. Every element of H will have an inverse in H since
H is a group. Every element of H is also an element of G and as such
will also have an inverse. Since inverses are unique, these inverses
will be equal. In other words, a 1 is the inverse of a in H as well as
in G.
(= Assume both properties hold. We must show H is a group, that is
check the four conditions of a group are satis…ed.
Closure. This is property 1).
Associativity. It is inherited from G
Identity. G being a group has an inverse e. We show e 2 H. Since
H 6= ?, pick x 2 H. By property 2), x 1 2 H. By property 1),
xx 1 = e 2 H.
Inverse. This is property 2).
Let us summarize how to prove or disprove a subset H of a group G is a
subgroup of G.
To prove a nonempty subset H of a group G is a subgroup of G, do all the
following:
1. Identify the property P which distinguishes the elements of H from those
of G.
2. Show e satis…es property P , this shows H 6= ?.
3. Assume a and b have property P and show that ab has property P .
4. Assume a has property P and show a
1
has property P .
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CHAPTER 3. FINITE GROUPS AND SUBGROUPS
To prove a nonempty subset H of a group G is not a subgroup of G, do one
of the following:
1. Show e 2
= H,
2. Or …nd an element a in H for which a
1
is not in H,
3. Or …nd two elements a and b of H for which ab is not in H.
We look at some examples.
Example 166 SL (2; R) is a subgroup of GL (2; R) under matrix multiplication.
The property which characterizes the elements of SL (2; R) is that their determinant is 1.
SL (2; R) 6= ? since the identity,
1
0
0
1
belongs to SL (2; R).
SL (2; R) is closed under matrix multiplication. If A and B are elements of SL (2; R) then AB also belongs to SL (2; R) since det (AB) =
det (A) det (B).
SL (2; R) is closed under inverses since det A
1
=
1
.
det (A)
Example 167 (R; +) is a group. (Z; +) is a subgroup of (R; +). It is clearly
not empty, closed under addition and inverses.
Example 168 Let G be the group of nonzero real numbers under multiplication.
Let H = fx 2 G j x = 1 or xpis irrationalg.
p Hpis not a subgroup of G, it is not
closed under multiplication. 2 2 H but 2
2=22
= H.
In the case H is a …nite subset of a group G, there is an easier subgroup
test.
Theorem 169 (Finite Subgroup Test) Let H be a nonempty, …nite subset
of a group G. H is a subgroup of G if and only if H is closed under the operation
of G.
Proof. We prove both directions.
=) Assume H is a subgroup of G. Then, H is itself a group under the operation of G, hence closed under the operation of G.
(= Suppose that H is closed under the operation of G. By theorem 165, it
is enough to show that a 1 2 H whenever a 2 H. Let a 2 H, we prove
a 1 2 H. If a = e, then a 1 = a = e 2 H. If a 6= e, since H is closed
under the operation of G, the sequence of elements a, a2 , a3 , ::: belongs
to H. Since H is …nite, not all these elements are distinct. There exists
two integers i > j such that ai = aj . That is ai j = e. Since a 6= e,
i j > 1. Therefore e = ai j = aai j 1 . Therefore, a 1 = ai j 1 and
since i j > 1, i j 1 1 and hence ai j 1 2 H.
3.3. IMPORTANT EXAMPLES
3.3
51
Important Examples
3.3.1
Subgroup Generated by an Element of a Group
We now look at some special subgroups and introduce new de…nitions in the
process.
De…nition 170 (Subgroup Generated by an Element) Let G be a group
and a 2 G. The subgroup generated by a, denoted hai is de…ned to be
hai = ak j k 2 Z
In other words, hai is the set consisting of all the powers of a for positive,
negative integers as well as 0 where a0 = e. Remember that ak means ka if
the operation is addition. We call it the subgroup generated by a because it is
indeed a subgroup. We are going to prove it.
De…nition 171 (Cyclic Group) A group G is called cyclic if there exists a 2
G such that G = hai. a is called a generator of G.
We will study cyclic groups in the next chapter in more details.
Theorem 172 Let G be a group and a 2 G. hai is a subgroup of G.
Proof. Clearly a 2 hai so that hai 6= ?. We need to verify hai is closed under
the operation of G and inverses.
Closure under the operation of G: Let an and am be two elements of hai.
Then, an am = an+m 2 hai.
Closure under inverses: Let an 2 hai. Then, (an )
1
=a
n
2 hai.
Remark 173 Though the list :::a 2 , a 1 , a0 , a1 , a2 , ::: has in…nitely many
entries, hai may have a …nite number of elements.
Example 174 U (10) is cyclic. h3i = f3; 9; 7; 1g = U (10) since 31 mod 10 = 3,
32 mod 10 = 9, 33 mod 10 = 7 and 34 mod 10 = 1.
Example 175 In Z10 , h2i = f2; 4; 6; 8; 0g since 21 = 2, 22 = (2 + 2) mod 10 =
4, 23 = (2 + 2 + 2) mod 10 = 6, 24 = (2 + 2 + 2 + 2) mod 10 = 8, 25 = (2 + 2 + 2 + 2 + 2) mod 10 =
0.
Example 176 Z with addition is cyclic. In fact, Z = h1i. Recall that with
addition an = na. 10 = 0 1 = 0, 11 = 1, 12 = 2 1 = 2, 1 2 = ( 2) 1 = 2, :::
Example 177 Z = h 1i. So, a cyclic group can have several generators.
We list some important properties regarding the order of an element which
will be proven in the problems.
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CHAPTER 3. FINITE GROUPS AND SUBGROUPS
Proposition 178 Let G be a group and a 2 G. The following is true:
1. jaj
jGj
2. If ak = e and jaj = n then n j k.
3. If jaj = n and d is a positive divisor of n, then ad =
n
.
d
4. If a has in…nite order then ai = aj () i = j
5. If jaj = n then ai = aj with i > j () n j (i
3.3.2
j)
Center of a Group
De…nition 179 (Center of a Group) Let G be a group. The center of G,
denoted Z (G), is the set of elements of G that commute with every element of
G in other words,
Z (G) = fa 2 G j ax = xa for all x 2 Gg
Remark 180 Z (G) 6= ? since e 2 Z (G) since ex = xe = x for any x 2 G.
Theorem 181 Let G be a group. Z (G) is a subgroup of G.
Proof. By de…nition Z (G) is a subset of G. Clearly, e 2 Z (G) since ex = xe
for all x 2 G therefore Z (G) 6= ?. We need to show that Z (G) is closed under
the operation of G and under inverses.
Closure under the operation of G: Let a and b be elements of Z (G). Show
that ab 2 Z (G) that is (ab) x = x (ab) for all x 2 G.
(ab) x
= a (bx) (associativity)
= a (xb) (b 2 Z (G) )
=
(ax) b (associativity)
=
(xa) b (a 2 Z (G) )
= x (ab) (associativity)
Closure under inverses: Let a 2 Z (G), we need to show a 1 2 Z (G) that
is a 1 x = xa 1 for all x 2 G. Since a 2 Z (G), we know that ax = xa.
ax
=
()
()
()
1
xa () a
a
1
a
e xa
xa
1
xa
1
(ax) a
1
= a
=a
1
= a
1
x e
x
Remark 182 We make a few additional remarks.
1
1
=a
x
1
aa
(xa) a
1
1
3.3. IMPORTANT EXAMPLES
53
1. Clearly, if G is Abelian then Z (G) = G.
2. Z (G) is Abelian.
Example 183 The center of GL (2; R) is = f I2 j
2 Rg. See problems.
Example 184 Z (D4 ) = fR0 ; R180 ; g
3.3.3
Centralizer of a Group
De…nition 185 (Centralizer of an Element) Let G be a group and a 2 G.
The centralizer of a in G, denoted C (a) is the set of elements of G which
commute with a in other words
C (a) = fg 2 G j ag = gag
Theorem 186 Let G be a group and a 2 G. C (a) is a subgroup of G.
Proof. See problems.
Example 187 Find C (H) in D4 . From the table, we see that C (H) = fR0 ; R180 ; H; V g.
1 1
1 1
By de…nition, we are looking for matrices in GL (2; R) which commute with
1 1
that is which satisfy
1 1
Example 188 Let G = GL (2; R) with matrix multiplication. Find C
a
c
and ad
b
d
1
1
1
1
=
1
1
1
1
a
c
=
a+c b+d
a+c b+d
b
d
bc 6= 0. This is equivalent to
a+b
c+d
This gives us the system
a+b
c+d
8
a+b=a+c
>
>
<
a+b=b+d
c
+d=a+c
>
>
:
c+d=b+d
Which is equivalent to
b=c
a=d
Hence
C
1
1
1
1
=
a
b
b
a
j a2
b2 6= 0
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CHAPTER 3. FINITE GROUPS AND SUBGROUPS
3.4
Problems
1. Prove that every cyclic group is Abelian.
2. If H and K are subgroups of a group G, prove that H \ K is also a
subgroup of G. What can be said about H [ K?
3. Prove theorem 186.
4. Prove that if G is a group and a 2 G then Z (G)
C (a).
5. Let G be a group. Prove that G is Abelian if and only if C (a) = G for
every a 2 G.
6. Find Z (G) for G = GL (2; R).
7. Do # 1, 2, 3, 4, 5, 6, 7, 8, 9, 18, 19, 20, 21, 23, 26, 35, 36, 45, 55 on pages
64 - 69