CHAPTER 9
Normal Subgroups and Factor Groups
Normal Subgroups
If H  G, we have seen situations where aH 6= Ha 8 a 2 G.
Definition (Normal Subgroup).
A subgroup H of a group G is a normal subgroup of G if aH = Ha 8 a 2 G.
We denote this by H C G.
Note. This means that if H C G, given a 2 G and h 2 H, 9 h0, h00 2 H
3 ah = h0a and ah00 = ha. and conversely. It does not mean ah = ha for all
h 2 H.
Recall (Part 8 of Lemma on Properties of Cosets).
aH = Ha () H = aHa 1.
Theorem (9.1 — Normal Subgroup Test).
If H  G, H C G () xHx
1
✓ H for all x 2 G.
Proof.
(=)) H C G =) 8 x 2 G and 8 h 2 H, 9 h0 2 H 3 xh = h0x =)
xhx 1 = h0 2 H. Thus xHx 1 ✓ H.
((=) Suppose xHx 1 ✓ H 8 x 2 G. Let x = a. Then aHa 1 ✓ H =)
aH ✓ Ha. Now let x = a 1. Then a 1H(a 1) 1 = a 1Ha ✓ H =) Ha ✓
aH.
By mutual inclusion, Ha = aH =) H C G.
117
⇤
118
9. NORMAL SUBGROUPS AND FACTOR GROUPS
Example.
(1) Every subgroup of an Abelian group is normal since ah = ha for all a 2 G
and for all h 2 H.
(2) The center Z(G) of a group is always normal since ah = ha for all a 2 G
and for all h 2 Z(G).
Theorem (4). If H  G and [G : H] = 2, then H C G.
Proof.
If a 2 H, then H = aH = Ha. If a 62 H, aH is a left coset distinct from H and
Ha is a right coset distinct from H. Since [G : H] = 2, G = H [aH = H [Ha
and H \ aH = ; = H \ Ha =) aH = Ha. Thus H C G.
⇤
Example. An C Sn since [Sn : An] = 2.
Note, for example, that for (1 2) 2 Sn and (1 2 3) 2 An,
(1 2)(1 2 3) 6= (1 2 3)(1 2),
but
(1 2)(1 2 3) = (1 3 2)(1 2)
and (1 3 2) 2 An.
Example. hR360/ni C Dn since [Dn : R360/n] = 2.
Example. SL(2, R) C GL(2, R).
Proof.
1
= (det(x)) 1. Then, for all
det(x)
Let x 2 GL(2, R). Recall det(x 1) =
h 2 SL(2, R),
det(xhx 1) = (det(x))(det(h))(det(x))
so xhx
1
2 SL(2, R) =) x SL(2, R)x
1
=
(det(x))(det(x)) 1(det(h)) = 1 · 1 = 1,
1
✓ SL(2, R). Thus SL(2, R) C GL(2, R).
⇤
9. NORMAL SUBGROUPS AND FACTOR GROUPS
119
Example. Consider A4, with group table from page 111 shown below:
Let H = {↵1, ↵2, ↵3, ↵4}  A4, A = {↵5, ↵6, ↵7, ↵8} ✓ A4,
and B = {↵9, ↵10, ↵11, ↵12}. 8 a 2 A, a
1
2 B, and 8 b 2 B, b
Case 1: x 2 H. Then xH ✓ H. Since x
1
2 H, xHx
Let x 2 A4:
Case 2: x 2 A. Then xH ✓ A =) xHx
Case 3: x 2 B. Then xH ✓ B =) xHx
Thus, 8 x 2 A4, xHx
1
1
1
✓ H.
✓ H.
✓ H.
✓ H =) H C A4 by Theorem 9.1.
Now let K = {↵1, ↵5, ↵9}  A4. Now ↵5 2 K, but
so K 6C A4.
1
↵2↵5↵2 1 = ↵2↵5↵2 = ↵2↵8 = ↵7 62 K,
1
2 A.
120
9. NORMAL SUBGROUPS AND FACTOR GROUPS
Factor Groups
Theorem (9.2 — Factor Groups). Let G be a group and H C G. The
set G/H = {aH|a 2 G} is a group under the operation (aH)(bH) = abH.
This group is called the factor group or quotient group of G by H.
Proof.
We first show the operation is well-defined. [Our product is determined by the
coset representatives chosen, but is the product uniquely determined by the
cosets themselves?]
Suppose aH = a0H and bH = b0H. Then a0 = ah1 and b0 = bh2 for some
h2, h2 2 H =) a0b0H = ah1bh2H = ah1bH = ah1Hb = aHb = abH by
associativity in G, the Lemma on cosets (page 145), and the fact that H C G.
Thus the operation is well-defined.
Associativity in G/H follows directly from associativity in G:
(aHbH)cH = (abH)cH = (ab)cH = a(bc)H = aH(bcH) = aH(bHcH).
The identity is eH = H, and the inverse of aH is a 1H.
Thus G/H is a group.
⇤
9. NORMAL SUBGROUPS AND FACTOR GROUPS
121
Example. Let 3Z = {0, ±3, ±6, ±9, . . . }. Then 3Z C Z since Z is abelian.
Consider the following 3 cosets:
0 + 3Z = 3Z = {0, ±3, ±6, ±9, . . . },
1 + 3Z = {1, 4, 7, . . . ; 2, 5, 8, . . . },
2 + 3Z = {2, 5, 8 . . . ; 1, 4, 7, . . . }.
For k 2 Z, k = 3q + r where 0  r < 3. Thus
k + 3Z = r + 3q + 3Z = r + 3Z.
So we have all the cosets. A Cayley table for Z/3Z:
0 + 3Z
0 + 3Z 0 + 3Z
1 + 3Z 1 + 3Z
2 + 3Z 2 + 3Z
Therefore, Z/3Z ⇡ Z3.
1 + 3Z
1 + 3Z
2 + 3Z
0 + 3Z
2 + 3Z
2 + 3Z
0 + 3Z
1 + 3Z
In general, for n > 0, nZ = {0, ±n, ±2n, ±3n, . . . }, and Z/nZ ⇡ Zn.
122
9. NORMAL SUBGROUPS AND FACTOR GROUPS
Example. Consider the multiplication table for A4 below, where i represents the permutation ↵i on page 117 of these notes.
Let H = {1, 2, 3, 4}. The 3 cosets of H are H, 5H = {5, 6, 7, 8}, and 9H =
{9, 10, 11, 12}. Notice how the above table is divided into coset blocks. Since
H C A4, when we replace the various boxes by their coset names, we get the
Cayley table below for A4/H.
The factor group collapses all the elements of a coset to a single group element
of A4/H.
When H C G, one can always arrange a Cayley table so this happens. When
H 6C G, one cannot.
9. NORMAL SUBGROUPS AND FACTOR GROUPS
Example. Is U (30)/U5(30) isomorphic to Z2
Z4?
123
Z2 (the Klein 4-group) or
Solution.
U (30) = {1, 7, 11, 13, 17, 19, 23, 29}, U5(30) = {1, 11}. 7U5(30) = {7, 17},
13U5(30) = {13, 23}, 19U5(30) = {19, 29}.
Thus U (30/U5(30) = {U5(30), 7U5(30), 13U5(30), 19U5(30)}.
(7U5(30))2 = 19U5(30), so |7U5((30)| 6= 2 =) |7U5((30)| = 4 =)
U (30)/U5(30) ⇡ Z4.
⇤
Note. |aH| has two possible interpretations:
(1) The order of aH as an element of G/H.
(2) The size of the set aH.
The appropriate interpretation will be clear from the context.
Note. When we take a group and factor out by a normal subgroup H, we
are essentially defining every element in H to be the identity.
In the example above, 7U5(30) = 17U5(30) since 17 = 7 · 11 in U (30) and going
to the factor group makes 11 the identity.
124
9. NORMAL SUBGROUPS AND FACTOR GROUPS
Problem (Page 201 # 25). Let G = U (32) and H = {1, 31}. Which of
Z8, Z4 Z2, or Z2 Z2 Z2 is G/H isomorphic to?
Solution.
G = U (32) = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31}. The cosets
in G/H are:
H = {1, 31}, 3H = {3, 29}, 5H = {5, 27}, 7H = {7, 25}, 9H = {9, 23},
11H = {11, 21}, 13H = {13, 19}, 15H = {15, 17}.
(3H)2 = 9H 6= H, do 3H has order at least 4, ruling out Z2
Z2
Z2.
(3H)4 = 17H 6= H, so |3H| 6= 4 =) |3H| = 8.
Thus G/H ⇡ Z8.
⇤
Applications of Factor Groups
Why are factor groups important? When G is finite and H 6= {e}, G/H is
smaller than G, yet simulates G in many ways. One can often deduce properties
of G from G/H.
Theorem (9.3 — G/Z Theorem). Let G be a group and let Z(G) be the
center of G. If G/Z(G) is cyclic, then G is Abelian.
Proof.
Let gZ(G) be a generator of G/Z(G), and let a, b 2 G. Then 9 i, j 2 Z 3
aZ(G) = (gZ(G))i = g iZ(G) and
bZ(G) = (gZ(G))j = g j Z(G). Thus
a = g ix and b = g j y for some x, y 2 Z(G). Then
ab = (g ix)(g j y) = g i(xg j )y = g i(g j x)y = (g ig j )(xy) =
(g j g i)(yx) = (g j y)(g ix) = ba.
Thus G is Abelian.
⇤
9. NORMAL SUBGROUPS AND FACTOR GROUPS
125
Note.
(1) From the proof above, we have a stronger e↵ect:
Theorem (9.30). Let G be a group and H  Z(G). If G/H is cyclic,
then G is Abelian.
(2) Contrapositive:
Theorem (9.300). If G is non-Abelian, then G/Z(G) is not cyclic.
Example. Consider a non-Abelian group of order pq, where p and q are
primes. Then, since G/Z(G) is not cyclic, |Z(G)| 6= p and |Z(G)| 6= q, so
|Z(G)| = 1 and Z(G) = {e}.
(3) If G/Z(G) is cyclic, it must be trivial (only the identity).
Theorem (9.4 — G/Z(G) ⇡ Inn(G)). For any group G, G/Z(G) is
isomorphic to Inn(G).
Proof.
Consider T : G/Z(G) ! INN(G) defined by T (gZ(G)) =
g.
[To show T is a well-defined function.] Suppose gZ(G) = hZ(G) =) h 1g 2
Z(G). Then, 8 x 2 G,
h 1gx = xh 1g =) gx = hxh 1g =) gxg
1
= hxh
1
=)
g (x)
=
h (x).
Thus g = h. Reversing the above argument shows T is 1–1. T is clearly
onto. For operation preservation, suppose gZ(G), hZ(G) 2 G/Z(G). Then
T (gZ(G) · hZ(G)) = T (ghZ(G)) =
Thus T is an isomorphism.
gh
=
g h
= T (gZ(G)) · T (hZ(G)).
⇤
126
9. NORMAL SUBGROUPS AND FACTOR GROUPS
Problem (Page 204 # 60). Find |Inn(Dn)|.
Solution. By Example 14, page 67, Z(Dn) = {R0, R180} for n even and
Z(Dn) = {R0} for n odd. Thus, for n odd, Dn/Z(Dn) = Dn ⇡ Inn(Dn) by
Theorem 9.4, |Inn(Dn)| = |Dn| = 2n for n odd.
Now suppose n is even. Then
|Dn|
2n
=
= n.
|Z(Dn)|
2
Then, since Dn/Z(Dn) ⇡ Inn(Dn) by Theorem 9.4,
|Dn/Z(Dn)| = [Dn : Z(Dn)] =
|Inn(Dn)| = |Dn/Z(Dn)| = n.
Further, n = 2p, p a prime. Then, by Theorem 7.3, Inn(Dn) ⇡ Zn or
Inn(Dn) ⇡ Dp. If Inn(Dn) were cyclic, Dn/Z(Dn) would also be by Theorem 9.4 =) Dn is Abelian by Theorem 9.3, an impossibility.
Thus Inn(D2p) ⇡ Dp.
⇤
9. NORMAL SUBGROUPS AND FACTOR GROUPS
127
Theorem (9.5 — Cauchy’s Theorem for Abelian groups). Let G be a finite
Abelian group and let p be a prime such that p |G|. Then G has an element
of order p.
Proof.
Clearly, the theorem is true if |G| = 2. We use the Second Principle of Induction
on |G|. Assume the staement is true for all Abelian groups with order less than
|G|. [To show, based on the induction assumption, that the statement holds
for G also.]
Now G must have elements of prime order: if |x| = m and m = qn, where q is
prime, then |xn| = q. Let x be an element of prime order q. If q = p, we are
finished, so assume q 6= p.
Since every subgroup of an Abelian group is normal, we may construct G =
|G|
G/hxi. Then G is Abelian and p |G|, since |G| =
. By induction, then, G
q
has an element – call it yhxi – of order p.
For the conclusion of the proof we use the following Lemma:
⇤
Lemma (Page 204 # 67). Suppose H C G, G finite. If G/H has an
element of order n, G has an element of order n.
Proof.
Suppose |gH| = n. Suppose |g| = m. Then (gH)m = g mH = eH = H, so by
Corollary 2 to Theorem 4.1, n|m. [We just proved Page 202 # 37.] Then
so, by Theorem 4.2,
9 t 2 Z 3 m = |g| = nt = |gH|t
|g t| =
m
m
=
= n.
gcd(m, t)
t
⇤
Example. Consider, for k 2 Z, hki C Z. 1 + hki 2 Z/hki with |1 + hki| =
k, but all elements of Z have infinite order, so the assumption that G must be
finite in the Lemma is necessary.
128
9. NORMAL SUBGROUPS AND FACTOR GROUPS
Internal Direct Products
Our object here is to break a group into a product of smaller groups.
Definition. G is the internal direct product of H and K and we write
G = H ⇥ K if H C G, K C G, G = HK, and H \ K = {e}.
Note.
(1) For an internal direct product, H and K must be di↵erent normal subgroups
of the same group.
(2)For external direct products, H and K can be any groups.
(3) One forms an internal direct product by starting with G, and then finding
two normal subgroups H and K within G such that G is isomorphic to the
external direct product of H and K.
9. NORMAL SUBGROUPS AND FACTOR GROUPS
129
Example. Consider Z35 where, as we recall, the group operation is addition.
h5i C Z35 and h7i C Z35 since Z35 is Abelian.
Since gcd(5, 7) = 1, 9 s, t 2 Z 3 1 = 5s + 7t. In fact, 1 = ( 4)5 + (3)7.
Thus, for m 2 Z35, m = ( 4m)5 + (3m)7 2 h5i + h7i. So Z35 = h5i + h7i.
Also, h5i \ h7i = {0}, so Z35 = h5i ⇥ h7i.
We also know h5i ⇡ Z7 and h7i ⇡ Z5, so that
Z35 ⇡ Z7
Z5 ⇡ h5i
h7i.
Example. S3 = {(1), (1 2 3), (1 3 2), (2 3), (1 2), (1 3)}.
Let H = h(1 2 3)i and K = h(1 2)i. Then
H = {(1), (1 2 3), (1 3 2)} and K = {(1), (1 2)}.
Since (1 2 3)(1 2) = (1 3) and (1 3 2)(1 2) = (2 3), S3 = HK.
Also, H \ K = h(1)i.
But S3 6⇡ H ⇥ K since S3 6⇡ H K because S3 is not cyclic and H
cyclic since |H| and |K| are relatively prime.
K is
What is the problem here?
K 6C S3 since (1 3)(1 2)(1 3)
1
= (1 3)(1 2)(1 3) = (2 3) 62 K.
Definition (Internal Direct Product H1 ⇥ H2 ⇥ · · · ⇥ Hn). !
Let H1, H2, . . . , Hn be a finite collection of normal subgroups of G. We say
that G is the internal direct product of H1, H2, . . . , Hn and write
G = H1 ⇥ H2 ⇥ · · · ⇥ Hn if
(1) G = H1H2 · · · Hn = {h1h2 · · · hn|hi 2 Hi},
(2) (H1H2 · · · Hi) \ Hi+1 = {e} for i = 1, 2, . . . , n
1.
130
9. NORMAL SUBGROUPS AND FACTOR GROUPS
Theorem (9.6 — H1H2 · · · Hn ⇡ H1 H2 · · · Hn). If a group G is
the internal direct product of a finite number of subgroups H1, H2, . . . , Hn,
then G = H1 H2 · · · Hn, i.e.,
H1 ⇥ H2 ⇥ · · · ⇥ Hn ⇡ H1
H2
···
Hn.
Proof.
[To show h’s from di↵erent Hi’s commute.] Let hi 2 Hi and hj 2 Hj with
i 6= j. Then, since Hi C G and Hj C G,
(hihj hi 1)hj 1 2 Hj hj 1 = Hj and hi(hj hi 1hj 1) 2 hiHi = Hi.
Thus hihj hi 1hj 1 2 Hi \ Hj = {e} by Page 200 # 5. [WLOG, suppose i < j.
Then, if h 2 Hi \Hj , h 2 H1H2 · · · Hi · · · Hj 1 \Hj = {e} from the definition
of internal direct product.] Thus hihj = hj hi.
[To show each element of G can be expressed uniquely in the form h1h2 · · · hn
where hi 2 Hi.] From the definition of internal direct product, there exist
h1 2 H1, . . . , hn 2 Hn such that g = h1h2 · · · hn for g 2 G. Suppose also
g = h01h02 · · · h0n where h0i 2 Hi. Using the commutative property shown above,
we can solve
(?) h1h2 · · · hn = h01h02 · · · h0n
to get
h0nhn 1 = (h01) 1h1(h02) 1h2 · · · (h0n 1) 1hn 1.
Then
h0nhn 1 2 H1H2 · · · Hn
1
\ Hn = {e} =) h0nhn 1 = e =) h0n = hn.
We can thus cancel hn and h0n from opposite sides of (?) and repeat the preceding
to get h0n 1 = hn 1. continuing, we eventually get hi = h0i for i = 1, 2, . . . , n.
9. NORMAL SUBGROUPS AND FACTOR GROUPS
Now define
: G ! H1
From the above,
H2
···
131
Hn by
(h1h2 · · · hn) = (h1, h2, · · · , hn).
is well-=defined. If (h1h2 · · · hn) = (h01h02 · · · h0n),
(h1, h2, · · · , hn) = (h01, h02, · · · , h0n) =)
so
is 1–1. That
hi = h0i for i = 1, . . . , n =) h1h2 · · · hn = h01h02 · · · h0n,
is onto is clear.
[To show operation preservation.]
Now let h1h2 · · · hn, h01h02 · · · h0n 2 G. Again, using the commutativity shown
above,
⇥
⇤
⇥
⇤
(h1h2 · · · hn)(h01h02 · · · h0n) = (h1h01)(h2h02) · · · (hnh0n) =
(h1h01, h2h02, · · · , hnh0n) = (h1h2 · · · hn) · (h01h02 · · · h0n) =
so operations are preserved and
Note. If G = H1
H2
···
H i = {e}
Clearly, each H i ⇡ Hi.
(h1h2 · · · hn) (h01h02 · · · h0n),
is an isomorphism.
Hn, then for
Hi
{e}, i = 1, ..n,
G = H 1 ⇥ H 2 ⇥ · · · H n.
⇤
132
9. NORMAL SUBGROUPS AND FACTOR GROUPS
Theorem (9.7 – Classification of Groups of Order p2). Every group of
order p2, where p is a prime, is isomorphic to Zp2 or Zp Zp.
Proof.
Let G be a group of order p2, p a prime. If G has an element of order p2, then
G ⇡ Zp2 . Otherwise, by Corollary 2 of Lagrange’s Theorem, we may assume
every non-identity element of G has order p.
[To show that 8 a 2 G, hai is normal in G.] Suppose this is not the case. Then
9 b 2 G 3 bab 1 62 hai. Then hai and hbab 1i are distinct subgroups of order
p. Since hai \ hbab 1i is a subgroup of both hai and hbab 1i,
hai \ hbab 1i = {e}. Thus the distinct left cosets of hbab 1i are
Since b
1
hbab 1i, ahbab 1i, a2hbab 1i, . . . , ap 1hbab 1i.
must lie in one of these cosets,
b
1
= ai(bab 1)j = aibaj b
for some i and j. Cancelling the b
1
1
terms, we get
e = aibaj =) b = a
i j
2 hai,
a contradiction. Thus, 8 a 2 G, hai is normal in G.
Now let x be any non-identity element in G and y any element of G not in hxi.
Then, by comparing orders and from Theorem 9.6,
G = hxi ⇥ hyi ⇡ Zp
Zp.
⇤
Corollary. If G is a group of order p2, where p is a prime, then G is
Abelian.
9. NORMAL SUBGROUPS AND FACTOR GROUPS
133
Example. We use Theorem 8.3, its corollary, and Theorem 9.6 for the
following.
If m = n1n2 · · · nk where gcd(ni, nj ) = 1 for i 6= j, then
U (m) = Um/n1 (m)⇥Um/n2 (m)⇥· · ·⇥Um/nk (m) ⇡ U (n1) U (n2) · · · U (nk ).
We use the “=” sign for the internal direct product since the elements are all
within U (m).
U (105) = U (15) · U (7) = U15(105) ⇥ U7(105)
= {1, 16, 31, 46, 61, 76} ⇥ {1, 8, 22, 29, 43, 64, 71, 92}
⇡ U (7) U (15)
U (105) = U (5 · 21) = U5(105) ⇥ U21(105)
= {1, 11, 16, 26, 31, 41, 46, 61, 71, 76, 86, 101} ⇥ {1, 22, 43, 64}
⇡ U (21) U (5)
U (105) = U (3 · 5 · 7) = U35(105) ⇥ U21(105) ⇥ U15(105)
= {1, 71} ⇥ {1, 22, 43, 64} ⇥ {1, 16, 31, 46, 61, 76}
⇡ U (3) U (5) U (7)