normal subgroups and factor groupsfacstaff.cbu.edu/wschrein/media/m402 notes/m402c9.pdfchapter 9...

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 H3�� ah = h0a and ah00 = ha. and conversely. It does not mean ah = ha for allh 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 Gand for all h 2 H.

(2) The center Z(G) of a group is always normal since ah = ha for all a 2 Gand 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 andHa is a right coset distinct from H. Since [G : H] = 2, G = H[aH = H[Haand 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.

Let x 2 GL(2, R). Recall det(x�1) =1

det(x)= (det(x))�1. Then, for all

h 2 SL(2, R),

det(xhx�1) = (det(x))(det(h))(det(x))�1 =

(det(x))(det(x))�1(det(h)) = 1 · 1 = 1,

so xhx�1 2 SL(2, R) =) x SL(2, R)x�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�1 2 A.

Let x 2 A4:

Case 1: x 2 H. Then xH ✓ H. Since x�1 2 H, xHx�1 ✓ H.

Case 2: x 2 A. Then xH ✓ A =) xHx�1 ✓ H.

Case 3: x 2 B. Then xH ✓ B =) xHx�1 ✓ H.

Thus, 8 x 2 A4, xHx�1 ✓ H =) H C A4 by Theorem 9.1.

Now let K = {↵1,↵5,↵9} A4. Now ↵5 2 K, but

↵2↵5↵�12 = ↵2↵5↵2 = ↵2↵8 = ↵7 62 K,

so K 6C A4.


Factor Groups

Theorem (9.2 — Factor Groups). Let G be a group and H C G. Theset 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 thecoset representatives chosen, but is the product uniquely determined by thecosets themselves?]

Suppose aH = a0H and bH = b0H. Then a0 = ah1 and b0 = bh2 for someh2, h2 2 H =) a0b0H = ah1bh2H = ah1bH = ah1Hb = aHb = abH byassociativity 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. ⇤


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 1 + 3Z 2 + 3Z0 + 3Z 0 + 3Z 1 + 3Z 2 + 3Z1 + 3Z 1 + 3Z 2 + 3Z 0 + 3Z2 + 3Z 2 + 3Z 0 + 3Z 1 + 3Z

Therefore, Z/3Z ⇡ Z3.

In general, for n > 0, nZ = {0,±n,±2n,±3n, . . . }, and Z/nZ ⇡ Zn.


Example. Consider the multiplication table for A4 below, where i repre-sents 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. SinceH C A4, when we replace the various boxes by their coset names, we get theCayley table below for A4/H.

The factor group collapses all the elements of a coset to a single group elementof A4/H.

When H C G, one can always arrange a Cayley table so this happens. WhenH 6C G, one cannot.


Example. Is U(30)/U5(30) isomorphic to Z2 � Z2 (the Klein 4-group) orZ4?

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, weare 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 goingto the factor group makes 11 the identity.


Problem (Page 201 # 25). Let G = U(32) and H = {1, 31}. Which ofZ8, 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 cosetsin 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 issmaller than G, yet simulates G in many ways. One can often deduce propertiesof G from G/H.

Theorem (9.3 — G/Z Theorem). Let G be a group and let Z(G) be thecenter 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 = giZ(G) and

bZ(G) = (gZ(G))j = gjZ(G). Thus

a = gix and b = gjy for some x, y 2 Z(G). Then

ab = (gix)(gjy) = gi(xgj)y = gi(gjx)y = (gigj)(xy) =

(gjgi)(yx) = (gjy)(gix) = ba.

Thus G is Abelian. ⇤


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 areprimes. 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) isisomorphic 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 2Z(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 clearlyonto. For operation preservation, suppose gZ(G), hZ(G) 2 G/Z(G). Then

T (gZ(G) · hZ(G)) = T (ghZ(G)) = �gh = �g�h = T (gZ(G)) · T (hZ(G)).

Thus T is an isomorphism. ⇤


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) byTheorem 9.4, |Inn(Dn)| = |Dn| = 2n for n odd.

Now suppose n is even. Then

|Dn/Z(Dn)| = [Dn : Z(Dn)] =|Dn|

|Z(Dn)|=

2n

2= n.

Then, since Dn/Z(Dn) ⇡ Inn(Dn) by Theorem 9.4,

|Inn(Dn)| = |Dn/Z(Dn)| = n.

Further, n = 2p, p a prime. Then, by Theorem 7.3, Inn(Dn) ⇡ Zn orInn(Dn) ⇡ Dp. If Inn(Dn) were cyclic, Dn/Z(Dn) would also be by Theo-rem 9.4 =) Dn is Abelian by Theorem 9.3, an impossibility.

Thus Inn(D2p) ⇡ Dp. ⇤


Theorem (9.5 — Cauchy’s Theorem for Abelian groups). Let G be a finiteAbelian group and let p be a prime such that p

��|G|. Then G has an elementof order p.

Proof.

Clearly, the theorem is true if |G| = 2. We use the Second Principle of Inductionon |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 holdsfor G also.]

Now G must have elements of prime order: if |x| = m and m = qn, where q isprime, then |xn| = q. Let x be an element of prime order q. If q = p, we arefinished, so assume q 6= p.

Since every subgroup of an Abelian group is normal, we may construct G =

G/hxi. Then G is Abelian and p��|G|, since |G| =

|G|q

. By induction, then, G

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 anelement of order n, G has an element of order n.

Proof.

Suppose |gH| = n. Suppose |g| = m. Then (gH)m = gmH = eH = H, so byCorollary 2 to Theorem 4.1, n|m. [We just proved Page 202 # 37.] Then

9 t 2 Z 3�� m = |g| = nt = |gH|tso, by Theorem 4.2,

|gt| =m

gcd(m, t)=

m

t= n.

⇤

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 befinite in the Lemma is necessary.


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 writeG = 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 subgroupsof 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 findingtwo normal subgroups H and K within G such that G is isomorphic to theexternal direct product of H and K.


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 �K iscyclic since |H| and |K| are relatively prime.

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 saythat 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.


Theorem (9.6 — H1H2 · · ·Hn ⇡ H1 �H2 � · · ·�Hn). If a group G isthe 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 withi 6= j. Then, since Hi C G and Hj C G,

(hihjh�1i )h�1

j 2 Hjh�1j = Hj and hi(hjh

�1i h�1

j ) 2 hiHi = Hi.

Thus hihjh�1i h�1

j 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 definitionof internal direct product.] Thus hihj = hjhi.

[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 existh1 2 H1, . . . , hn 2 Hn such that g = h1h2 · · ·hn for g 2 G. Suppose alsog = h01h

02 · · ·h0n where h0i 2 Hi. Using the commutative property shown above,

we can solve(?) h1h2 · · ·hn = h01h

02 · · ·h0n

to geth0nh

�1n = (h01)

�1h1(h02)�1h2 · · · (h0n�1)

�1hn�1.

Then

h0nh�1n 2 H1H2 · · ·Hn�1 \Hn = {e} =) h0nh

�1n = e =) h0n = hn.

We can thus cancel hn and h0n from opposite sides of (?) and repeat the precedingto get h0n�1 = hn�1. continuing, we eventually get hi = h0i for i = 1, 2, . . . , n.


Now define � : G ! H1 �H2 � · · ·�Hn by

�(h1h2 · · ·hn) = (h1, h2, · · · , hn).

From the above, � is well-=defined. If �(h1h2 · · ·hn) = �(h01h02 · · ·h0n),

(h1, h2, · · · , hn) = (h01, h02, · · · , h0n) =)

hi = h0i for i = 1, . . . , n =) h1h2 · · ·hn = h01h02 · · ·h0n,

so � is 1–1. That � is onto is clear.

[To show operation preservation.]

Now let h1h2 · · ·hn, h01h02 · · ·h0n 2 G. Again, using the commutativity shown

above,

�⇥�(h1h2 · · ·hn)(h

01h02 · · ·h0n)

⇤= �

⇥(h1h

01)(h2h

02) · · · (hnh

0n)

⇤=

(h1h01, h2h

02, · · · , hnh

0n) = (h1h2 · · ·hn) · (h01h

02 · · ·h0n) =

�(h1h2 · · ·hn)�(h01h02 · · ·h0n),

so operations are preserved and � is an isomorphism. ⇤

Note. If G = H1 �H2 � · · ·�Hn, then for

Hi = {e}�Hi � {e}, i = 1, ..n,

G = H1 ⇥H2 ⇥ · · ·Hn.

Clearly, each Hi ⇡ Hi.


Theorem (9.7 – Classification of Groups of Order p2). Every group oforder 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, thenG ⇡ Zp2. Otherwise, by Corollary 2 of Lagrange’s Theorem, we may assumeevery 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. Then9 b 2 G 3�� bab�1 62 hai. Then hai and hbab�1i are distinct subgroups of orderp. 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

hbab�1i, ahbab�1i, a2hbab�1i, . . . , ap�1hbab�1i.Since b�1 must lie in one of these cosets,

b�1 = ai(bab�1)j = aibajb�1

for some i and j. Cancelling the b�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 isAbelian.


Example. We use Theorem 8.3, its corollary, and Theorem 9.6 for thefollowing.

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 allwithin 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)

normal subgroups and factor groupsfacstaff.cbu.edu/wschrein/media/m402 notes/m402c9.pdfchapter 9...

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