22 Quotient Groups

Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan

Let’s look closely on the construction of the group (ZZn, ⊕). We know that the elements in ZZn are equivalence classes of the equivalence relation deﬁned on ZZ by a ∼ b if and only if n|(a − b).

Also, an element of ZZn is a right coset. Indeed, if 0 ≤ k < n then < n > +k = [k] = {nq + k : q ∈ ZZ}. The operation ⊕ is defined by [a] ⊕ [b] = [a + b] or using right cosets (< n > +a) ⊕ (< n > +b) =< n > +(a + b). Note that the cyclic group < n > is a normal subgroup of ZZsince ZZis Abelian (See Example 21.4). We are going to use the above ideas to construct new groups where G is a group replacing ZZand N is a normal subgroup of G playing the role of < n > . More precisely, we have the following theorem. Theorem 22.1 Let N be a normal subgroup of a group G and let G/N be the set of all right cosets of N in G. Define the operation on G/N × G/N by (Na)(Nb) = N(ab). Then (G/N, ·) is a group, called the quotient group of G by N. Proof. · is a well-defined operation We must show that if (Na1, Nb1) = (Na2, Nb2) then N(a1b1) = N(a2b2). To see this, since (Na1, Nb1) = (Na2, Nb2) then Na1 = Na2 and Nb1 = Nb2. Since a1 = ea1 ∈ Na1 then a1 ∈ Na2 so that a1 = na2 for some n ∈ N. Sim- 0 0 0 ilarly, b1 = n b2 for some n ∈ N. Therefore, a1b1 = na2n b2. Since N/G 0 −1 0 −1 00 0 00 then a2n a2 ∈ N, say a2n a2 = n ∈ N. Hence, a2n = n a2 so that 00 00 a1b1 = nn a2b2. But nn ∈ N so that a1b1 ∈ N(a2b2). Since N(a2b2) is an equivalence class and a1b1 ∈ N(a2b2) then N(a1b1) = N(a2b2) (Theorem 9.2).

· is associative Let a, b, c ∈ G. Then Na(NbNc) = Na(Nbc) = N(a(bc)) = N((ab)c) = N(ab)Nc = (NaNb)(Nc)

1 where we used the fact that multiplication in G is associative.

Ne = N is the identity element If a ∈ G then (Na)(Ne) = N(ae) = Na and (Ne)(Na) = N(ea) = Na.

Every element of G/N is invertible If a ∈ G then (Na)(Na−1) = N(aa−1) = Ne and (Na−1)(Na) = N(a−1a) = Ne so that (Na)−1 = Na−1.

Remark 22.1 Note that for a ﬁnite group G, the number of elements of G/N is just the index of N in G, i.e. [G : N]. That is, |G/N| = [G : N]. By Lagrange’s theorem, |G| |G| |G| = [G : N]|N| so that [G : N] = |N| . Hence, |G/N| = |N| .

Remark 22.2 If N is not a normal subgroup of G, then the operation deﬁned in the previous theorem will not be well-deﬁned. To see this, consider the subgroup −1 N =< (12) >= {(1), (12)} of S3. Since (123)(12)(123) = (23) 6∈ N then N is not a normal subgroup of S3. However, N(123) = N(23) = {(123), (23)} and N(132) = N(13) = {(13), (132)}. But N(123)(132) 6= N(23)(13) since N(123)(132) = N and N(23)(13) = N(123).

Example 22.1 Let G = S3 and N =< (123) >= {(1), (123), (132)}. One can verify easily that N / G, G/N = {N,N(12)}.

Example 22.2 Quotient groups can be used to show that A4 has no subgroup of order 6 and thus showing that the converse of Lagrange’s Theorem is false in general. To see this, assume that N is a subgroup of A4 of order 6. Then [G : N] = 2 and therefore N/A4 (See Exercise 21.7). Hence, A4/N makes sense. Moreover, for each a ∈ N, (Na)−1 = Na so that Na2 = N and hence a2inN. One can show that (123)2 = (132), (132)2 = (123), (124)2 = (142), (142)2 = (124), (134)2 = (143), (143)2 = (134), (234)2 = (243), and (243)2 = (234). This yields more than six diﬀerent elements of A4 in N. That is, |N| > 6, a contradiction. Thus, A4 has no subgroup of order 6.

By Theorem 21.1(ii), the kernel of a homomorphism is a normal subgroup. Conversely, quotient groups enable us to show that every normal subgroup is the kernel of some homomorphism.

Theorem 22.2 Let G be a group and let N / G. The mapping η : G −→ G/N deﬁned by η(a) = Na for all a ∈ G is a homomorphism such that Ker η = N. We call η the natural homomorphism from G onto G/N.

2 Proof. First we show that η is well-deﬁned. Indeed, if a = b then Na = Nb, i.e. η(a) = η(b). Since η(ab) = N(ab) = NaNb = η(a)η(b) then η is a homomorphism. η is onto since any member of G/N is of the form Na for some a ∈ G. That is, η(a) = Na. It remains to show that Ker η = N. The proof is by double inclusions. If a ∈ Ker η then η(a) = Ne. Thus, Na = Ne so that a = ne = n for some n ∈ N. Hence, a ∈ N and Ker η ⊆ N. Conversely, for all n ∈ N we have η(n) = Nn = N = Ne so that N ⊆ Ker η.

The operation of multiplication on the quotient group was deﬁned in terms of right cosets. The following theorem shows that when working with cosets of a normal subgroup N, it is immaterial whether we use Na or aN.

Example 22.3 If G = ZZand N =< n > then G/N = ZZn and η : ZZ −→ ZZn is given by η(a) =< n > +a = [a] and Ker η =< n > .

Theorem 22.3 Let N be a normal subgroup of a group G. Then aN = Na for all a ∈ G.

Proof. Let x ∈ aN. Then x = an for some n ∈ N. Since N is normal then ana−1 ∈ N. Thus, ana−1 = n0 ∈ N and therefore an = n0a. That is, x ∈ Na. Hence, aN ⊆ Na. A similar argument shows that Na ⊆ aN.

We close this section with some properties of quotient groups. More properties of quotient groups are discussed in the exercises.

Lemma 22.1 If G is Abelian and N/G then G/N is also Abelian.

Proof. We have to show that for any Na, Nb ∈ G/N we have (Na)(Nb) = (Nb)(Na). Now, since G is Abelian then ab = ba. Thus,

(Na)(Nb) = Nab = Nba = (Nb)(Na).

That is, G/N is Abelian.

Lemma 22.2 If G is a group such that G/Z(G) is cyclic then G is Abelian, where

Z(G) = {g ∈ G : xg = gx ∀x ∈ G}.

Proof. First we show that Z(G) / G. Indeed, if g ∈ G and x ∈ Z(G) then gxg−1 = gg−1x = x ∈ Z(G). Since G/Z(G) is a cyclic group then G/Z(G) =< Z(G)g >

3 for some g ∈ G. If a, b ∈ G then Z(G)a and Z(G)b belong to G/Z(G). Thus, Z(G)a = Z(G)gn and Z(G)b = Z(G)gm for some integers n and m. Hence, a = xgn and b = ygm for some x, y ∈ Z(G). This implies that ab = (xgn)(ygm) = xygn+m = yxgmgn = ygmxgn = ba. Hence, G is Abelian.

Lemma 22.3 Let G be a group of order p2 where p is a prime number. Assuming that the center of G is nontrivial, then G is Abelian.

Proof. Let G be a group such that |G| = p2. Since e ∈ C(G) then C(G) 6= ∅. By Lagrange’s theorem either |C(G)| = p2 or |C(G)| = p. If |C(G)| = p2 then C(G) = G and so G is Abelian. If |C(G)| = p then |G/C(G)| = p and so G/C(G) is cyclic (See Corollary 3). By the previous lemma, G is Abelian.

Review Problems

Exercise 22.1 Find the order of each of the following quotient groups.

(i) ZZ8/ < [4] > . (ii) < 2 > / < 8 > .

Exercise 22.2 Let G = A4 and H = {(1), (12)(34), (13)(24), (14)(23)} / G. Write out the dis- tinct elements of G/H and make a Cayley table for G/H.

Exercise 22.3 Construct the Cayley table for ZZ12/ < [4] > .

Exercise 22.4 Let G be a cyclic group. Prove that for every subgroup H of G, G/H is cyclic.

Exercise 22.5 Assume N / G.

(a) Prove that if [G : N] is prime, then G/N is cyclic. (b) Prove or disprove the converse of the statement in part (a).

Exercise 22.6 Determine the order of ZZ12 × ZZ4/ < [3], [2] > .

Exercise 22.7 Prove that if N/G and a ∈ G then o(Na)|o(a).

Exercise 22.8 Prove that G/N is Abelian if and only if aba−1b−1 ∈ N for all a, b ∈ G.

4 Exercise 22.9 Let N be a normal subgroup of G and a ∈ G. Prove that the order of Na in G/N is the smallest positive integer n such that an ∈ N.

Exercise 22.10 Let θ : G → H be a homomorphism and K = Ker θ. Show that φ : G/K → θ(G) deﬁned by φ(Ka) = θ(a) is an isomorphism. Also, show that φ ◦ η = θ, where η is the canonical homomorphism.

Exercise 22.11 Let H and K be subgroups of a group G such that K / G. Prove that H/H ∩ K and HK/K make sense, where HK = {hk : h ∈ H and k ∈ K}.

Exercise 22.12 Let H and K be as in the previous exercise. Show that φ : H → HK/K deﬁned by φ(h) = Kh an epimorphism with Ker φ = H ∩ K.

Exercise 22.13 Assume that H,K4G and K4H. Prove that H/K / G/K.

Exercise 22.14 Let H,K, and G as in the previous exercise. Show that φ : G/K → G/H, deﬁned by φ(Kg) = Hg is an epimorphism with Ker φ = H/K.

Exercise 22.15 If N is a subgroup of G such that the product of two right cosets of N in G is again a right coset of N in G, prove that N / G.