Basic Algebra (Solutions)
by Huah Chu
Exercises (§1.8, p.57)
1. Determine addition tables for (Z/Z3, +) and (Z/Z6, +). Determine all the sub- groups of (Z/Z6, +).
Sol. The subgroups of (Z/Z6, +) are {¯0}, {¯0, ¯2, ¯4}, {¯0, ¯3} and Z/Z6. Addition table for (Z/Z6, +):
¯0 ¯1 ¯2 ¯3 ¯4 ¯5
¯0 ¯0 ¯1 ¯2 ¯3 ¯4 ¯5
¯1 ¯1 ¯2 ¯3 ¯4 ¯5 ¯0
¯2 ¯2 ¯3 ¯4 ¯5 ¯0 ¯1
¯3 ¯3 ¯4 ¯5 ¯0 ¯1 ¯2
¯4 ¯4 ¯5 ¯0 ¯1 ¯2 ¯3
¯5 ¯5 ¯0 ¯1 ¯2 ¯3 ¯4
2. Determine a multiplication table for (Z/Z6, ·).
Sol. Multiplication table for (Z/Z6, ·):
¯0 ¯1 ¯2 ¯3 ¯4 ¯5
¯0 ¯0 ¯0 ¯0 ¯0 ¯0 ¯0
¯1 ¯0 ¯1 ¯2 ¯3 ¯4 ¯5
¯2 ¯0 ¯2 ¯4 ¯0 ¯2 ¯4
¯3 ¯0 ¯3 ¯0 ¯3 ¯0 ¯3
¯4 ¯0 ¯4 ¯2 ¯0 ¯4 ¯2
¯5 ¯0 ¯5 ¯4 ¯3 ¯2 ¯1
3. Let G be the group of pairs of real numbers (a, b), a 6= 0, with the product (a, b)(c, d) = (ac, ad + b) (exercise 4. p.36). Verify that K = {(1, b)|b ∈R} is a normal subgroup of G. Show that G/K ' (R∗, ·, 1) the multiplicative group of non-zero reals.
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Proof. The isomorphism from G/K to (R∗, ·) is K(a, b) → a. All verifications are
routine. ¤
4. Show that any subgroup of index two is normal. Hence prove that An is normal in Sn.
Proof. Let H be a subgroup of index two. Its right cosets are {H, Hx}. If x /∈ H, its left cosets must be {H, xH}. Hence Hx = xH for all x /∈ H, i.e. x−1Hx ⊂ H. For x ∈ H, x−1Hx ⊂ H holds trivially. Hence H is normal. ¤ Remark. This exercise is a special case of the following:
Let H be a subgroup a finite group G with index p. If p is the smallest prime dividing |G|. Then H is a normal subgroup of G. (See §1.12, exercise 5).
5. Verify that the intersection of any set of normal subgroups of a group is a normal subgroup. Show that if H and K are normal subgroups, then HK is a normal subgroup.
Proof. The second statement follows from g−1HKg = (g−1Hg)(g−1Kg). ¤
6. Let G1 and G2 be simple groups. Determine the normal subgroups of G1× G2. Proof. In this problem, we use some terminology and some results which will be proved in sections 1.9 and 1.10. Then answer to this problem is the following: (1) When G1 6' G2 or G1 ' G2 6'Zp, the only normal subgroups of G1× G2 are {1} × {1}, G1 × {1}, {1} × G2, and G1 × G2; (2) When G1 ' G2 ' Zp, all the subgroups of G1× G2 are normal, and there are p + 3 subgroups in G1× G2 'Zp×Zp.
Let N be any normal subgroup of G1× G2 with N 6= G1× G2.
Let πi : G1× G2 → Gi, i = 1, 2, be the projection onto the i-th coordinate. It is straight-forward to verify that πi(N) < Gi, i = 1, 2. Hence πi(N) = {1} or Gi since Gi is simple. It is easy to deduce the followings
π1(N) = π2(N) = {1} ⇒ N = {(1, 1)}
π1(N) = 1 and π2(N) = G2 ⇒ N = {1} × G2 π1(N) = G1 and π2(N) = {1} ⇒ N = G1× {1}.
It remains to establish the following:
π1(N) = G1, π2(N) = G2, N 6= G1× G2 ⇒ G1 ' G2 'Zp. Step 1. N ∩ (G1× {1}) = {(1, 1)}, N ∩ ({1} × G2) = {(1, 1)}.
Since N, G1 × {1} are normal, it follows N ∩ (G1 × {1}) / (G1 × {1}). Thus N ∩(G1×{1}) = {(1, 1)} or G1×{1}. If N ∩(G1×{1}) = G1×{1}, then N ⊃ G1×{1}.
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Then N/G1 × {1} is a normal subgroup of G1× G2/G1× {1} ' G2 by Theorem 1.8.
(Page 63). Hence N = G1× G2 or G1 × {1}. But N 6= G1× G2 by assumption. If N = G1× {1}, then π2(N) 6= G2; again a contradiction.
Step 2. For any h ∈ G1, there is a unique k ∈ G2 such that (h, k) ∈ N. Similarly for any k ∈ G2, there is a unique h ∈ G1 such that (h, k) ∈ N.
For any h ∈ G1, there is a k ∈ G2 with (h, k) ∈ N since π1(N) = G1. Now for the uniqueness: if (h, k1), (h, k2) ∈ N, then (1, k1k−12 ) = (h, k1) · (h, k2)−1 ∈ N. By Step 1, (1, k1k−12 ) ∈ N ∩ ({1} × G2) = {(1, 1)}. Hence k1 = k2.
Step 3. For any h ∈ G1, define φ(h) ∈ G2 such that (h, φ(h)) ∈ N. φ is a well- defined homomorphism from G1 into G2 by Step 2. Moreover, φ is onto by Step 2.
Since G1 is simple, φ is one to one. Hence φ : G1 → G2 is an isomorphism.
Step 4. We shall show that G1 (' G2) is abelian.
For any h ∈ G1, consider (h, k) ∈ N. For any g ∈ G1, (g, 1)−1· (h, k) · (g, 1) ∈ N.
Hence (h, k) ∈ N. By Step 2, h = g−1hg. Thus gh = hg for all g, h ∈ G1.
Step 5. The only simple abelian groups are Zp, p, a prime. Choose any nonzero element g. Since G is abelian, hgi is normal in G. Since G is simple, hgi = G. Thus G is cyclic. ButZ and Zn (n: a composite number) cannot be simple.
Step 6. Any nontrivial proper subgroup in Zp×Zp, p, a prime, is cyclic of order p;
and there are p + 1 such subgroups.
Since |Zp×Zp| = p2, any nontrivial subgroup is of order p. Any nonzero element inZp×Zp generates such a subgroup. There are pp−12−1 = p + 1 such subgroups.
7. Let ≡ be an equivalence relation on a monoid M. Show that ≡ is a congruence if and only if the subset of M × M defining ≡ (p.10) is a submonoid of M × M.
Proof. Let S be the subset of M × M defining ≡. Then (1, 1) ∈ S obviously. The equivalence ≡ is a congruence ⇔ a ≡ a0 and b ≡ b0 imply ab ≡ a0b0 ⇔ (a, b) ∈ S and (a0, b0) ∈ S imply (aa0, bb0) ∈ S ⇔ S is a monoid. ¤
8. Let {≡i} be a set of congruences on M. Define the intersection as the intersection of the corresponding subsets of M × M. Verify that this is a congruence on M.
Proof. The reader first verify that it is an equivalence by definition and then apply
exercise 7. ¤
9. Let G1 and G2 be subgroups of a group G and let α be the map of G1× G2 into G defined by α(g1, g2) = g1g2. Show that the fiber over g1g2 — that is, α−1(g1g2) — is the set of pairs (g1k, k−1g2) where k ∈ K = G1∩ G2. Hence show that all fibers have the same cardinality, namely, that of K. Use this to show that if G1 and G2 are finite then
|G1G2| = |G1||G2|
|G1∩ G2|. 3
Proof. (1) Let (h1, h2) ∈ α−1(g1g2), i.e. h1h2 = g1g2. Then g1−1h1 = g2h−12 ∈ G1∩ G2 = K. Set k = g−1h1. We have h1 = g1k and h2 = k−1g2.
(2) Let α be the map: G1 × G2 → G defined by α(g1, g2) = g1g2. The image α(G1×G2) = G1G2. Then |G1×G2| =P
g1g2∈G1G2|α−1(g1g2)| = P
g1g2∈G1G2|G1∩G2| =
|G1G2||G1 ∩ G2|. Hence the result. ¤
10. Let G be a finite group. A and B non-vacuous subsets of G. Show that G = AB if |A| + |B| > |G|.
Proof. Suppose |A| + |B| > |G|. For any g ∈ G, let A−1g def= {a−1g ∈ G|a ∈ A}. Then
|A−1g| = |A|. Since |A−1g| + |B| > |G|, A−1g ∩ B 6= ∅. Thus a−1g = b for some a ∈ A
and b ∈ B. So g = ab. ¤
11. Let G be a group of order 2k where k is odd. Show that G contains a subgroup of index 2.
Proof. Suppose |G| = 2k, k is odd. The permutation group GL of left translations is a subgroup of S2k and isomorphic to G. It suffices to show that GL contains a subgroup of index 2.
G contains an element a of order 2 by [Chapter 1, §1.2. Exercise 13, p.36] or by Sylow’s theorem (p/78). Since ag 6= g for all g ∈ G, aL has no fixed point in {1, 2, . . . , 2k}. Regarding aL as an element of S2k, its cycle decomposition must be (12)(34) · · · (2k − 1, 2k) by suitable change the notation. α is an odd permutation because k is odd.
Set H = A2k∩ GL. For any odd permutation β ∈ GL, βα−1 ∈ H. Hence β ∈ Hα and GL= H ∪ Hα. Hence GL contains the subgroup H of index 2. ¤
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