Basic Algebra (Solutions)
by Huah Chu
Exercises (§1.12, pp.76–78)
1. Let γ = (12 · · · n) in Sn. Show that the conjugacy class of γ in Sn has cardinality (n − 1)!. Show that the centralizer C(γ) = hγi.
Proof. (1) Let γ = (12 · · · n). A permutation is conjugate to γ if and only if it has the form (i1i2· · · in). Then n permutations (i1i2· · · in), (i2i3· · · ini1), (i3· · · ini1i2), . . ., (ini1· · · in−1) are equal. Hence all such permutations has cardinality n!/n = (n − 1)!
(2) Since the conjugacy class of γ has cardinality [G : C(γ)], hence |C(γ)| = n, hγi ⊆ C(γ) is obvious. Moreover, |hγi| = n. Then hγi = C(γ). ¤ 2. Determine representatives of the conjugacy classes in S5 and the number of elements in each class. Use this information to prove that the only normal subgroups of S5 are 1, A5, S5.
Sol. representative cardinality parity 1
(12) (123) (12)(34) (1234) (12)(345) (12345)
1 10 20 15 30 20 24
even odd even even odd odd even
A normal subgroup H is a union of some conjugacy classes and one of them must be {1}
Case 1. H ⊂ A5. Then |H|||A5| = 60.
Hence the possible order of H is 1, 1 + 20 + 15 + 24. Thus H = {1} or A5.
Case 2. H 6⊂ A5. Then H ∩ A5/ S5 and H/H ∩ A5 ' H · A5/A5 ' S5/A5. Hence [H : H ∩ A5] = 2. By Case 1, H ∩ A5 = {1} or A5. If H ∩ A5 = A5, then H = S5. If H ∩ A5 = {1}, then |H| = 2. But a subgroup of order 2 in S5 cannot be normal by inspecting the table of conjugacy classes constructed above.
3. Let the partition associated with a conjugacy class be (n1, n2, . . . , nq) where n1 =
· · · = nq1 > nq1+1 = · · · = nq1+q2 > nq1+q2+1 = · · · . Show that the number of elements in this conjugacy class is n!/Q
qi!Q nj.
Proof. Let S = {(i11· · · i1n1)(i21· · · i2n2) · · · (iq11· · · iqn1q1)(iq11+1· · · iqn1q1+1+1 ) · · · (in1q· · · innqq)|1 ≤ ijk≤ n and all ijk are distinct}. Then |S| = n!.
In S, we define an equivalence relation:
(1) For any cyclic (ij1· · · ijnj), · · · (ij1· · · ijnj) · · · ∼ · · · (ij2· · · ijnji−1j) · · · ∼ · · · (ij3· · · ijnj ij1ij2) · · · ∼ · · · and so on. For any element α in S, α is equivalent to Qq
j=1nj elements under this relation.
(2) For the first q1cycles (i11· · · i1n1), . . . , (iq11· · · iqn1q1), any permutation of these cycles are equivalent: (i11· · · i1n1)(i12· · · i1n2) · · · ∼ (i12· · · i1n2)(i11· · · i1n1) · · · , and so on. The same equivalence also defined for the second q2 cycles, . . .. Hence any element in S is equivalent to Q
qi! elements.
A equivalence class under this two relations determine a partition in this conjugacy class. Hence the number of partitions is n!/Q
qi!Q
nj. ¤
4. Show that if a finite group G has a subgroup H of index n then H contains a normal subgroup of G of index a divisor of n!.
Proof. Let H be a subgroup of index n. Consider the action of G on G/H by left translations, T : G → Sym(G/H).
The kernel K of this action is a normal subgroup of G contained in H. And G/K ∼ Im T is a subgroup of Sym(G/H) = Sn. Hence |G/K| is a divisor of n!. ¤
5. Let p be the smallest prime dividing the order of a finite group. Show that any subgroup H of G of index p is normal.
Proof. Let H be a subgroup of index p. Applying exercise 4, H contains a subgroup K which is normal in K and [G : K]|p!. The relation p = [G : H]|[G : K]|p! implies that [G : K] = p since p is the smallest prime dividing |G|. Thus H = K and H is normal
in G. ¤
6. Show that every group of order p2, p a prime, is abelian. Show that up to isomor- phism there are only two such groups.
Proof. (1) Let G be a group of order p2 and C(G) the center of G. Suppose G is not abelian, then |C(G)| = p by Theorem 1.11. Take any g ∈ G − C(G). Since [G : C(G)] = p, g and C(G) generate G. Any element of G can be written in the form gih for h ∈ C(G), 0 ≤ i ≤ p − 1. Thus for any two elements gih1, gihg ∈ G, h1, h2 ∈ C(G), (gih1)(gjh2) = gigjh2h1 = (gjh2)(gih1). G is abelian. A contradiction.
(2) Let G be a group of order p2, then G must be a cyclic group or an elementary abelian group ha, b|ap = bp = 1, ab = bai:
Case 1. G has an element with order p2. Then G is cyclic.
Case 2. All non-identity element of G has order p. Take any 1 6= a ∈ G. Then
|hai| = p. Choose b ∈ G−hai. Then a, b generate a group of order p2, hence G = ha, bi.
¤
Remark. ¿From the Sylow’s Theorems (§1.13), a group G with |G| = pe11· · · penn con- tains, for each i, a subgroup of order peii and all subgroups of this order are isomorphic.
Thus the problem of constructing finite groups may be regarded as having two parts:
(1) constructing p-groups, and (2) combining p-groups to form a group of order n.
Neither of these problems is solved in general.
The known results about first problem are (1) If |G| = p, G is cyclic,
(2) If |G| = p2, G is abelian,
(3) If |G| = p3, there are five such groups up to isomorphism. (See M. Hall: The theory of groups, pp.49–53.)
(4) If |G| = p4, there are 15 groups for p ≥ 3 and 16 groups for p = 2.
(5) There are 51 groups with order 25 and 267 groups with order 26. (M. Hall and J. K. Senior.)
(6) Rodemich claimed that there are 2356 groups with order 27. For more details, we refer to Huppert: Endlich. Gruppen Chap. 3.
7. Let H be a proper subgroup of a finite group G, show that G 6=S
g∈GgHg−1. 8. Let G act on S, H act on T , and assume S ∩ T = ∅. Let U = S ∪ T and define for g ∈ G, h ∈ H, s ∈ S, t ∈ T , (g, h)s = gs, (g, h)t = ht. Show that this defines an actio of G × H on U.
Proof. Omitted. ¤
9. A group H is said to act on a group K by automorphisms if we have an action of H on K and for every h ∈ H the map k → hk of K is an automorphism. Suppose this is the case and let G be the product set K × H. Define a binary composition in K × H by
(k1, h1)(k2, h2) = ((h−12 k1)k2, h1h2)
and define 1 = (1, 1) — the units of K and H respectively. Verify that this defines a group such that h → (1, h) is a monomorphism of H into K × H and k → (k, 1) is a monomorphism of K into K ×H whose image is a normal subgroup. G is called a semi- direct product of K and H. Note that if H and K are finite then |K × H| = |K||H|.
Proof. (1) The composition (k1, h1)(k2, h2) = ((h−12 k1)k2, h1h2) on H × K defines a group.
(i) The associativity:
((k1, h1)(k2, h2))(k3, h3) = ((h−12 k1)k2, h1h2)(k3, h3)
= ((h−13 ((h−12 k1)k2)k3, h1h2h3) (k1, h1)((k2, h2)(k3, h3)) = (k1, h1)((h−13 k2)k3, h2h3)
= ((h2h3)−1k1(h−13 k2)k3, h1h2h3).
We have
(h2h3)−1k1(h−13 k2)k3
= (h−13 h−12 )k1(h−13 k2)k3
= (h−13 (h−12 k1))(h−13 k2)k3 (By the definition (ii) of actions)
= h−13 ((h−12 k1)k2)k3 (k → hk is an automorphism).
Hence the associative law holds.
(ii) (1,1) is the unit and ((h1k1)−1, h−11 ) is the inverse of (k1, h1). All the verifications are left to the reader.
(2) From (1, h1)(1, h2) = (1, h1h2) and (k1, 1)(k2, 1) = (k1k2, 1), we know that h → (1, h) and k → (k, 1) are monomorphisms.
(3) K is normal subgroup of K × H: Since
(k1, h1)(k2, 1)((h1k1)−1, h−11 ) = ((h1(k1k2))(h1k1)−1, 1). ¤ Remark. The Jordan-H¨older Theorem claims: for any finite group G admits a com- position series
G ∈ G1. G2. · · · . Gk = 1,
the composition factor Gi/Gi+1= Qi is simple and is uniquely determined by G (§4.6, p.241). The inverse question is: given the factor group Qi, how can we recapture G?
We want to construct G inductively. That is, given Qi and Gi+1, we want to determine Gi such that Gi+1 is normal in Gi and Gi/Gi+1 ' Qi. This problem is called “The extension problem”. Where Gi is called an extension of Gi+1 by Qi.
Given K and H, the most simple extension of K by Q is the direct product G = K × H. A natural generalization of it is semidirect product G of K by H: G contains subgroup K and H such that K / G, KH = G and K ∩ H = 1. It is not difficult to see that this definition is the same as that given in exercise.
The extension problem was solved by O. Schreier in 1926.
10. Let G be a group, H a transformation group acting on a set S and let GS denote the set of maps of S into G. Then GS is a group (the S-direct power of G) if we define (f f )(s) = f (s)f (s), f ∈ GS, s ∈ S. If h ∈ H and f ∈ GS define hf by
(hf )(s) = f (h−1s). Verify that this defines an action of H on GS by automorphisms.
The semi-direct product of H and GS is called the (unrestricted) wreath product G o H of G with H.
Proof. If h ∈ H, f ∈ GS define hf by (hf )(s) = f (h−1s).
(1) we first check that this is an action:
(1f )(s) = f (1−1s) = f (s) ⇒ 1f = f.
(h1h2, f )(f ) = f ((h1h2)−1s) = f (h−12 (h−11 s)) = (h2f )(h−11 s)
= h1(h2f )(s) ⇒ (h1h2)f = h1(h2f ).
(2) The map f → hf is an automorphism on GS:
(i) (h(f1f2))(s) = (f1f2)(h−1s) = f1(h−1s)f2(h−1s) = hf1(s) · hf2(s).
Hence f → hf is a homomorphism.
(ii) Note that the unit in GS is the map 1 : s → 1.
Suppose that hf = 1, that is, hf (s) = f (h−1s) = 1 for all s ∈ S. Since H is a transformation group acting on S, this implies that f (t) = 1 for all t ∈ S.
Thus f = 1 and f → hf is injective.
(iii) For any g ∈ GS, set f (s) = g(h(s)). Then (hf )(s) = f (h−1s) = g(h(h−1s)) =
g(s). Hence hf = g and f → hf is surjective. ¤
11. Let G, H, S be as in exercise 10 and suppose G acts on a set T . Let (f, h) ∈ G o H where f is a map of S into G. If (f1, h1), (f − 2, h2) are two such elements, the product in G o H is ((h−12 f1)f2, h1h2). If (t, s) ∈ T × S define (f, h)(t, s) = (f (s)t, hs). Verify that this defines an action of G o H on T × S. Note that if everything is finite then
|G o H| = |G||S||H| and the degree of the action, defined to be the cardinality of the set on which the action takes place, is the product of the degrees of the actions of H and of G.
Proof. For (f, h) ∈ G o H, (t, s) ∈ T × S, define (f, h)(t, s) = (f (s)t, hs). We verify that this defines an action.
(1) (1,1) is the unit of G o H where the first 1 is the unit 1(s) = 1 in GS, the second 1 is the unit in H. Then
(1, 1)(t, s) = (1(s)t, 1s) = (1t, 1s) = (t, s).
(2) (f1, h1)((f2, h2)(t, s)) = (f1, h1)(f2(s)t, h2s) = (f1(h2s)f2(s)t, h1h2s) On the other hand
((f1, h1)(f2, h2))(t, s) = ((h−12 f1)f2, h1h2)(t, s)
= (((h−12 f1)f2)(s)t, h1h2h3)
= ((h−12 f1(s)f2(s))t, h1h2s) (by the multiplication in GS)
= (f1(h2s))f2(s)t, h1h2s)
= (f1, h1)((f2, h2)(t, s)).
¤ Remark. An example of wreath product is the Sylow p-subgroups of symmetric group Sn (see exercise 16, 17 in §1.13).
12. Let G act on S. Then the action is called k-fold transitive for k = 1, 2, 3, . . ., if given any two elements (x1, . . . , xk), (y1, . . . , yk) in S(k), where the xi and the yi are distinct, there exists a g ∈ G such that gxi = yi, 1 ≤ i ≤ k. Show that if the action of G is doubly transitive then it is primitive.
Proof. Let the action of G on S be doubly transitive and π(S) be any nontrivial partition. Hence there is A ∈ π(S) such that |A| ≥ 2 and A 6= S. Choose x, y ∈ A and z ∈ S − A. By the hypothesis on G, there exists g ∈ G such that g(x) = x and g(y) = z. Thus π(S) is not stabilized by G and the action is not primitive. ¤
13. Show that if the action of G on S is effective and primitive then the induced action on S by any normal subgroup N 6= 1 of G is transitive.
Proof. Suppose that N (6= 1) is not transitive on S. Then the set of orbits of N, {Ns}s∈S, forms a partition π(S) of S with |π(s)| ≥ 2. For all g ∈ G and Ns ∈ π(S), gNs = Ngs ∈ π(S) since N is normal. Thus π(S) is stabilized by G. Since G is not primitive, π(S) = {{S}|s ∈ S}. Hence hs = s for all h ∈ N, s ∈ S. Since the action of
G is effective, it follows that N = {1}. ¤
Remark. If G is not effective on S, the above exercise is not true. In fact, let G1 be any primitive action on S, G2 any group. We define a group action of G1× G2 on S by (g1, g2) · s = g1s. Clearly {1} × G2 is normal in G1× G2. But {1} × G2 is transitive only when |S| = 1.
