As applications of group actions on sets and Sylow theorems, we give some elementary results on solvable groups and p-groups.
Theorem 2.3.1 A finite p-group is solvable.
Proof: Let G be a finite p-group and |G| =pn. Use induction on n.
If n = 0, the conclusion is trivial. Now assume that n > 0. Take a maximal subgroup M of G. Then |M| = pn−1 and M EG, by Corol- lary 2.2.7. By inductive hypothesis, M is solvable. Moreover, G/M is cyclic and of order p, and hence solvable. By Proposition 1.4.3(1), G
is solvable. ¤
The following are some sufficient conditions for a group to be solv- able.
Theorem 2.3.2 For any two distinct primes pand q, a finite group of order pq is solvable.
Proof: Without any loss of generality, one may assume that p < q. Let Q be a Sylow q-subgroup of G. Then |Q| = q, and the number of conjugates of Q is kq + 1, and kq+ 1 | p. Since p < q, we have
k = 0. ThusGhas only one Sylowq-subgroup, and hence QEG. Since
|G/Q| = p and |Q| = q, G/Q and Q are solvable, implying that G is
Theorem 2.3.3 Let p > q > r be primes. Then a group G of order
pqr is solvable.
Proof: Let G havenp subgroups of order p, nq of order q, and nr of
order r. All those subgroups are Sylow subgroups of G. Suppose that none of np, nq and nr is equal to 1. Then by the third Sylow Theorem
we have np > p, nq > q, and nr > r, and np | qr, nq | pr, nr | pq. It
follows thatnp =qr, nq≥p and nr ≥q. Thus the number of elements
of order p in G is np(p−1) = qr(p−1) = pqr−qr, the number of
elements of orderq isnq(q−1)≥p(q−1) = pq−p, and the number of
elements of order r is nr(r−1)≥q(r−1) =qr−q. Hence we have
|G| ≥ 1 +pqr−qr+pq−p+qr−q
= |G|+ (p−1)(q−1),
a contradiction. This shows that at least one ofnp, nqandnr is 1. Then
G has a normal Sylow subgroup N. Since the factor group G/N has order a product of two distinct primes and is solvable by Theorem 2.3.2,
we have that G is solvable. ¤
Note that the alternating group A5 with |A5| = 5 · 22 · 3 is not solvable because A0
5 =A5. See Exercise 1.4.4.
However, Theorems 2.3.2 and 2.3.3 can be generalized to the fol- lowing: A finite group G is solvable if all of its Sylow subgroups are cyclic. In particular, a finite group of square-free order is solvable; see Corollary 2.5.4.
Theorem 2.3.4 Let p, q be any two distinct primes and n a positive integer. Then a group G of orderpnq is solvable.
Proof: By induction onn, it suffices to prove thatGhas a nontrivial normal subgroup.
Let P be a Sylow p-subgroup of G. Assume that P 5 G and that
P = P1, . . . , Ps are all Sylow p-subgroups of G. Since s | q and s > 1
we have s =q. Now we distinguish two cases: (i) Pi∩Pj = 1 for anyi6=j: In this case,
Sq
i=1Picontains (pn−1)q+1
p-groups and solvable groups 73
nontrivial q-elements. By Sylow theorem, G has a subgroup of order q which consists of q-elements, and hence G has only one subgroup of order q, which is normal in G.
(ii) Pi ∩Pj 6= 1 for some i 6= j: In this case we choose i, j with
i 6= j such that D = Pi ∩Pj has the maximum possible order.
Since D < Pi, we have D < NPi(D) := Hi ≤ Pi. Similarly,
D < NPj(D) := Hj ≤Pj. Thus we have DEhHi, Hji:=T.
(1) AssumeT is ap-subgroup. Then there is a Sylowp-subgroup
Pk of G such that T ≤ Pk. Since Pk ∩Pi ≥ Hi > D and
Pk∩Pj ≥Hj > D, by the maximality ofD, we havePk =Pi
and Pk =Pj. So Pi =Pj, a contradiction.
(2) Assume |T|= ptq with t ≥1. Let Q be a subgroup of T of
order q. Then G = QPi. Let N = DG, the normal closure
of D. We have N EG. We shall prove that N is a proper subgroup of G, and complete the proof of the theorem. For any g ∈ G, we let g = xy where x ∈ Q and y ∈ Pi. Then
Dg =Dxy =Dy ≤P
i, andDG ≤Pi < G, we are done. ¤
The following famous theorem due to Burnside generalizes Theo- rem 2.3.4.
Theorem 2.3.5 (Burnside paqb-Theorem) Let p, q be any two dis-
tinct primes and a, b positive integers. Then any group of order paqb is
solvable.
Its proof can be found in most of group theory texts, and the short- est and very beautiful proof is due to Burnside, in which he used the character theory of finite groups. We give this proof in Chapter 5. (See page 216.)
The following Feit-Thompson Theorem shows that most of finite groups are solvable. Its proof took up an entire journal issue (Feit and Thompson 1963).
Theorem 2.3.6 (Feit-Thompson (1963)) Every group of odd order is solvable.
By Feit-Thompson Theorem and Theorem 2.4.7, every group of or- der 2nwithnodd is also solvable. Hence, every finite simple nonabelian group is doubly even, i.e., divisible by 4.
Next we discuss some elementary properties of finite p-groups.
Theorem 2.3.7 For any nontrivial p-group G, |Z(G)|>1.
Proof: Let
G=C1∪C2∪ · · · ∪Cs
be the decomposition ofG into conjugacy classes withC1 ={1}. Since
|G| = pn and |C
i| = |G : CG(xi)| for xi ∈ Ci, |Ci| is a power of p for
eachi. Since |C1|= 1, there exists at least onei6= 1 such that|Ci|= 1.
Hence |Z(G)|>1. (Note that|Ci|= 1 if and only if Ci ⊆Z(G).) ¤
Theorem 2.3.8 Let G be a finite p-group and N a normal subgroup of G of order p. Then N ≤Z(G).
Proof: By N/C-theorem,
G/CG(N) = NG(N)/CG(N).Aut(N).
Since Aut(N) is cyclic of orderp−1 and|G/CG(N)|is a power ofp, we
have|G/CG(N)|= 1. It follows thatG=CG(N) and henceN ≤Z(G).
¤
Now, one may classify the p-groups of order ≤ p2. Obviously, any group of order pis cyclic. For groups of order p2, we have
Theorem 2.3.9 Any group of order p2 is abelian.
Proof: If G has an element of order p2, then G is cyclic and hence abelian. If Ghas no element of order p2, then every nontrivial element is of orderp. By Theorem 2.3.7,Z(G)>1. Take an elementa∈Z(G),
a 6= 1, then hai is a subgroup of Z(G) of order p. Take an element
b /∈ hai. Then G = ha, bi. Since a ∈ Z(G), ab = ba. Hence G is an
Transitive permutation representations 75
From Theorems 1.2.9 and 2.3.9, one can see that there are exactly two types of groups of order p2, with type invariants (p2) and (p, p).
Exercises
2.3.1. Let p be the smallest prime divisor of |G|. If N EG with |N|= p, thenN ≤Z(G).
2.3.2. Letp6= 2 be the smallest prime divisor of|G|. IfNEGwith|N|=p2, then|G:CG(N)| ≤p.
2.3.3. Let G be a finite p-group andN EG. Assume that |N|=p2. Then N Z(G)/Z(G)≤Z(G/Z(G)).
2.3.4. Prove that any group of order 2·3·5·11 is solvable. 2.3.5. Justify the following.
(1) Solvable groups: abelian groups; Dihedral groups;p-groups; groups of odd orders; their direct products and subgroups; Ttheir ho- momorphism images.
(2) Unsolvable groups: any nonabelian simple groups, e.g., An, Sn
(n ≥ 5) and PSL(n, q) ((n, q) 6= (2,2) or (2,3); their products; all groups containing a section which is a nonabelian simple sub- group. Asectionof a groupGis a quotient group of a subgroup ofG.