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Sylow's TheoremsWe now come to the three theorems of Sylow, which are among the most important results in the study of finite groups. Recall that earlier we proved Lagrange's theorem, which asserts that if G is a finite group and H is a subgroup of G, then H divides the order of G. Note, however, that the converse of Lagrange's theorem is false. That is, if G has order n and if mn, then there does not necessarily exist a subgroup of G of order m. For example, A_{4} is a group of order 12 which has no subgroups of order 6. Sylow's theorem comes about as close as one can come to providing a converse to Lagrange's theorem. Throughout this section, let G be a group of order n, let p be a prime dividing n, and let Definition 1: A subgroup of G of order p^{a} is called a pSylow subgroup. Our main results are the following theorems of the Norwegian mathematician Sylow, which we will denote SI, SII, and SII. Theorem 2 (SI): G contains a pSylow subgroup.
Theorem 3 (SII): If P and Q are any two pSylow subgroups of G, then there exists
Theorem 4 (SIII): Suppose that G has r pSylow subgroups. Then rn and In order to prove these remarkable results, a number of preliminaries must be introduced. If G is any group, then the center of G, denoted Z_{G}, is defined as
Z_{G} = {g G  gx = xg for all x G}.
Then If S is any subset of G, then the normalizer of S, denoted
N_{G}(S) = {g G gSg^{1} S}.
Then N_{G}(S) < G. Moreover, if S is a subgroup of G, then Let
C = {xgx^{1}  x G}.
Moreover, if C_{1},...,C_{h} are conjugacy classes of G, then
G = C_{1} ... C_{h}
and C_{i} C_{j} = if i j.
Proposition 5: Let
N_{G}(S) = n/h_{g}.
Proof: Let C denote the conjugacy class containing g,
G = g_{1}N_{G}(S) ... g_{h}N_{G}(S)
and (2)
g_{i}N_{G}(S) g_{j}N_{G}(S) = (i j).
This will suffice to prove the proposition. For g_{i}N_{G}(S) contains N_{G}(S) elements for i = 1,...,h and by (2), none of these sets overlap. Therefore, by (1),
x = g_{i}m = g_{j}m'.
However,
N_{G}(S) = {x G  xgx^{1} = g}
= {x G  xg = gx}.
But, by (3), g_{j}^{1}g_{i} = m'm^{1} N(S), so that
g_{j}^{1}g_{i}g = gg_{j}^{1}g_{i} g_{i}gg_{i}^{1} = g_{j}gg_{j}^{1},
which contradicts the choice of g_{i}'s. Thus (2) is proved. Let
g_{i}^{1}xgx^{1}g_{i} = g (g_{i}^{1}x)g(g_{i}^{1}x)^{1} = g
g_{i}^{1}x N_{G}(S)
x g_{i}N_{G}(S).
Thus (1) is proved. Let us now give proofs of the Sylow theorems. Case A: p(Z_{G}). In this case, by Proposition 7 of the section on abelian groups, Z_{G} contains an element x of order p. Let Case B: p(Z_{G}) and 1 < Z_{G} < n_{0}. In this case, G/Z_{G} is of order p^{a}n' for some Case C: Z_{G} = n_{0} or 1. Let C_{1},...,C_{t} denote the conjugacy classes of G. Note that
n = Z_{G} + C_{s+1} + ... + C_{t}
n n_{0} or 1 (mod p)
n_{0} or 1 0 (mod p) (since pn).
But Proof of SII: Let S_{1} and S_{2} be two pSylow subgroups of G, and let
G = S_{1}g_{1}S_{2} ... S_{1}g_{t}S_{2}
be the decomposition of G relative to the two subgroups S_{1} and S_{2}. Then since
n = (p^{a} · p^{a})/u_{i},
where u_{i} = the order of
n_{0} = p^{a}/u_{i}.
Since U_{i} < S_{2},
g_{i}^{1}S_{1}g_{i} S_{2}.
But both g_{i}^{1}S_{1}g_{i} and S_{2} have order p^{a}, so that we conclude that
g_{i}^{1}S_{1}g_{i} = S_{2}.
Corollary 6: Let S be a pSylow subgroup of G. Then S is the only pSylow subgroup of G if and only if Proof: Let
Let S' be any pSylow subgroup of G. Then by SII,
Proposition 7: Let S be a pSylow subgroup of G, Proof: Let
G = g_{1}N g_{2}N ... g_{i}N, t = [G:N]
be a coset decomposition. Then (4)
g_{1}Sg_{1}^{1},g_{2}Sg_{2}^{1},...,g_{i}Sg_{i}^{1}
are pSylow subgroups of G. These are distinct, since if
g_{i}Sg_{i}^{1} = g_{j}Sg_{j}^{1},
we see that Proof of SIII: Let t be the number of pSylow subgroups of G. Then, by Proposition 7,
G = Sg_{1}N Sg_{2}N ... Sg_{v}N
be the decomposition of G with respect to the two subgroups S and N. Without loss of generality, assume that
n = p^{a}n_{0} = p^{a}r/c_{i}
where
n_{0} = r 1/c_{i} = m_{0} p^{a}/c_{i}
(5)
= m_{0}(1 + p^{a}/c_{i})
n_{0}/m_{0} = 1 + p_{a}/c_{i}
t = n/r = 1 + p^{a}/c_{i}
Note that since Let us now give a few examples of applications of the Sylow theorems.
Proposition 8: Let p and q be distinct primes, Proof: By the fundamental theorem of abelian groups, it suffices to show that G is abelian. Let P and Q, be respectively a pSylow and a qSylow subgroup of G. Since every element of P except the identity has order p and every element of Q except the identity has order q, we see that Thus, for example, since Example 1: No group of order 20 is simple. Let G be a group of order 20. The number of 5Sylow subgroups is one of 1,2,4,5,10,20. However, the number of 5Sylow subgroups is congruent to 1(mod 5), so that this number is 1. Thus, if P is the unique 5Sylow subgroup, then 
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