Abstract Algebra: Theory and Applications, Remixed!

Proof.

We will use induction on the order of \(G\text{.}\) If \(|G|=p\text{,}\) then clearly \(G\) itself is the required subgroup. We now assume that every group of order \(k\text{,}\) where \(p \leq k \lt n\) and \(p\) divides \(k\text{,}\) has an element of order \(p\text{.}\) Assume that \(|G|= n\) and \(p \mid n\) and consider the class equation of \(G\text{:}\)

We have two cases.

Proof.

We induct on the order of \(G\) once again. If \(|G| = p\text{,}\) then we are done. Now suppose that the order of \(G\) is \(n\) with \(n \gt p\) and that the theorem is true for all groups of order less than \(n\text{,}\) where \(p\) divides \(n\text{.}\) We shall apply the class equation once again:

First suppose that \(p\) does not divide \([G:C(x_i)]\) for some \(i\text{.}\) Then \(p^r \mid |C(x_i)|\text{,}\) since \(p^r\) divides \(|G| = |C(x_i)| \cdot [G:C(x_i)]\text{.}\) Now we can apply the induction hypothesis to \(C(x_i)\text{.}\)

Hence, we may assume that \(p\) divides \([G:C(x_i)]\) for all \(i\text{.}\) Since \(p\) divides \(|G|\text{,}\) the class equation says that \(p\) must divide \(|Z(G)|\text{;}\) hence, by Cauchy's Theorem, \(Z(G)\) has an element of order \(p\text{,}\) say \(g\text{.}\) Let \(N\) be the group generated by \(g\text{.}\) Clearly, \(N\) is a normal subgroup of \(Z(G)\) since \(Z(G)\) is abelian; therefore, \(N\) is normal in \(G\) since every element in \(Z(G)\) commutes with every element in \(G\text{.}\) Now consider the factor group \(G/N\) of order \(|G|/p\text{.}\) By the induction hypothesis, \(G/N\) contains a subgroup \(H\) of order \(p^{r- 1}\text{.}\) The inverse image of \(H\) under the canonical homomorphism \(\phi : G \rightarrow G/N\) is a subgroup of order \(p^r\) in \(G\text{.}\)

Proof.

Certainly \(x \in N(P)\text{,}\) and the cyclic subgroup, \(\langle xP \rangle \subset N(P)/P\text{,}\) has as its order a power of \(p\text{.}\) By the Correspondence Theorem there exists a subgroup \(H\) of \(N(P)\) containing \(P\) such that \(H/P = \langle xP \rangle\text{.}\) Since \(|H| = |P| \cdot |\langle xP \rangle|\text{,}\) the order of \(H\) must be a power of \(p\text{.}\) However, \(P\) is a Sylow \(p\)-subgroup contained in \(H\text{.}\) Since the order of \(P\) is the largest power of \(p\) dividing \(|G|\text{,}\) \(H=P\text{.}\) Therefore, \(H/P\) is the trivial subgroup and \(xP = P\text{,}\) or \(x \in P\text{.}\)

Proof.

We define a bijection between the conjugacy classes of \(K\) and the right cosets of \(N(K) \cap H\) by \(h^{-1}Kh \mapsto (N(K) \cap H)h\text{.}\) To show that this map is a bijection, let \(h_1, h_2 \in H\) and suppose that \((N(K) \cap H)h_1 = (N(K) \cap H)h_2\text{.}\) Then \(h_2 h_1^{-1} \in N(K)\text{.}\) Therefore, \(K = h_2 h_1^{-1} K h_1 h_2^{-1}\) or \(h_1^{-1} K h_1 = h_2^{-1} K h_2\text{,}\) and the map is an injection. It is easy to see that this map is surjective; hence, we have a one-to-one and onto map between the \(H\)-conjugates of \(K\) and the right cosets of \(N(K) \cap H\) in \(H\text{.}\)

Proof.

Let \(P\) be a Sylow \(p\)-subgroup of \(G\) and suppose that \(|G|=p^r m\) with \(|P|=p^r\text{.}\) Let

consist of the distinct conjugates of \(P\) in \(G\text{.}\) By Lemma 14.6, \(k = [G: N(P)]\text{.}\) Notice that

Since \(p^r\) divides \(|N(P)|\text{,}\) \(p\) cannot divide \(k\text{.}\)

Given any other Sylow \(p\)-subgroup \(Q\text{,}\) we must show that \(Q \in {\mathcal S}\text{.}\) Consider the \(Q\)-conjugacy classes of each \(P_i\text{.}\) Clearly, these conjugacy classes partition \({\mathcal S}\text{.}\) The size of the partition containing \(P_i\) is \([Q :N(P_i) \cap Q]\) by Lemma 14.6, and Lagrange's Theorem tells us that \(|Q| = [Q :N(P_i) \cap Q] |N(P_i) \cap Q|\text{.}\) Thus, \([Q :N(P_i) \cap Q]\) must be a divisor of \(|Q|= p^r\text{.}\) Hence, the number of conjugates in every equivalence class of the partition is a power of \(p\text{.}\) However, since \(p\) does not divide \(k\text{,}\) one of these equivalence classes must contain only a single Sylow \(p\)-subgroup, say \(P_j\text{.}\) In this case, \(x^{-1} P_j x = P_j\) for all \(x \in Q\text{.}\) By Lemma 14.5, \(P_j = Q\text{.}\)

Proof.

Let \(P\) be a Sylow \(p\)-subgroup acting on the set of Sylow \(p\)-subgroups,

by conjugation. From the proof of the Second Sylow Theorem, the only \(P\)-conjugate of \(P\) is itself and the order of the other \(P\)-conjugacy classes is a power of \(p\text{.}\) Each \(P\)-conjugacy class contributes a positive power of \(p\) toward \(|{\mathcal S}|\) except the equivalence class \(\{ P \}\text{.}\) Since \(|{\mathcal S}|\) is the sum of positive powers of \(p\) and \(1\text{,}\) \(|{\mathcal S}| \equiv 1 \pmod{p}\text{.}\)

Now suppose that \(G\) acts on \({\mathcal S}\) by conjugation. Since all Sylow \(p\)-subgroups are conjugate, there can be only one orbit under this action. For \(P \in {\mathcal S}\text{,}\)

by Lemma 14.6. But \([G : N(P)]\) is a divisor of \(|G|\text{;}\) consequently, the number of Sylow \(p\)-subgroups of a finite group must divide the order of the group.