Let \(C(x)\) be the set of elements of \(G\) which are conjugate in \(G\) to \(x\), we call this the conjugacy class of \(x\). Bhattacharya shows that the set of all conjugacy classes form a partition of \(G\) [ 2 , p.112 ] . Now consider the following rearranged class equation of \(G\), where \(S\) is a subset of \(G\) containing exactly one element from each conjugacy class not contained in \(Z(G)\).

Since \(|G| = p^m\), each subgroup of \(G\) is of order \(p^k\) for some \(k \leq m\). In particular each \(N_G(x)\) has order \(p^k\) and is strictly contained in \(G\) since \(x \not\in Z(G)\) by assumption. Thus each \([G:N_G(x)] {\gt} 1\), and are therefore divisible by \(p\). Since \(p\) divides the left hand side of (1), it must also divide the right, thus \(p\) divides \(|Z(G)|\).

See [ .

(i) Let \(x\) be chosen arbitrarily from \(G \! \setminus \! Z\). Then by Corollary 5.24, \(C_L(x)\) is abelian. By definition, \(C_G(x) = C_L(x) \cap G\), and using the elementary fact that the intersection of 2 groups is itself a group, we have \(C_G(x) {\lt} C_L(x)\). Now since every subgroup of an abelian group is abelian, \(C_G(x)\) is also abelian.

Now let \(J\) be a maximal abelian subgroup of \(G\) containing \(C_G(x)\). Since \(J\) is abelian and \(x \in C_G(x) \subset J\), we have \(jx=xj\), \(\forall j \in J\), thus \(J \subset C_G(x)\). Therefore \(J=C_G(x)\) and \(C_G(x) \in \mathfrak {M}\).

(ii) Consider \(x \in A \cap B\). Since both \(A\) and \(B\) are abelian, \(x\) commutes with each \(a \in A\) and \(b \in B\) and thus \(C_G(x)\) contains both \(A\) and \(B\). If \(x \in G \setminus Z\), then \(C_G(x) \in \mathfrak {M}\) by (i) and because \(A\) and \(B\) are distinct we have \(A \subsetneq A \cup B \subset C_G(x)\). This contradicts the fact that \(A\) is maximum abelian and thus \(x \in Z\). Finally, note that Z is contained in every maximal abelian subgroup, since otherwise we would have the contradiction that \(\langle A, Z \rangle \) would generate a larger abelian subgroup than \(A\). Hence \(A \cap B = Z\).

(iii) First consider the trivial case of \(G=Z\). Here \(G\) is the only element of \(\mathfrak {M}\). If \(p \neq 2\) then \(|G|=2\) and \(G\) is a cyclic group whose order is relatively prime to \(p\). If \(p=2\) then \(G = I_G\) which is trivially a \(S_p\)-subgroup.

Now assume \(G \neq Z\). Since \(Z \not\in \mathfrak {M}\), each \(A \in \mathfrak {M}\) contains at least one \(x \not\in Z\). By Proposition 5.18 this \(x\) is conjugate to either \(d_\omega \) or \(\pm t_\lambda \) in \(L\). It suffices to only consider these cases:

\(\pmb {x}\) conjugate to \(\pmb {d_\omega }\) in \(\pmb {L}\). There is a \(y \in L\) such that \(x = y d_\omega y^{-1}\). Since \(x \not\in Z\), we have \(d_\omega \not\in Z\), because otherwise we get the contradiction,

Thus \(\omega \neq \pm 1\). Let \(A = C_G(x)\), since \(C_G(x) \in \mathfrak {M}\) by part (i). Observe that

Since \(A\) is conjugate to \(C_G(d_\omega )\) by Proposition 5.23, we have that \(A\) is isomorphic to a finite subgroup of \(F^*\) and by Lemma 6.5, \(A\) is cyclic. By Lagrange’s Theorem any finite subgroup of \(F^*\) has an order which divides \(p^m - 1\) for some \(m \in \mathbb {Z}^+\), and since \(p \nmid (p^m - 1)\), \(|A|\) is relatively prime to \(p\).

\(\pmb {x}\) conjugate to \(\pmb {\pm t_\lambda }\) in \(\pmb {L}\). Again let \(A = C_G(x) \in \mathfrak {M}\). \(A\) is conjugate to \(C_G({\pm t_\lambda })\) in \(L\) by Proposition 5.23. Since \(x \notin Z\), we have \(\lambda \neq 0\). Observe that

So \(A\) is isomorphic to a finite subgroup of \(F \times Z\), call it \(Q \times Z\). Now \(A = Q \times Z \cong QZ\) by Corollary 3.21, which means that an arbitrary element of \(A\) is of the form \(q_1z_1\), where \(q_1 \in Q\), \(z_1 \in Z\).

Thus \(Q\) is also abelian. Recall from the proof of Proposition 5.18(ii) that all non-trivial elements of \(T\) have order \(p\), so each non-trivial element of \(Q\) has order \(p\) which means that \(Q\) is elementary abelian. Thus \(Q\) has order \(p^m\), for some \(m \in \mathbb {Z}^+\).

Now let \(S\) be a Sylow \(p\)-subgroup containing \(Q\). We apply Lemma 6.4 to determine that \(p\) divides \(|Z(S)|\), moreover \(|Z(S)| \geq p\).

If \(p=2\), then \(Z=I_L\) by Lemma 5.16. So \(|Z| = 1\) and hence \(|Z(S)| \geq 2 {\gt} |Z|\).
If \(p {\gt} 2\), then \(Z = \langle - I_L \rangle \) also by Lemma 5.16. So \(|Z| = 2\) and again we get \(|Z(S)| {\gt} 2 = |Z|\).

So \(Z(S)\) must contain at least one element which is not in \(Z\), let \(y\) be one such element. Let \(s_1z_1\) be an arbitrary element of \(S \times Z\).

Thus \(s_1z_1 \in C_G(y)\) and since it was chosen arbitrarily, \(S \times Z \subset C_G(y)\). Also since \(y \in G \! \setminus \! Z\) we have \(C_G(y) \in \mathfrak {M}\) by part (i).

Since \(A\) and \(C_G(y)\) are both in \(\mathfrak {M}\) it must be that \(A = C_G(y)\). This means \(Q = S\) and \(Q\) is a Sylow \(p\)-subgroup of G.

(iv) If \(|A| \leq 2\) then \(A=Z=G\). So \(A\) is trivially normal in \(G\) and \([N_G(A): A] = 1\).

Now assume that \(|A| {\gt} 2\). Since \(|A|\) is relatively prime to \(p\), we have that \(A\) is a cyclic group conjugate to a finite subgroup of \(D\) in \(L\) by the proof of part (iii), call this subgroup \({\widetilde{A}}\). Thus both \({\widetilde{A}}\) and \(D\) have orders greater than 2. Applying Proposition 5.22 we observe that

Since \(A\) and \({\widetilde{A}}\) are conjugate in \(L\), there exists an element \(z \in L\) such that \(zAz^{-1} = {\widetilde{A}}\). This \(z\) determines an inner automorphism of \(L\) defined by

Let \(i_z(G) = {\widetilde{G}}\) denote the image of \(G\) under \(i_z\). Since \(A\) is a maximal abelain subgroup of \(G\) it’s a simple task to show that \({\widetilde{A}}\) is a maximal abelian subgroup of \({\widetilde{G}}\) and I will leave this to the reader to verify. We now show that \(i_z(N_G(A)) = N_{\widetilde{G}}({\widetilde{A}})\) . Take an arbitrary \(g \in N_G(A)\).

So \(z g z^{-1} = i_z(g) \in N_{\widetilde{G}}({\widetilde{A}})\) and since it was chosen arbitrarily, \(i_z(N_G(A)) \subset N_{\widetilde{G}}({\widetilde{A}})\). Now take an arbitrary \(z h z^{-1} \in N_{\widetilde{G}}({\widetilde{A}})\).

Now multiplication on the left by \(z^{-1}\) and right by \(z\) gives:

so \(h \in N_G(A)\). Furthermore, \(z h z^{-1}\) and indeed the whole of \(N_{\widetilde{G}}({\widetilde{A}})\) is contained in \(i_z(N_G(A))\). Thus \( i_z(N_G(A)) = N_{\widetilde{G}}({\widetilde{A}})\). In particular, we have,

Since \({\widetilde{G}} {\lt} L\), the normaliser of \({\widetilde{A}}\) in \({\widetilde{G}}\) is simply the normaliser of \({\widetilde{A}}\) in \(L\) restricted to \({\widetilde{G}}\), thus \(N_{\widetilde{G}}({\widetilde{A}}) {\lt} N_L({\widetilde{A}}) = N_L(D)\) by (2). Now since \(D \vartriangleleft N_L(D)\), the Second Isomorphism Theorem shows that,

Clearly \({\widetilde{A}} \subset {\widetilde{G}} \cap D\). We show that this inclusion is infact an equality. Assume that there exists some \(d_\omega \in {\widetilde{G}} \cap D\) which is not in \({\widetilde{A}}\). The group \(\langle d_\omega , {\widetilde{A}} \rangle \) is thus an abelian subgroup of \({\widetilde{G}}\), strictly larger than \({\widetilde{A}}\) and contradicting the fact that \({\widetilde{A}}\) is maximal abelian in \({\widetilde{G}}\). Thus \({\widetilde{A}} = {\widetilde{G}} \cap D\). It is trivial to see that \({\widetilde{A}} \subset N_{\widetilde{G}}({\widetilde{A}}) \cap D\). Also \(N_{\widetilde{G}}({\widetilde{A}}) \cap D \subset {\widetilde{G}} \cap D = {\widetilde{A}}\). So,

Observe also that,

Now we piece the preceding results together to give the desired result.

We have shown that \(N_{\widetilde{G}}({\widetilde{A}}) / {\widetilde{A}}\) is isomorphic to a subset of \(\mathbb {Z}_2\). Thus by (3) we have established that,

For the second part, if \([N_G(A): A] = 2\), then the above argument shows that \(N_{\widetilde{G}}({\widetilde{A}}) / {\widetilde{A}} \; \cong \; \mathbb {Z}_2\). Thus \(DN_{\widetilde{G}}({\widetilde{A}}) = N_L(D) = \langle D, w \rangle \). This means that \(N_{\widetilde{G}}({\widetilde{A}})\) contains some element \(wd_\omega \). In fact, since \(w d_\omega \not\in D\), we have \(w d_\omega \in N_{\widetilde{G}}({\widetilde{A}}) \! \setminus \! {\widetilde{A}}\). Take any element \(x \in A\). Since \({\widetilde{A}} = zAz^{-1}\), \(zxz^{-1} \in {\widetilde{A}}\), call it \(d_\sigma \). Let \(y = z^{-1}w d_\omega z\). Since \(wd_\omega \in N_{\widetilde{G}}({\widetilde{A}}) \! \setminus \! {\widetilde{A}}\) it follows that \(y \in N_G(A)\! \setminus \! A\). We show that this \(y\) inverts \(x\):

(v) By part (iii), \(Q\) is conjugate to a finite subgroup of \(T\) in \(L\). In fact, without loss of generality we can assume that \(Q \subset T\), moreoever \(Q \subset T \cap G\). We show that this is in fact an equality by showing that the reverse inclusion also holds. Let \(t_\lambda \) be an arbitrary element of \(T \cap G\). Then \(\langle t_\lambda , Q \rangle \) is a \(p\)-group of \(G\) which must be equal to \(Q\) since it is a Sylow \(p\)-subgroup of \(G\). Thus \(t_\lambda \in Q\) and

Since \(|Q| {\gt} 1\), Proposition 5.21 gives that \(N_G(Q) \subset N_L(Q) \subset H\). So \(N_G(Q) \subset H \cap G\). Now take an arbitrarily chosen \(d_\omega t_\lambda \in H \cap G\) and \(t_\mu \in Q\).

Since it is a product of elements of \(G\), \(t_\sigma \in T \cap G = Q\) by (7). Thus \(d_\omega t_\lambda \in N_G(Q)\) and indeed the whole of \(H \cap G\) is contained in \(N_G(Q)\) and

We now define a map \(\phi \) by,

Next we determine the kernel of \(\phi \).

We show that \(\phi \) is a group homomorphism. Take \(d_\omega t_\lambda \), \(d_\rho t_\mu \) from \( N_G(Q)\).

Thus by the First Isomorphism Theorem,

Since \(N_G(Q)\) is a finite group, it’s image under \(\phi \) is thus a finite subgroup of \(D\). Furthermore, since \(D \cong F^*\) (by Lemma ??), \(\phi (N_G(Q))\) is a cyclic group whose order divides \(p^m-1\) and is therefore relatively prime to \(p\), and by 9, so too is \(N_G(Q) / Q\).

Let \(r\) be the order of \(N_G(Q) / Q\). Since it is cyclic, \(N_G(Q)/Q\) is generated by a single element, namely a coset of \(Q\) in \(N_G(Q)\), call it \(kQ\). So \(|kQ| = r\). Observe that,

Since \(Q\) is elementary abelian, each of it’s non-trivial elements has order \(p\), so \(k\) has order \(r\) or \(rp\). In either case, since gcd\((r,p)=1\), the order of \(k^p\) is \(r\). Let \(K = \langle k^p \rangle \). Now \(|K| = r\) and

Thus,

Now assume \(|K| {\gt} |Z|\). Since \(K\) is abelian, it must be contained in some maximal abelian group \(A \in \mathfrak {M}\). By part (iii), \(A\) must also be a cyclic group whose order is relatively prime to \(p\).

Since \(A\) is conjugate in \(L\) to a subgroup of \(D\), each non-central element of \(A\) has exactly 2 fixed points on the projective line \(\mathscr {L}\) by Proposition 5.28. Let \(A = \langle x \rangle \) and let \(P_1\) and \(P_2\) be the points fixed by \(x\). We show by induction on \(n\) that \(x^n\) also fixes \(P_1\) and \(P_2\), for all \(n \in \mathbb {Z^+}\). We do this by assuming first that \(x^{n-1}\) fixes \(P_i\).

The importance of this is that since each element of \(A\) can be expressed as some power of \(x\), they must have the same two fixed points, namely \(P_1\) and \(P_2\). In other words,

By Proposition 5.28(ii), each element of \(T\) has a common fixed point \(P\) and Stab\((P) = H\). Since \(K \subset H\), each element in \(K\) fixes \(P\). Also, since \(K \subset A\), this \(P\) must be equal to either \(P_1\) or \(P_2\). Therefore by (11), \(A \subset \text{Stab}(P) = H\). We arrive at the following result:

Furthermore, we get,

Thus \(K \in \mathfrak {M}\).

(i) Define a relation \(\sim \) on \(\mathfrak {M}^*\) as follows:

If we choose \(x \in A_i^*\), then clearly \(A_i^* = A_i^*xx^{-1} = xA_i^*x^{-1}\), thus \(A_i^* \sim A_i^*\) and \(\sim \) is reflexive.

If \(A_i^* \sim A_j^*\), then \(\exists \; x \in G\) such that,

Thus \(A_j^* \sim A_i^*\) and \(\sim \) is symmetric.

If \(A_i^* \sim A_j^*\) and \(A_j^* \sim A_k^*\), then \(\exists \; x, y \in G\) such that,

Thus \(A_i^* \sim A_k^*\) (since \(xy \in G\)), which shows that \(\sim \) is transitive and moreover an equivalence relation on \(\mathfrak {M}^*\).

The equivalence class of \(A_i^*\) in \(\mathfrak {M}^*\) therefore coincides with the set \(\mathcal{C}_i^* = \{ xA_i^*x^{-1} : x \in G \} \). Furthermore, this tells us that each \(A_i^*\) belongs to exactly one conjugacy class. Thus the conjugacy classes \(\mathcal{C}_i^*\) form a partition of \(\mathfrak {M}^*\),

Since the set of \(\mathcal{C}_i^*\) are pairwise disjoint, it follows that the set of \(C_i^*\) are also pairwise disjoint and we get the desired result,

(ii) Let \(x A_i x^{-1} \in \mathcal{C}_i\) and \(x A_i^* x^{-1} \in \mathcal{C}_i^*\). Since \(x A_i x^{-1} \! \setminus \! Z = x A_i^* x^{-1}\), it is quite clear that,

Thus \(|\mathcal{C}_i^*| = |\mathcal{C}_i|\) as desired.

(iii) Now we define a map \(\phi \) by:

Clearly \(\phi \) is trivially surjective. We now show that it is both well-defined and injective.

Hence \(\phi \) is well-defined and injective. This shows that \(\phi \) is a bijection proving that \(|\mathcal{C}_i| = [G:N_G(A_i)]\). This is a crucial result which shows that the number of maximal abelian subgroups conjugate to \(A_i\) is equal to the index of the normaliser of \(A_i\) in \(G\).

(iv) This follows directly from parts (i), (ii) and (iii) and 12.

(iii) Let \(x \in N_G(A^*)\). Take an arbitary \(a \in A = A^* \cup Z\). If \(a \in A^*\), then since \(x \in N_G(A^*)\), we have \(xax^{-1} \in A^* \subset A\). If \(a \in Z\), then \(xzx^{-1} = zxx^{-1} = z \in A\). Therefore \(x\) is in the normaliser of \(A\) and \(N_G(A^*) \subset N_G(A)\).

Conversely, take \(y \in N_G(A)\) and \(a \in A^*\). \(yay^{-1} \in A = A^* \cup Z\). If \(yay^{-1} \in Z\), then

This contradicts the fact that \(a \in A^*\). Therefore \(yay^{-1} \in A^*\) and \(y \in N_G(A^*)\). Since \(y\) was chosen arbitrarily we get \(N_G(A) \subset N_G(A^*)\) and hence \(N_G(A) =N_G(A^*)\).

If \(p= 2\) then \(Z = I_G\) and the result is trivial. Now assume \(p \neq 2\). Thus \(|Z| = 2\). Let \(x\) and \(q_1\) be arbitrarily chosen elements of \(N_G(Q)\) and \(Q\) respectively.

Thus any element \(x\) which is in \(N_G(Q)\) is also in \(N_G(Q \times Z)\) so we have \(N_G(Q) \subset N_G(Q \times Z)\).

Let \(q_1 z_1\) be an arbitrarily chosen element of \(Q \times Z\) such that \(q_1 \in Q\) and \(z_1 \in Z\). Now let \(y\) be an arbitrarily chosen element of \(N_G(Q \times Z)\).

Consider now the order of \(q_1z_1\) in \(G\). Since \(p \neq 2\), \(Q \cap Z = I_G\) and \(|q_1 z_1| = |q_1| |z_1|\). Note that \(q_1 z_1\) and \(q_2 z_2\) are conjugate in \(G\), and thus their orders are equal. This means that \(|z_1| = |z_2|\), because otherwise 2 would divide one of them and not the other. Thus \(z_1 = z_2\) and,

Hence \(y \in N_G(Q)\). Furthermore, since \(y\) was chosen arbitrarily, any element which is in \(N_G(Q \times Z)\) is also in \(N_G(Q)\), so \(N_G(Q \times Z) = N_G(Q)\) as desired.