Tag Archives: subgroup index

The order of conductor f in the ring of quadratic integers is a subring and has index f

Let D be a squarefree integer, and let O be the ring of integers in the quadratic field \mathbb{Q}(\sqrt{D}). (I.e. O = \mathbb{Z}[\omega].) For any positive integer f prove that the set O_f = \mathbb{Z}[f\omega] = \{ a + bf\omega \ |\ a,b \in \mathbb{Z} \} is a subring of O containing the identity. Prove that [O : O_f] = f. (The index as an abelian group.) Prove conversely that a subring of O containing the identity and having (as a subgroup) finite index f is equal to O_f. (The ring O_f is called the order of conductor f in the field \mathbb{Q}(\sqrt{D}). The ring of integers O is called the maximal order in \mathbb{Q}(\sqrt{D}).)


First we show that O is a subgroup of \mathbb{Z}[\omega]. To see this, note that 0 + 0f\omega \in O. Now let x = a_1 + a_2 f \omega and y = b_1 + b_2 f \omega be in O. Then x-y = (a_1 - b_1) + (a_2 - b_2)f \omega \in O, so that by the subgroup criterion, O is an additive subgroup of \mathbb{Z}[\omega].

Now it is easy to see that \omega^2 = D or \omega + (D-1)/4, depending on whether D is not or is 1 mod 4. Moreover, if D is 1 mod 4, then (D-1)/4 is an integer. Thus we have xy = (a_1b_1 + a_2b_2f^2D) + (a_1b_2 + a_2b_1)f \omega or (a_1b_1 + a_2b_2f^2(D-1)/4) + (a_1b_2 + a_2b_1 + a_2b_2)f \omega, depending on whether D is not or is 1 mod 4. In either case, xy \in \mathbb{Z}[f \omega]. Thus \mathbb{Z}[f \omega] \subseteq \mathbb{Z}[\omega] is a subring.

Now we will show that [\mathbb{Z}[\omega] : \mathbb{Z}[f \omega]] = f. Define a mapping \Phi : \mathbb{Z}[\omega] \rightarrow \mathbb{Z} by \Phi(a + b\omega) = b. We claim that \Phi is an additive group homomorphism. To see this, note that \Phi((a_1 + a_2 \omega)(b_1 + b_2\omega)) = \Phi((a_1 + b_1) + (a_2 + b_2)\omega) = a_2 + b_2 = \Phi(a_1 + a_2 \omega) + \Phi(b_1 + b_2 \omega). Moreover, \Phi is surjective since for all n \in \mathbb{Z}, \Phi(0 + n \omega) = n. Consider now \pi \circ \Phi : \mathbb{Z}[\omega] \rightarrow \mathbb{Z}/(f), where \pi denotes the natural projection. We claim that \mathsf{ker}(\pi \circ \Phi) = \mathbb{Z}[f\omega], as we show. (\subseteq) If \Phi(a+b\omega) = b \equiv 0 mod f, then b = b^\prime f for some integer b^\prime. Thus a + b\omega = a + b^\prime f \omega \in \mathbb{Z}[f\omega]. (\supseteq) Clearly \Phi(a+bf\omega) = bf \equiv 0 mod f. By the First Isomorphism Theorem for groups, then, \mathbb{Z}[\omega]/\mathbb{Z}[f\omega] \cong \mathbb{Z}/(f), hence [\mathbb{Z}[\omega] : \mathbb{Z}[f\omega]] = f.

Suppose now that S \subseteq \mathbb{Z}[\omega] is a subring containing 1 and having finite index f. If f = 1, then trivially, \mathbb{Z}[\omega] = \mathbb{Z}[1\omega]. If f \geq 2, then since 1 \in S, \mathbb{Z} \subseteq S, and in particular, (f-1)a \in S for all integers a.

Let a,b \in \mathbb{Z}. Since S has index f, fxS = S for all x \in \mathbb{Z}[\omega]. (Here, fx denotes the f-fold sum of x.) Specifically, f(a+b\omega)S = (fa + bf \omega)S = (f-1)aS + (a + bf\omega)S = (a+bf \omega)S; thus a + bf \omega \in S for all a,b \in \mathbb{Z}, and we have \mathbb{Z}[f\omega] \leq S.

Now f = [\mathbb{Z}[\omega] : \mathbb{Z}[f\omega]] = [\mathbb{Z}[\omega] : S] \cdot [S : \mathbb{Z}[f\omega]] = f \cdot [S : \mathbb{Z}[f\omega]]. Thus [S : \mathbb{Z}[f\omega]] = 1, and we have S = \mathbb{Z}[f\omega].

A sufficient condition for Sylow intersections of bounded index

Prove that if n_p \not\equiv 1 mod p^k, then there are distinct Sylow p-subgroups P and Q in G such that P \cap Q has index at most p^{k-1} in both P and Q.


(We will follow the strategy of this previous exercise.)

Let P \leq G be a Sylow p-subgroup. Now P acts on \mathsf{Syl}_p(G) by conjugation; note that if Q \in \mathsf{Syl}_p(G) is distinct from P, then using the Orbit-Stabilizer theorem and Lemma 4.19, we have |P \cdot Q| = [P : N_P(Q)] = [P : P \cap N_G(Q)] = [P : Q \cap P]. Since the orbit containing P itself has order 1, then n_p(G) = 1 + \sum [P : Q_i \cap P], where Q_i ranges over a set of orbit representatives. If every [P : Q_i \cap P] is divisible by p^k, then n_p(G) \equiv 1 mod p^k, a contradiction. Thus some [P : Q_i \cap P] is a power of p and is at most p^{k-1}.

We can see then that Q_i \cap P has index at most p^{k-1} in both P and Q_i.

No simple groups of order 2205, 4125, 5103, 6545, or 6435 exist

Prove that there are no simple groups of order 2205, 4125, 5103, 6545, or 6435.


We begin with a lemma.

Lemma 1: Let G be a finite simple group. If |G| does not divide k!, then for no subgroup H \leq G do we have [G:H] \leq k. Proof: Suppose to the contrary that some H exists, with [G:H] \leq k. This affords a left permutation representation G \rightarrow S_{G/H}, and the kernel of this representation is contained in H. Since the kernel is also normal in G, the kernel is trivial and the representation is injective; thus G \leq S_{G/H} \leq S_k. But by Lagrange this means that |G| divides k!, a contradiction. \square

  1. Note that 2205 = 3^2 \cdot 5 \cdot 7^2. Let G be a simple group of order 2205. Note that |G| does not divide 13! since the largest power of 7 dividing 13! is 7^1; by the lemma, G has no subgroups of index less than or equal to 13. By Sylow’s Theorem, we have n_7(G) = 15. Note that 15 \not\equiv 1 mod 49, so that by Lemma 13, there exist P_7, Q_7 \in \mathsf{Syl}_7(G) such that P_7 \cap Q_7 \neq 1. Then P_7 \cap Q_7 = P_0 \cong Z_7. Moreover, P_0 is normal in P_7 and Q_7 since it is maximal in both; thus P_7, Q_7 \leq N_G(P_0). Note that both P_7 and Q_7 are Sylow 7-subgroups of N_G(P_0), so that |N_G(P_0)| = 7^2 \cdot k where k \geq 8 (using Sylow). By Lagrange and because G is simple, we have k \in \{9,15\}. But if k=9, then N_G(P_0) has index 5, and if k=15, then N_G(P_0) has index 3, both cases yielding a contradiction.
  2. Note that 4125 = 3 \cdot 5^3 \cdot 11. Let G be a simple group of order 4125. By Sylow’s Theorem, we have n_5(G) \in \{1,11\}. Note that |G| does not divide 14!, since the largest power of 5 that divides 14! is 5^2. By the lemma, no subgroup of G has index less than 15. However, we have n_5 = 11, so that [G:N_G(P_5)] = 11 for each Sylow 5-subgroup P_5; this is a contradiction.
  3. Note that 5103 = 3^6 \cdot 7. Let G be a simple group of order 5103. By Sylow’s Theorem we have n_3(G) \in \{1,7\}. Note that |G| does not divide 8!, since the highest power of 3 which divides 8! is 3^2. By the lemma, no subgroup of G has index 7, a contradiction since [G:N_G(P_3)] = 7 for each Sylow 3-subgroup P_3.
  4. Note that 6545 = 5 \cdot 7 \cdot 11 \cdot 17. Let G be a simple group of order 6545. By Sylow’s Theorem, n_5(G) \in \{1,11\}. Now |G| does not divide 16!, so that by the lemma G contains no subgroups of index 11. This is a contradiction since [G:N_G(P_5)] = 11 for each Sylow 5-subgroup P_5.
  5. Note that 6435 = 3^2 \cdot 5 \cdot 11 \cdot 13. Let G be a simple group of order 6435. By Sylow’s Theorem, we have n_5(G) \in \{1,11\}. Now |G| does not divide 12!, so that by the lemma no subgroup has index 11. This is a contradiction since [G:N_G(P_5)] = 11 for each Sylow 5-subgroup P_5.

The center of a subgroup of index 2 in a 2-group with cyclic center has rank at most 2

Use the preceding exercise to prove that if P is a 2-group which has a cyclic center and M is a subgroup of index 2 in P, then Z(M) has rank at most 2.


Note that M \leq P is normal and that Z(M) \leq M is characteristic; thus Z(M) is normal in P. Moreover, letting Z(M)_2 = \{ z \in Z(M) \ |\ z^2 = 1 \}, we know that Z(M)_2 \leq Z(M) is characteristic, so that Z(M)_2 \leq P is normal. Moreover, Z(M)_2 is elementary abelian by construction. Note that, by §5.2 #7, Z(M) and Z(M)_2 have the same rank.

If w \in P, then conjugation by w is an automorphism of Z(M)_2. If we choose w \notin M, and let z be fixed by w– that is, wz = zw, then we have z \in C_P(M) (since z \in Z(M)) and z \in C_P(\langle w \rangle). Since P = M \langle w \rangle, in fact w \in Z(P). Thus the subgroup of Z(M)_2 which is fixed under conjugation by w is contained in Z(P), and thus has rank 1.

By this previous exercise, Z(M)_2 has rank at most 2, so that Z(M) also has rank at most 2.

Every maximal subgroup of a finite solvable group has prime power index

Prove that every maximal subgroup of a finite solvable group has prime power index.


Let G be a finite solvable group. We will proceed by induction on the breadth k of G; that is, the number of prime factors in the factorization of |G|, including multiplicity.

For the base case, if k = 1, we have G \cong Z_p for some prime p. Clearly then the (unique) maximal subgroup of G, namely 1, has prime power index.

For the inductive step, suppose that every finite solvable group of breadth at most k satisfies the conclusion. Let G be a finite solvable group of breadth k+1.

If G is simple, then G^\prime = 1 (since G is solvable). Thus G is an abelian simple group, and we have G \cong Z_p for some prime p; this is a contradiction.

Suppose now that G is not simple, so that nontrivial normal subgroups exist. H \leq G be a maximal subgroup and let M \leq G be a minimal nontrivial normal subgroup. Now G/M is a finite solvable group of breadth at most k. There are two cases:

If M \leq H, then H/M \leq G/M is maximal by the Lattice Isomorphism Theorem. By the Third Isomorphism Theorem and the induction hypothesis, [G:H] = [G/M : H/M] is a prime power.

If M \not\leq H, then G = MH. Now [G:H] = [HM:H] = |HM|/|H| = (|H| \cdot |M|/|H \cap M|)/|H| = |M|/|H \cap M| is a prime power, since M is a p-group.

Maximal subgroups of a finite nilpotent group have prime index

Prove that a maximal subgroup of a finite nilpotent group has prime index.


Let G be a finite nilpotent group and M \leq G a maximal subgroup. By Theorem 3, M < N_G(M) is a proper subgroup, so that (since M is maximal) N_G(M) = G. Thus M \leq G is normal. Now G/M is a finite nilpotent group.

Suppose [G:M] is composite, with a proper prime divisor q. By this previous exercise, there is a normal subgroup \overline{H} \leq G/M. By the Lattice Isomorphism Theorem, \overline{H} = H/M for some normal subgroup H \leq G with M \leq H. Since M is maximal, either H = M or H = G, so that q is not a proper divisor of [G:M]– a contradiction.

Thus [G:M] is prime.

Properties of the image and kernel of the p-power map on a finite abelian group

Let A be a finite abelian group (written multiplicatively) and let p be a prime. Let A^p = \{ a^p \ |\ a \in A \} and A_p = \{ x \in A \ |\ x^p = 1 \}. (I.e. A^p and A_p are the image and kernel of the p-power map, respectively.

  1. Prove that A/A^p \cong A_p. (Show that they are both elementary abelian and have the same order.)
  2. Prove that the number of subgroups of A of order p equals the number of subgroups of index p. (Reduce to the case where A is an elementary abelian p-group.)

  1. We begin with some lemmas.

    Lemma 1: Let A and B be abelian groups and let \theta : A \rightarrow B be an isomorphism. Then there exists an isomorphism \theta^\prime : A/A^p \rightarrow B/B^p. Proof: Let \pi denote the natural projection from Z to Z/Z^p (for Z \in \{A,B\}). If x \in A^p, then x = y^p for some y \in A. Then (\pi \circ \theta)(x) = (\pi \circ \theta)(y^p) = \pi(\theta(y)^p) = B^p, so that A^p \leq \mathsf{ker}\ (\pi \circ \theta). By the remarks on page 100 in D&F, there exists a unique group homomorphism \theta^\prime : A/A^p \rightarrow B/B^p such that \theta^\prime \circ \pi = \pi \circ \theta. (\theta^\prime is injective) Suppose xA^p \in \mathsf{ker}\ \theta^\prime. Then \theta^\prime(\pi(x)) = 1, thus (\theta^\prime \circ \pi)(x) = 1, so that (\pi \circ \theta)(x) = 1, hence \theta(x) \in B^p. Now since \theta is surjective, \theta(x) = \theta(y)^p for some y \in A. Thus x = y^p, so that x \in A^p. Thus \mathsf{ker}\ \theta^\prime = 1, so that \theta^\prime is injective. (\theta^\prime is surjective) Let xB^p \in B/B^p. Now xB^p = (\pi \circ \theta)(y) for some y \in A, so that xB^p = \theta^\prime(\pi(y)). Hence \theta^\prime is surjective. \square

    Lemma 2: Let A and B be abelian groups and let \theta : A \rightarrow B be an isomorphism. Then \theta[A_p] = B_p. Proof: (\subseteq) If x \in \theta[A_p], then x = \theta(y) where y \in A_p. Now y^p = 1, so that \theta(y)^p = 1, hence x^p = 1. Thus x \in B_p. (\supseteq) Let x \in B_p. Then x^p = 1$. Since \theta is surjective, x = \theta(y) for some y \in A. Now \theta(y^p) = \theta(y)^p = x^p = 1, and since \theta is injective, y^p = 1. Thus y \in A_p, hence x \in \theta[A_p]. \square

    Lemma 3: Let G be a finite abelian group of order n, let m be an integer, and let \varphi : G \rightarrow G denote the m-power homomorphism. If \mathsf{gcd}(m,n) = 1, then \varphi is an isomorphism. Proof: Since G is finite, it suffices to show that \varphi is injective. Let x \in \mathsf{ker}\ \varphi. Then x^m = 1, so that |x| divides m. By Lagrange, |x| divides n. Thus |x| = 1, hence x = 1, and \varphi is injective. \square

    Now to the main result. By Theorem 5 in the text, A \cong A_1 \times A_2 where A_1 is a p-group of rank t and p does not divide |A_2|. By a lemma to the previous exercise, \varphi = \varphi_1 \times \varphi_2, where each map is the p-power map on A_1 \times A_2, A_1, and A_2, respectively.

    We have \mathsf{ker}\ \varphi = \mathsf{ker}(\varphi_1 \times \varphi_2) = (\mathsf{ker}\ \varphi_1) \times (\mathsf{ker}\ \varphi_2) \cong Z_{p^t} \times 1 \cong Z_{p^t} by the previous exercise.

    Similarly, (A_1 \times A_2)/\mathsf{im}\ \varphi = (A_1 \times A_2)/(\mathsf{im}\ \varphi_1 \times \mathsf{im}\ \varphi_2) \cong (A_1/\mathsf{im}\ \varphi_1) \times (A_2/\mathsf{im}\ \varphi_2) \cong Z_{p^t} \times 1 \cong Z_{p^t}, again using the previous exercise.

    Now the first conclusion follows using Lemmas 1 and 2.

  2. We begin with some more lemmas.

    Lemma 4: Let E_{p^k} be the elementary abelian group of order p^k, and suppose E_{p^k} = \langle S \rangle where |S| = k. Finally let G be a group. If \overline{\varphi} : S \rightarrow G is a mapping such that |\overline{\varphi}(s)| = p for each s \in S and \overline{\varphi}(s) \overline{\varphi}(t) = \overline{\varphi}(t) \overline{\varphi}(s) for all s,t \in S, then \overline{\varphi} extends uniquely to a group homomorphism \varphi : E_{p^k} \rightarrow G. Proof: Every element of E_{p^k} can be written uniquely as x = s_1^{a_1}s_2^{a_2} \cdots s_k^{a_k}, where s_1 \in S and 0 \leq a_k < p. Define \varphi(x) = \overline{\varphi}(s_1)^{a_1}\overline{\varphi}(s_2)^{a_2} \cdots \overline{\varphi}(s_k)^{a_k}. \varphi is well defined since the S-expansion of x is unique, and is a homomorphism since the \overline{\varphi}(s) commute with one another. To see uniqueness, suppose \psi : E_{p^k} \rightarrow G is a group homomorphism such that \psi(s) = \varphi(s) for all s \in S. Then \psi(x) = \psi(s_1^{a_1}s_2^{a_2} \cdots s_k^{a_k}) = \psi(s_1)^{a_1} \psi(s_2)^{a_2} \cdots \psi(s_k)^{a_k} = \overline{\varphi}(s_1)^{a_1}\overline{\varphi}(s_2)^{a_2} \cdots \overline{\varphi}(s_k)^{a_k} = \varphi(x). Thus \varphi is unique. \square

    Note that every subgroup of A of order p is contained in \mathsf{ker}\ \varphi, so that in fact the order p subgroups of A and \mathsf{ker}\ \varphi coincide. Now because \mathsf{ker}\ \varphi \cong Z_{p^t} is elementary abelian, every nonidentity element has order p and thus generates an order p subgroup. Each such subgroup is generated by p-1 elements; thus there are (p^t-1)/(p-1) order p subgroups in \mathsf{ker}\ \varphi, thus in A.

    Let B \leq A be a subgroup of index p. Now let x \in A^p, say x = y^p, and suppose x \notin B. Then y \notin B, so that A = \langle y \rangle B. Now A/B is a group of order p, so that (yB)^p = xB = B; then x \in B, a contradiction. Thus A^p \leq B. Moreover, by the Third Isomorphism Theorem, [A/A^p : B/A^p] = [A:B] = p. By the Lattice Isomorphism Theorem, the index p subgroups of A/A^p \cong Z_p^t correspond precisely to the index p subgroups of A. In particular, it suffices to count the order and index p subgroups in an elementary abelian p-group.

    Now \mathsf{Aut}(Z_p^t) acts on \mathcal{M} = \{ M \leq Z_p^t \ |\ [Z_p^t : M] = p \} by \varphi \cdot M = \varphi[M]. We claim that this action is transitive. To that end, let M_1, M_2 \leq Z_p^t be index p subgroups. Now M_1 \cong M_2 \cong Z_p^{t-1} since M_1 and M_2 are elementary abelian, thus M_1 = \langle S_1 \rangle and M_2 = \langle S_2 \rangle where |S_1| = |S_2| = t-1. Let \overline{\psi} : S_1 \rightarrow S_2 be a bijection, and choose some t_1 \in Z_p^t \setminus M_1 and t_2 \in Z_p^t \setminus M_2. Now \overline{\psi} \cup \{(t_1,t_2)\} extends by Lemma 4 to a homomorphism \psi : Z_p^t \rightarrow Z_p^t. Clearly \psi is surjective, hence an isomorphism, so that \psi \in \mathsf{Aut}(Z_p^t). Moreover, we see that \psi[M_1] = M_2. Thus this action is transitive, and we have \mathsf{Aut}(Z_p^t) \cdot M = \mathcal{M} for all index p subgroups M.

    Now let M \leq Z_p^t be an arbitrary index p subgroup and let \varphi be an automorphism of Z_p^t which stabilizes M. As above, M = \langle S \rangle for some set S with |S| = t-1. Moreover, if x \in Z_p^t \setminus M then Z_p^t = \langle S \cup \{x\} \rangle since M is maximal. By Lemma 4, \varphi is determined uniquely by is action on S and x. We see also that since \varphi stabilizes M, we must have \varphi(s) \in M for each s \in S. Since M \cong Z_p^{t-1} and \varphi|_M \in \mathsf{Aut}(M), there are \prod_{i=0}^{t-2}(p^{t-1} - p^{t-2-i}) distinct choices we can make for \varphi[S]. Moreover, \varphi(x) may be chosen arbitrarily so long as \varphi(x) \notin M; thus there are p^t - p^{t-1} = p^{t-1}(p-1) choices for \varphi(x). Thus in total \mathsf{Stab}(M) contains (p-1)p^{t-1} \prod_{i=0}^{t-2}(p^{t-1} - p^{t-2-i}) elements.

    By the Orbit-Stabilizer Theorem and Lagrange's Theorem, |\mathcal{M}| = (\prod_{i=0}^{t-1} (p^t - p^{t-1-i}))/((p-1)p^{t-1} \prod_{i=0}^{t-2}(p^{t-1} - p^{t-2-i})) = (\prod_{i=1}^t (p^t - p^{t-i}))/((p-1) \prod_{i=1}^{t-1}(p^{t} - p^{t-i})) = (p^t-1)/(p-1).

    Thus the theorem is proved.

For n at least 5, Alt(n) is the only subgroup of index less than n in Sym(n)

Prove that A_n is the only proper subgroup of index less than n in S_n for n \geq 5.


Let n \geq 5 and let H \leq S_n be a proper subgroup. Now S_n acts on the left cosets of H by left multiplication. Let \varphi : S_n \rightarrow S_{[S_n:H]} be the permutation representation induced by this action, and let K be its kernel. Note that K \leq H is proper in S_n and that K is normal in S_n. By this previous exercise, there are two possibilities.

If K = A_n, then since A_n is maximal, H = A_n.

If K = 1, then \varphi is injective and we have S_n \leq S_{[S_n : H]}. If [S_n : H] < n, this yields a contradiction.

Thus for n \geq 5, S_n has precisely one proper subgroup of index less than n, namely A_n.

For n at least 5, the index of a subgroup of Alt(n) is at least n

Prove that A_n does not have a proper subgroup of index less than n for all n \geq 5.


We begin with a lemma.

Lemma: If n \geq 3, then n!/2 > (n-1)!. Proof: For the base case, note that 3!/2 = 3 > 2 = 2!. For the inductive step, suppose n!/2 > (n-1)!. Now n+1 > n, so that (n+1)!/2 > n!. \square

Let n \geq 5, and let H < A_n be a proper subgroup. Note that A_n acts on the left cosets of H by left multiplication; let \varphi : G \rightarrow S_{[A_n : H]} be the permutation representation induced by this action, and let K be the kernel of this action. Note that K \leq \mathsf{stab}(H) = H < A_n. Since A_n is simple and K is normal, we have K = 1. Thus \varphi is injective.

If [A_n : H] < n, we have A_n \leq S_{n-1}. By the lemma this is a contradiction. Hence no proper subgroup has order less than n.

A congruence condition on the index of subgroups containing Sylow normalizers

Let P be a Sylow p-subgroup of G and let M be any subgroup of G which contains N_G(P). Prove that [G : M] \equiv 1 \mod p.


We prove a slightly stronger result by induction. First a definition.

Let G be a finite group and H \leq G a subgroup. The distance from H to G is the length d of the longest chain of proper subgroups K_i such that H = K_0 < K_1 < \ldots < K_d.

Lemma: Let G be a finite group and P \leq G a Sylow p-subgroup. If H,K \leq G are subgroups such that N_G(P) \leq H \leq K, then [K : H] \equiv 1 \mod p. Proof: We proceed by induction on the distance d from N_G(P) to K. For the base case, if d = 0, then K = N_G(P) so that H = N_G(P), and we have [K : H] = 1. For the inductive step, suppose that for some d \geq 0, if the distance from N_G(P) to K is at most d then for all H such that N_G(P) \leq H \leq K we have [K : H] \equiv 1 \mod p. Now let K \leq G be a subgroup containing N_G(P) such that the distance from N_G(P) to K is d+1, and suppose further that N_G(P) \leq H \leq K. First, if H = K, we have [K : H] = 1 so that the conclusion holds. Suppose now that the inclusion H \leq K is proper. Note that [K : N_G(P)] = [K : H] \cdot [H : N_G(P)]. By the induction hypothesis, we have [H : N_G(P)] \equiv 1 \mod p. Recall that N_K(P) = K \cap N_G(P), so that [K : N_G(P)] = [K : N_K(P)] = n_p(K) \equiv 1 \mod p by Sylow's Theorem. Modulo p, we thus have [K:H] \equiv 1. \square

In particular, if K = G and N_G(P) \leq M \leq G, we have [G:M] \equiv 1 \mod p.