Automorphism Group of the Symmetric Group of Order n (4.4.18)

Dummit and Foote Abstract Algebra, section 4.4, exercise 18:

This exercise shows that for n≠6 every automorphism of S_n is inner. Fix an integer n > 1 with n≠6.
(a) Prove that ∀σ∈Aut(G) and each conjugacy class K of G the set σ(K) is also a conjugacy class of G, ie σ permutes the conjugacy classes of G.
(b) Let K be the conjugacy class of transpositions in S_n and let K' be the conjugacy class of any element of order 2 in S_n that is not a transposition. Prove that |K|≠|K'|. Deduce that any automorphism of S_n maps transpositions to transpositions [see exercise 33 in section 3].
(c) Prove that for any σ∈Aut(S_n), we have σ: ( 1 2 ) → ( a b₂ ), ( 1 3 ) → ( a b₃ ), ..., ( 1 n ) → ( a b_n ), for distinct integers a, b₂, b₃, ..., b_n.
(d) Show that the transpositions moving 1 generate S_n and deduce that any automorphism of S_n is determined by its action on these elements. Use (c) to show that S_n has at most n! automorphisms and conclude that Aut(S_n) = Inn(S_n) for n≠6.

Proof: (a) Take a representative x of K. We have that for any y∈K, y = gxg^-1 for some g, so that σ(y) = σ(gxg^-1) = σ(g)σ(x)σ(g)^-1∈σ(K), thus every element of σ(K) is within the same conjugacy class. Furthermore, for any g, σ(g)σ(x)σ(g)^-1 = σ(gxg^-1)∈σ(K), so that σ(K) contains the conjugacy class of σ(x). Therefore, σ(K) is exactly one conjugacy class.

(b) For σ a transposition S_n, we have that its cycle type consists of n-2 1s and exactly one 2. For z∈S_n of order 2 but not a transposition, we have that its cycle type consists of n-2x 1s and x 2s for some integer x > 1. By the aforementioned exercise, we can calculate their conjugacy class sizes in general form: σ's conjugacy class is of order n!/((n-2)!*2), and z's conjugacy class is of order n!/((n-2x)!*x!*2^x). The two being equal would imply that (n-2)!*2 = (n-2x)!*x!*2^x. Prove that this is impossible for n > 6 using induction: If for all 2 ≤ x ≤ n/2 we have (n-2)!*2 > (n-2x)!*x!*2^x, then for all 2 ≤ x ≤ (n+2)/2 we have ((n+2)-2)!*2 = (n-2)!*2*(n-1)*n > ((n+2)-2x)!*x!*2^x = (n-2x)!*x!*2^x*(n-2x+1)*(n-2x+2). This is evident for 2 ≤ x ≤ n/2 since (n-1)*n > (n-2x+1)*(n-2x+2) for x > 2. There is only one other possibility for x, then, namely x = (n+2)/2 if n is even, or x = (n+1)/2 if n is odd. Even if x is maximized to (n+2)/2 regardless of n's evenness or oddness, we have, after the substitution and cleaning up, (n-2)! > (n/2 + 1)!*2^n/2, since it's true for the base cases n = 7 and n = 8, and if it's true for n > 6, then it's true for n+2, as the left side of the inequality gets multiplied by (n-1)*n, and the right side only by (n/2+2)*2. Now, since the original inductively attacked equation has base cases true for n = 7 and n = 8 (easily checked by force), we have the complete result proven after routinely checking that the equation does not hold for n = 2, 3, 4, or 5 and any possible values of x.

(c) Keep in mind isomorphisms preserve the order of the elements they permute. Take σ( ( m k ) ) = ( a b₂ ) and σ( ( m n ) ) = ( x bn ) for n≠1,2. If a = x and b₂ = bn, then we have σ( ( 1 2 )( 1 n) ) (aka σ( ( 1 n 2 ) ) ) is the identity permutation, which obviously violates the automorphism. If a ≠ x and b₂ ≠ bn , then σ( ( 1 2 )( 1 n) ) is the product of two disjoint transpositions, and thus of order two, despite how σ( ( 1 n 2 ) ) is of order 3. This implies that for every pair of transpositions moving one common integer, their images under an automorphism move one common integer (if a ≠ x and b₂ = bn then just "flip" the transposition around to obtain the proper form). Now, inductively prove that if two transpositions move exactly one common element both under S_n and σ(S_n), then all transpositions under these two groups move exactly one common element. Take the two that move exactly one common element under S_n, and take an arbitrary third that moves exactly one common element with the first and the second under S_n and exactly one with the first and then the second (proved by the preceding manipulations):

Note not that the three under the map need not move a common element

in unison by the above logic alone, however.

But, we can prove that they must by an additional progression. Assume that x distinct transpositions ordered ( 1 2 ), ( 1 3 ), ..., ( 1 x-1 ) move one common element before (the integer 1) and after the automorphism (the integer a). Assume ( 1 x ) does not move a. We know that ( 1 x )( 1 x-1 )...( 1 2) is an x-cycle, and yet since σ( ( 1 x ) ) moves a common element with every transposition before it and therefore fixes nothing not already moved by the preceding transpositions under the automorphism, we have that σ( ( 1 x )( 1 x-1 )...( 1 2) ) only moves x-1 indices and is thus something less than an x-cycle, not in the same conjugacy class as its premapped counterpart, and therefore a contradiction by part (a). So ( 1 x ) moves a. Using this progression, every mapped transposition obeys the sequence hypothesized above.

(d) Any transposition ( i j ) within σ(S_n) can be constructed by the transpositions permuted to the set of transpositions moving a (demonstrated in the previous part) as such: ( a j )( a i )( a j ). Therefore, since σ(S_n) = S_n is clearly generated by transpositions, we have that an automorphism's action on the transpositions moving 1 in S_n are all that's necessary to determine its action on the whole group. Since there are n-1 of these transpositions with n-1 available slots after determining one of the n common integers these transpositions move, we have | Aut(S_n) | ≤ n!. Since S_n / Z(S_n) ≅ Inn(S_n) ≤ Aut(S_n) and | S_n / Z(S_n) | = | S_n / 1 | = | S_n | = n!, we have that Inn(S_n) = Aut(S_n) for n ≠ 6.

∎

Groups of Prime-Power Orders' Subgroups (4.3.29)

Dummit and Foote Abstract Algebra, section 4.3, exercise 29:

Let G be a group of order p^a for some prime p. Prove that G has a subgroup of order p^b for 0 ≤ b ≤ a (apply theorem 8 and then use induction on a).

Proof: To start off, first observe that any subgroup contained in the center of its parent group is normal within its parent group. (1)

Part 1: By theorem 8, we have that Z(G) is nontrivial, and by the nature of G it is a group of order p^c for some c > 0. By Cauchy's Theorem, take x∈Z(G) such that | x | = p. Since Z(G) is abelian, we have that < x > is normal in Z(G) and thus Z(G) / < x > is an abelian group of order p^c-1 (and what's important is that there is now a group of such order in G contained in Z(G)). Inductively apply this reasoning on Z(G) / < x > until the basis is the identity subgroup, and thus every group of order p^d for 0 ≤ d ≤ c has been demonstrated to exist in G.

Part 2: By (1), we have that G / Z(G) is a group of order p^a-c, and refer its equivalent within G as H. Since Z(G) ≤ Z(H) for Z(G) ≤ H (2) and by (1), we have that J = H / Z(G) is a subgroup of order p^a-2c. Iterate this process to obtain the set of subgroups of order p^a-nc for integers n such that | Z(G) | = c ≤ a-nc. Now, by (2) we have that the subgroups constructed in part 1 are normal within the subgroups just now constructed, such that, taking the subgroup F of order p^a-mc and a subgroup I ≤ Z(G) ≤ Z(F) of order pⁱ for i ≤ c (there exists one for every i ≤ c by part 1), we have that F / I is a subgroup of order p^a-mc-i. This "bridges the gaps" for the integers between each a-mc and a-(m+1)c, such that every subgroup of order p^z for 0 ≤ z ≤ c has been constructed in part 1, and for c ≤ z ≤ a in part 2.

∎

Alternatively, I suppose one could do what Dummit and Foote actually suggested, and perform a similar induction on G where you take the group of prime-power order (G in the basis case) and, using theorem 8 to prove that the center is of nontrivial prime-power order, construct and use for the inductive step the prime-power-order quotient group G / A where A = < x > for x∈Z(G) and | x | = p by Cauchy's Theorem. Honestly, that would've been a bit less complicated...

Permutation Representations of Non-Abelian Groups of Semiprime Orders (4.3.28)

Dummit and Foote Abstract Algebra, section 4.3, exercise 28:

Let G be a non-abelian group of order pq, where p and q are prime and p < q. Prove that G has a nonnormal subgroup of index q, so that there exists an injective homomorphism into S_q. Deduce that G is isomorphic to a subgroup of the normalizer of < ( 1 2 ... q ) > in S_q.

Proof: Take x and y from G such that | x | = p and | y | = q by Cauchy's Theorem.

We have that < x, y > = G since | < x > | and | < y > | both divide | < x, y > | by Lagrange's Theorem, therefore | < x, y > | = pq.

We have that < y > is normal since it is of index p in G, and since p is the smallest prime dividing | G |, by a previous result we have our justification.

We have that < x > is nonnormal. If it weren't, then it would be the union of conjugacy classes, and since for every g∈G we have that the order of the conjugacy class of g is either 1, p, q, or pq (since the order of a conjugacy class divides the order of the group), and since 1∈< x > has a conjugacy class of order 1, it must be that every element in < x > has a conjugacy class order of 1, ie every element in < x > is in Z(G). This includes x, which implies xy = yx, which would imply G abelian by the fact that these two elements generate G.

Now, let < x > = H and let π_H be the left regular permutation representation of G into S_{| G : H |} = S_q. Since the kernel of the action is the largest normal subgroup of G contained in H, and H is nonnormal and without nontrivial proper subgroups by the primality of its order, we have that the kernel is the identity element; thus the mapping is injective. Thus we know that π_H(y) is of order q in S_q, so π_H(y) is a q-cycle as q is prime and therefore not the result of a combination of smaller cycles whose least common multiple of orders is q (and this is the desired q-cycle if the cosets of H that y permutes are numbered appropriately). Now, since g< y >g^-1 = < y > for all g∈G by the demonstrated normality of < y >, and since π_H is an isomorphism, we have that π_H(g)< π_H(y) >π_H(g)^-1 = < π_H(y) >, so that ∀g∈G, π_H(g)∈N_G(< π_H(y) >).

∎

Commuting Conjugacy Classes' Representatives (4.3.27)

Dummit and Foote Abstract Algebra, section 4.3, exercise 27:

Let g₁, g₂, ..., g_r be the representatives of the conjugacy classes of the finite group G and assume that these elements commute pairwise. Prove that G is abelian.

Proof: Since there are r conjugacy classes in G, we have that each conjugacy class is, on average, of size | G | / r. Since we have that r ≤ | C_G(g_i) | for all i ≤ r, we have that each conjugacy class is at the most of size | G : C_G(g) | ≤ | G | / r. In order to fulfill the average requirement, every group must be of the average size, and consequently of the same size. Since the conjugacy class of the identity element is of size 1, that implies that every conjugacy class is of size 1, i.e. C_G(g) = G for all g∈G, therefore G is abelian.

∎

Normal Subgroups' Orbits Under Transitive Actions (4.1.9)

Dummit and Foote Abstract Algebra, section 4.1, exercise 9:

Assume G acts transitively on the finite set A and let H be a normal subgroup of G. Let O₁, O₂, ..., O_r be the distinct orbits of H on A.

(a) Prove that G permutes the sets O₁, O₂, ..., O_r in the sense that for any g∈G, gO_i is an orbit (i.e. the orbits are blocks). Prove that G is transitive on the set of orbits. Deduce that all the orbits of H on A have the same cardinality.

(b) Prove that if a∈O₁, then | O₁ | = | H : H∩G_a | and thus r = | G : HG_a |.

Proof: (a) First of all, notice that if H is a normal subgroup, then for all f∈G, k₁∈H, fk₁f^-1 = k₂ for some k₂∈H, so fk₁ = k₂f, which is to say the two element can be rewritten in opposite order (while losing the original element in H).

Note that O₁ is of the same cardinality as gO₁, since gh*a = gj*a ⇒ h*a = j*a. Now, assume that gO₁∩O₁ ≠ O₁, so that there is some m∉O₁ such that m = gh*a for some h∈H. Now, if gO₁∩O₁ is not the null set, then gh₁*a = h₂*a for some h₁, h₂∈H, so h₂^-1gh₁ = z∈G_a. Rewrite the two elements so that h₂^-1h₃g = z, so h₄g = z, thus g = h₅z. Retrace steps: m = h₅zh*a. Transpose and rewrite z and h, and collapse the two h terms on the left into one. Thus, m = h₆z*a = h₆*a∈O₁, a contradiction. Thus, the orbits behave like blocks on A. Since G acts transitively on A, any element n of A for n = v*a, belongs to the orbit produced by vO₁. So G acts transitively on the orbits in that every element's orbit can be constructed from O₁ by the left multiplication of elements in G, so for any i, p∈G, iO₁ = (ip^-1)pO₁.

(b) Clearly, HG_a ≤ G_O1. Let x∈G_O1 and assume x∉HG_a. This means that xO₁ = O₁, so that x*a = h*a for some h∈H. This entails h^-1x = z for some z∈G_a, so that x = hz, which obviously violates x's assumption. Therefore HG_a = G_O1. Since wO₁ = qO₁ ⇔ q^-1w∈G_O1 ⇔ wG_O1 = qG_O1, we have r = | G : G_O1 | = | G : HG_a |.

As well, we have h*a = j*a ⇔ j^-1h ∈ H_a = H∩G_a ⇔ hH∩G_a = jH∩G_a, so that | O₁ | = | H : H∩G_a |.

∎

Symmetric Groups Embedded in Alternating Groups (3.5.12)

Dummit and Foote Abstract Algebra, section 3.5, exercise 12:

Prove that A_n contains a subgroup that is isomorphic to S_n-2 for all integers n > 2.

Proof: We shall work backward, defining an injective homomorphism φ : S_n-2 → img φ ≤ A_n. Define φ(x) = x if the parity of x is even, φ(x) = x ( n-1 n ) if the parity of x is odd.

Well defined : If x = y, then if the parity of x is odd, φ(x) = x ( n-1 n ) = y ( n-1 n ) = φ(y), and if the parity of x is even, then φ(x) = x = y = φ(y).

Homomorphism : Assume the parity of x is odd and the parity of y is odd. Then φ(x)φ(y) = x ( n-1 n ) y ( n-1 n) = xy ( n-1 n )² = xy = φ(xy).
Assume the parity of x is odd and the parity of y is even. Then φ(x)φ(y) = x ( n-1 n ) y = xy ( n-1 n ) = φ(xy).
Assume the parity of x is even and the parity of y is odd. Then φ(x)φ(y) = xy ( n-1 n ) = φ(xy).
Assume the parity of x is even and the parity of y is even. Then φ(x)φ(y) = xy = φ(xy).

Injective : If φ(x) = φ(y), then this implies that the two "heads" of φ(x) and φ(y) that consist of the permutations stabilizing n-1 and n are equivalent, which is to say x = y.

Since an injective homormorphism is always an isomorphism when its range is narrowed to its image, we have S_n-2 ≅ img φ. Since φ(x) will make the parity of x even, and since the image of a homormorphism is a group, we have S_n-2 ≅ img φ ≤ A_n.

∎

In retrospect, this was all a lot of rigorous mathematical fluff to prove the intuitive hunch that since φ operates by bottlenecking a parity-random permutation to an even parity through means of attaching an abelian tail to the permuation, and since the decision to apply this tail is a homomorphic one, φ must be our uniquely mapping homomorphism.

Generation of Prime Symmetric Groups (3.5.5)

Dummit and Foote Abstract Algebra, section 3.5, exercise 5:

Show that if p is prime, then S_p = { a, b | a,b∈S_p, a is a transposition, b is a p-cycle}.

Proof: Let a₁ and a₂ be the indices transposed by a. Let c₁ and c₂ be the indices that b permutes a₁ and a₂ to, respectively. Therefore, bab^-1 = ( c₁ c₂ ) = c. As we can see, bcb^-1 = ( d₁ d₂ ), where d₁ and d₂ are the indices that b permutes c₁ and c₂ to. In this fashion, we can construct all of the transpositions that are of the same distance away in b as a₁ is to a₂. Call this distance m.

By using induction, all of the linked indices above and alike are generated transpositions. Every transposition of distance m is within { a, b | a,b∈S_p, a is a transposition, b is a p-cycle}.

Take the two generated transpositions ( 1 x ) and ( 1 y ), where x is going one direction across b and y another. If x and y were the same element, that would imply that m ≡ -m mod p, so p | 2m, a contradiction since ( p,m ) = 1.

( 1 y ) ( 1 x ) ( 1 y ) = ( x y ), a transposition of distance 2m. Using the same method as demonstrated above, every transposition of distance 2m is within { a, b | a,b∈S_p, a is a transposition, b is a p-cycle}. Take ( z y ) as a transposition of distance m, where ( x z ) would be of distance 3m rather than m.

( z y ) ( x y ) ( z y ) = ( x z ), a transposition of distance 3m. Continuing this process, every transposition of distance jm is within { a, b | a,b∈S_p, a is a transposition, b is a p-cycle} for j∈Z⁺. Since ( p,m ) = 1, there exists integers k and i such that ip + km = 1, so for all integers g, (kg)m ≡ g mod p. So there exists f such that fm ≡ g mod p, for any integer g. Since two sets of transpositions of distances that are congruent to each other modulo p are in fact the same set, this implies that every transposition of any distance is contained in { a, b | a,b∈S_p, a is a transposition, b is a p-cycle}. Since every permutation in any symmetric group is rewritable as a product of transpositions, this implies that S_p = { a, b | a,b∈S_p, a is a transposition, b is a p-cycle}.

∎

Looking closely, one can see that the primality of p was not specially invoked in this proof. The only assumption is that the distance of the transposition a is relatively prime to p, which may be composite.

Feit-Thompson Theorem's Logically Equivalent Representation (3.4.12)

Dummit and Foote Abstract Algebra, section 3.4, exercise 12:

Prove (without invoking the validity of the Feit-Thompson Theorem) that the following are logically equivalent:
(i) every group of odd order is solvable
(ii) the only simple groups of odd order are those of prime order

Proof: (i) ⇒ (ii) Let T be a simple group with odd order pn, and 1 < n∈Z⁺. Since T is of odd order, T is solvable (i), but since T is simple, T has no proper normal subgroups. By the solvability of T, T/1 ≅ T is abelian. By Cauchy's theorem, ∃x∈T, | x | = p. Now, < x > < T, but since T is abelian, < x > is normal in T, which violates the simplicity of T. Modus tollens, no simple groups with composite odd orders exist.

(ii) ⇒ (i) Let T be a group of odd order. Take the composition series of T (existent by part 1 of the Jordan-Hölder Theorem). By Lagrange's Theorem, all of its component link groups must be of odd order, therefore all of its factors are of odd order. By the definition of a composition series, these factors are simple, and by (ii) they are of prime order. Thus they are cyclic, thus abelian, thus the composition series fulfills the requirement to render T solvable.

∎

Solvable Groups' Composition Representations and Factors (3.4.8)

Dummit and Foote Abstract Alegbra, section 3.4 exercise 8:

G a finite group. Prove that the following are logically equivalent:
(i) G is solvable
(ii) G has a chain of subgroups: 1 = H₀ ≤ H₁ ≤ H₂ ... ≤ H_s = G such that H_i / H_i-1 is a cyclic group, 0≤i≤s
(iii) All composition factors of G have prime order
(iv) G has a chain of subgroups: 1 = N₀ ≤ N₁ ≤ N₂ ... ≤ N_s = G such that N_i is normal in G, and N_i / N_i-1 is abelian, 0≤i≤s

Proof:

(ii) ⇒ (iii) : Since H_i / H_i-1 is cyclic, assume | H_i / H_i-1 | is of prime order. Since prime-ordered cyclic groups are without nontrivial, proper subgroups, H_i / H_i-1 would be simple. Therefore, assume | H_i / H_i-1 | = pn for some prime p and n∈Z⁺. Then by Cauchy's Theorem there exists xH_i-1∈H_i / H_i-1 such that | xH_i-1 | = p. Since H_i / H_i-1 is cyclic and thus abelian, < xH_i-1 > = < x > / H_i-1 is a normal subgroup of H_i / H_i-1, and 1 / H_i-1 = H_i-1 / H_i-1 is clearly a normal subgroup of < x > / H_i-1. By the Lattice Isomorphism Theorem, < x > is a normal subgroup of H_i, and H_i-1 is a normal subgroup of < x >. Furthermore, < x > / H_i-1 is cyclic, and H_i / < x > is cyclic (since H_i is cyclic), so the initial conditions can be applied to these two groups to seal an infinite descent-style induction proof that eventually only prime-ordered quotient groups will remain.

 

(iii) ⇒ (ii) : If G has such a composition series, then since H_i / H_i-1 is of prime order, it must be cyclic. We have proved (ii) ⇔ (iii)

(ii) ⇒ (i) : If these quotients are cyclic, then by nature they are abelian, and thus fulfill the requirement of a solvable group.

(i) ⇒ (ii) : If G has such a chain of subgroups, then obtain the composition series by splitting the non-simple factors thus: If G_i is normal in G_i-1, then nontrivial, proper N / G_i-1 normal in G_i / G_i-1 exists, so N normal in G_i exists by the Lattice Isomorphism Theorem. Similarly, since G_i-1 / G_i-1 = 1 / G_i-1 is clearly normal in N / G_i-1, G_i-1 is normal in N. Now to prove that these two new potential quotients are abelian: Since N / G_i-1 is a subgroup of a priori abelian G_i / G_i-1, N / G_i-1 is abelian. G_i / G_i-1 abelian implies:

∀x,y∈G_i, xG_i-1yG_i-1 = yG_i-1xG_i-1 ⇒
xy∈yxG_i-1 ⇒
xy∈yxN (since G_i-1 ≤ N) ⇒
xNyN = yNxN ⇒ G_i / N is abelian

Using the familiar infinite descent, we must eventually arrive at a composition series of simple abelian factors/quotients. By exercise number one, these are isomorphic to prime cyclic subgroups, and are thereby cyclic. We have proved (i) ⇔ (ii)

(iv) ⇒ (i) : Implied by definition of a solvable group.

(i) ⇒ (iv) : Let M be the minimal normal subgroup after the trivial group within the solvable series (in special cases it may be that M=G). Take the next smallest part of the link of the composition series of M (existent by part 1 of the Jordan-Hölder Theorem) as N, which is normal in M. Then since M is solvable due to being a subgroup of a solvable group (exercise 5), and since (i) ⇒ (iii), we have that | M / N | is prime. Thus M / N is cyclic, and therefore abelian. xNyN = yNxN ⇒ x^-1y^-1xy∈N, ∀x,y∈M.

Now assume that M isn't abelian, so there exist m,n∈M such that m^-1n^-1mn = [m,n] ≠ 1. Set V = < [x,y] | ∀x,y∈M >. Now, g[x,y]g^-1 = gx^-1g^-1gy^-1g^-1gxg^-1gyg^-1 = [gxg^-1,gyg^-1] = [x₁,y₁]∈V, so we have that V is a normal subgroup in G. Since we have proved that V ≤ N and N < M, we have V is a normal subgroup of strictly smaller order than M, which violates the minimality of M. Therefore, M is abelian, so M / 1 is abelian.

Now, the induction step. If Q is the next minimal normal subgroup of G after another normal subgroup T, then not only is Q / T a normal subgroup of G / T, but Q / T is the minimal normal subgroup after the trivial group of G / T by extrapolation of the Lattice Isomorphism Theorem. Therefore, Q / T is abelian. We have proved (i) ⇔ (iv), and thus the proof is complete.

∎

Sylow-p Subgroups' Intersections With Normal Subgroups (3.3.9)

Dummit and Foote, section 3.3, exercise 9:

Let p be prime with group G, and | G | = p^am, where p does not divide m. Assume P is a Sylow-p subgroup in G, ie | P | = p^a and N is a normal subgroup with | N | = p^bn, where p does not divide n. Prove that P∩N is a Sylow-p subgroup in N and | PN/N | = p^a-b.

Proof: Since N is normal, PN ≤ G. Now by the Second Isomorphism Theorem:

PN/N ≅ P/(P∩N) ⇒
| PN/N | = | P/(P∩N) | ⇒
| PN : N | = | P : P∩N | = x, a placeholder.

If we draw the lattice diagram, we obtain the following:

Note that the fact that | PN : P | = y₁ = | N : P∩N | = y₂ follows from exercise 3.2.11, which states | PN : P∩N | = xy₁ and | PN : P∩N | = xy₂, thus y₁ = y₂ ; this is actually a general case of the Second Isomorphism Theorem.

Now, w is a power of p since P∩N is a subgroup of the Sylow-p subgroup P. As well, y clearly divides yz = | G : P | = (p^am)/p^a = m. Since p does not divide m, this implies p does not divide y. Since wy = | N | = p^bn, this can only mean that w = | P∩N | = p^b. Since wx = | P | = p^a, this means x = | PN/N | = p^a-b.

∎

Normal Subgroup with Prime Index (3.3.3)

Dummit and Foote section 3.3, exercise 3:

Let G be a group, let H be a normal subgroup of G with | G : H | = p for some prime p. Then, for all K≤G:

(i) K≤H, or
(ii) HK=G and | K : H∩K | = p

Proof: If K is not a subgroup of H, then that implies that there exists k∈K, k∉H. This then implies that kH≠H within G/H. Since G/H forms a group itself, this implies that < kH > is a subgroup of G/H. But G/H is a group of prime order, p, so any subgroup of G/H must divide p, so | kH | = 1 or p. Since the former implies that kH=H, | kH | = p.

But G/H is only composed of p elements/cosets, and since these cosets partition G, that implies that G is formed in entirety by the union of the elements of < kH >; therefore, KH=HK=G.

Now, since K ≤ N_KH(H), the Second Isomorphism theorem implies:

(KH)/H ≅ K/(K∩H) ⇒
G/H ≅ K/(H∩K) ⇒
| G : H | = | K : H∩K | ⇒
| K : H∩K | = p

∎