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Abstract Algebra

Exercise 4.5.31 ($n_p(A_5)$ and $n_p(S_5)$)

For p = 2 , 3 and 5 find n p ( A 5 ) and n p ( S 5 ) [Note that A 4 ≤ A 5 .]

Answers

Proof. The elements of order 5 in A 5 are the 24 cycles of order 5 . Since any two distinct Sylow 5 -subgroups have trivial intersection, there are n 5 ( A 5 ) ⋅ 4 = 24 elements of order 5 in A 5 , thus

n 5 ( A 5 ) = 6 .

Since any two distinct Sylow 5 -subgroups have trivial intersection, there are n 5 ( A 5 ) ⋅ 4 = 24 elements of order 5 (no surprise : there are 24 cycles of order 5 in A 5 ).

There are N = 5 ⋅ 4 ⋅ 3 3 = 20 cycles of order 3 in A 5 , and the elements of order 3 in A 5 are 3 -cycles. Since any two distinct Sylow 3 -subgroups have trivial intersection, N = n 3 ( A 5 ) ⋅ 2 , thus

n 3 ( A 5 ) = 10 .

Since the 4 -cycles are odd permutations, there are no 4 -cycle in a Sylow 2 -subgroup of A 5 . The elements of any Sylow 2 -subgroup of A 5 are (except the identity) the commutative product of two disjoint transpositions.

Note that

K = ⟨ ( 1 2 ) ( 3 4 ) , ( 1 3 ) ( 2 4 ) ⟩ = { ( 1 2 ) ( 3 4 ) , ( 1 3 ) ( 2 4 ) , ( 1 4 ) ( 2 3 ) }

Since two Sylow 2 -subgroups of A 5 are conjugate, every Sylow 2 -subgroups of A 5 is of the form

K a , b , c , d = { ( a b ) ( c d ) , ( a c ) ( b d ) , ( a d ) ( b c ) } ( a , b , c , d ∈ [ [ 1 , 5 ] ] ) .

Each such subgroup is determined by the missing element, so there are 5 such subgroups:

n 2 ( A 5 ) = 5 .

(This is compatible with Sylow’s Theorem, which gives

n 2 ( A 5 ) ∈ { 1 , 3 , 5 , 15 } , n 3 ( A 5 ) ∈ { 1 , 4 , 10 } , n 5 ( A 5 ) ∈ { 1 , 6 } . )

The elements of order 5 in S 5 are the 24 cycles of order 5 . Since any two distinct Sylow 5 -subgroups have trivial intersection, there are n 5 ( S 5 ) ⋅ 4 = 24 elements of order 5 , thus

n 5 ( S 5 ) = 6 .

There are N = 5 ⋅ 4 ⋅ 3 3 = 20 cycles of order 3 in S 5 , and the elements of order 3 in S 5 are 3 -cycles. Since any two distinct Sylow 3 -subgroups have trivial intersection, N = n 3 ( S 5 ) ⋅ 2 , thus

n 3 ( S 5 ) = 10 .

Note that

H = ⟨ ( 1 2 3 4 ) , ( 1 3 ) ⟩ = { ( ) , ( 1 2 3 4 ) , ( 1 3 ) ( 2 4 ) , ( 1 4 3 2 ) , ( 1 3 ) , ( 2 4 ) , ( 1 2 ) ( 3 4 ) , ( 1 4 ) ( 2 3 ) }

is a Sylow 2 -subgroup of S 5 . Since two Sylow 2 -subgroup are conjugate, every Sylow 2 -subgroup is of the form

H a , b , c , d = ⟨ ( a b c d ) , ( a c ) ⟩ .

Since H a , b , c , d = H b , c , d , a (because ⟨ ( a b c d ) , ( a c ) ⟩ = ⟨ ( a b c d ) , ( b d ) ⟩ = ⟨ ( b c d a ) , ( b d ) ⟩ ), and H a , b , c , d = H d , c , b , a (because ⟨ ( a b c d ) , ( a c ) ⟩ = ⟨ ( a b c d ) − 1 , ( a c ) ⟩ ), we obtain 5 ⋅ 4 ⋅ 3 ⋅ 2 8 = 15 distinct Sylow 2 -subgroups of A 5 , so

n 2 ( S 5 ) = 15 .

(This is compatible with Sylow’s Theorem, which gives

n 2 ( S 5 ) ∈ { 1 , 3 , 5 , 15 } , n 3 ( S 5 ) ∈ { 1 , 4 , 10 , 40 } , n 5 ( S 5 ) ∈ { 1 , 6 } . )

□

Proof. The elements of order 5 in A 5 are the 24 cycles of order 5 . Since any two distinct Sylow 5 -subgroups have trivial intersection, there are n 5 ( A 5 ) ⋅ 4 = 24 elements of order 5 in A 5 , thus

n 5 ( A 5 ) = 6 .

Since any two distinct Sylow 5 -subgroups have trivial intersection, there are n 5 ( A 5 ) ⋅ 4 = 24 elements of order 5 (no surprise : there are 24 cycles of order 5 in A 5 ).

There are N = 5 ⋅ 4 ⋅ 3 3 = 20 cycles of order 3 in A 5 , and the elements of order 3 in A 5 are 3 -cycles. Since any two distinct Sylow 3 -subgroups have trivial intersection, N = n 3 ( A 5 ) ⋅ 2 , thus

n 3 ( A 5 ) = 10 .

Since the 4 -cycles are odd permutations, there are no 4 -cycle in a Sylow 2 -subgroup of A 5 . The elements of any Sylow 2 -subgroup of A 5 are (except the identity) the commutative product of two disjoint transpositions.

Note that

K = ⟨ ( 1 2 ) ( 3 4 ) , ( 1 3 ) ( 2 4 ) ⟩ = { ( 1 2 ) ( 3 4 ) , ( 1 3 ) ( 2 4 ) , ( 1 4 ) ( 2 3 ) }

Since two Sylow 2 -subgroups of A 5 are conjugate, every Sylow 2 -subgroups of A 5 is of the form

K a , b , c , d = { ( a b ) ( c d ) , ( a c ) ( b d ) , ( a d ) ( b c ) } ( a , b , c , d ∈ [ [ 1 , 5 ] ] ) .

Each such subgroup is determined by the missing element, so there are 5 such subgroups:

n 2 ( A 5 ) = 5 .

(This is compatible with Sylow’s Theorem, which gives

n 2 ( A 5 ) ∈ { 1 , 3 , 5 , 15 } , n 3 ( A 5 ) ∈ { 1 , 4 , 10 } , n 5 ( A 5 ) ∈ { 1 , 6 } . )

The elements of order 5 in S 5 are the 24 cycles of order 5 . Since any two distinct Sylow 5 -subgroups have trivial intersection, there are n 5 ( S 5 ) ⋅ 4 = 24 elements of order 5 , thus

n 5 ( S 5 ) = 6 .

There are N = 5 ⋅ 4 ⋅ 3 3 = 20 cycles of order 3 in S 5 , and the elements of order 3 in S 5 are 3 -cycles. Since any two distinct Sylow 3 -subgroups have trivial intersection, N = n 3 ( S 5 ) ⋅ 2 , thus

n 3 ( S 5 ) = 10 .

Note that

H = ⟨ ( 1 2 3 4 ) , ( 1 3 ) ⟩ = { ( ) , ( 1 2 3 4 ) , ( 1 3 ) ( 2 4 ) , ( 1 4 3 2 ) , ( 1 3 ) , ( 2 4 ) , ( 1 2 ) ( 3 4 ) , ( 1 4 ) ( 2 3 ) }

is a Sylow 2 -subgroup of S 5 . Since two Sylow 2 -subgroup are conjugate, every Sylow 2 -subgroup is of the form

H a , b , c , d = ⟨ ( a b c d ) , ( a c ) ⟩ .

Since H a , b , c , d = H b , c , d , a (because ⟨ ( a b c d ) , ( a c ) ⟩ = ⟨ ( a b c d ) , ( b d ) ⟩ = ⟨ ( b c d a ) , ( b d ) ⟩ ), and H a , b , c , d = H d , c , b , a (because ⟨ ( a b c d ) , ( a c ) ⟩ = ⟨ ( a b c d ) − 1 , ( a c ) ⟩ ), we obtain 5 ⋅ 4 ⋅ 3 ⋅ 2 8 = 15 distinct Sylow 2 -subgroups of A 5 , so

n 2 ( S 5 ) = 15 .

(This is compatible with Sylow’s Theorem, which gives

n 2 ( S 5 ) ∈ { 1 , 3 , 5 , 15 } , n 3 ( S 5 ) ∈ { 1 , 4 , 10 , 40 } , n 5 ( S 5 ) ∈ { 1 , 6 } . )

□

We check with Sagemath:

sage: G = AlternatingGroup(5)
sage: p = gap(’(1,2)(3,4)’); q = gap(’(1,3)(2,4)’)
sage: H = G.subgroup([p,q]); H.list()
[(), (1,3)(2,4), (1,2)(3,4), (1,4)(2,3)]
sage: n_2 = G.order()/ G.normalizer(H).order();n_2
5
sage: G = SymmetricGroup(5)
sage: p = gap(’(1,2)(3,4)’); q = gap(’(1,3)’)
sage: H = G.subgroup([p,q]); H.order()
8
sage: n_2 = G.order()/ G.normalizer(H).order();n_2
15

This confirms n 2 ( A 5 ) = 5 and n 2 ( S 5 ) = 15 .

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2026-03-25 12:23
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