Exercises Sylow theorem

1. Prove that if a group G has a unique Sylow p-subgroup, then is normal H ⊲̲ G.

2. Prove that the alternating group A₄ does not possess subgroups of order 6.

3. Prove that if p is a prime number and m is not divisible by p, then

   $$(\genfrac{}{}{0ex}{}{p^a\cdot m}{p^a})$$

   is not divisible by p.

4. Prove that a finite group G is a p-group iff every element of G has order a power of p.

5. Prove that a non-abelian group of order p³ has a center with order p.

6. Calculate all Sylow 2-subgroups and 3-subgroups of S₃ and S₄.

7. Prove that a group of order 65 has a non-trivial normal subgroup.

8. Let G a group with pq elements, with p and q primes such that p < q. Prove that G has one and only one subgroup of order q. Prove in addition that if p does not divide q − 1, then G is cyclic.

9. Let G a group with order 2m, with odd m and greater than 1. Prove that G possesses a non-trivial normal subroup and hence G is not simple.

10. Find all groups with order 33.

Solutions

1. gHg⁻¹ is still a Sylow p-subgroup as the cardinality of H is the same as gHg⁻¹ and gHg⁻¹ is a subgroup. Since is unique it must be gHg⁻¹ = H, that is gHg⁻¹ ⊲̲ H.  ■

2. Suppose the contrary and let N be such subgroup of order 6. By Lagranges's theorem the index |A₄|/|N| = 2 = [A₄: N]. If N is a subgroup of index 2, then N is normal (see Proposition 7.11.12). Let k be the number of Sylow 3-subgroups of N. Then by Sylow's theorems k | 6 and k ≡ 1 modulo 3: thus k = 1. Let S be such Sylow 3-subgroup of N. S is a Sylow 3-subgroup also of A₄ (12 = 2² ⋅ 3). Let T a Sylow 3-subgroup of A₄. Then S and T are conjugates, that is T = σSσ⁻¹ for some A₄ and then

   T = σSσ⁻¹ ⊆ σNσ⁻¹ = N

   T is a Sylow 3-subgroup of N and thus must be equal to S. S is thus the unique Sylow 3-subgroup of A₄. But then it must contain all elements of A₄ with order 3 (if not these elements would generate other Sylow 3-subgroup. See Proposition 7.2.5. Such elements are 8, which is absurd because S has 3 elements. Thus A₄ canonot contain a subgroup with order 6.  ■

3. First Proof.

   $$(\genfrac{}{}{0ex}{}{m{p}^a}{p^a})=\frac{m{p}^a}{p^a}\frac{\cdots }{\cdots }\frac{(m{p}^a-k)}{(p^a-1)}\frac{\cdots }{\cdots }\frac{(m{p}^a-p^a+1)}{1}=m\prod _{k=1}^{p^a-1}\frac{p^{a}m-k}{k}$$

   We want to show that the binomial coefficient and p are relatevely prime. Consider the factors (p^am − k)/k. There are two cases:

   1. (p, k) = 1. Then p does not divide p^a m − k.

   2. (p, k) > 1, say k = p^βs where 1 ≤ β < α and (p, s) = 1. Then,

      $$\frac{p^a-k}{k}=\frac{p^{\beta }(m{p}^{a-\beta }-s)}{p^{\beta }s}=\frac{m{p}^{a-\beta }-s}{s}$$

   and p does not divide m p^a−β − s. Hence, after cancellation, k is a rational fraction with no factor of p in the numerator. Thus, p cannot divide the binomial coefficient.   ■

   Proof 2. Consider the polynomial (1 + x)^p; The binomial coefficient theorem (1 + x)ⁿ is the sum from k = 0 to k = n of (n k) x^k applied this case says

   $$(1+x{)}^p=1+px+(\genfrac{}{}{0ex}{}{p}{2})x^2+(\genfrac{}{}{0ex}{}{p}{3})x^3+\dots +p{x}^{p-1}+x^p$$

   except for the first and the last term, all other terms of this expansion have a coefficient that is divisible by p. Hence

(1 + x)^pn ≡ 1 + x^pn   mod p

Applying this fact a second time, we see that (1 + x)^p2 ≡ (1 + x^p)^p ≡ (1 + x^p)^p2   mod p, and thus by induction on a this gives

(1 + x)^pa ≡ 1 + x^pa   mod p

Raising both sides of this equation to the mth power we get:

(1 + x)^pam ≡ (1 + x^pa)^m = 1 + mx^pa + binom (m 2) x^2pa + ...

Since these polynomials are congruent, the coefficients of corresponding term are congruent modulo p, and the result follow by considering the coefficient of x^pa on each side.

The coefficient on the right of x^pa is m. To calculate the coefficient on the left note that (1 + x)^pam is the sum from k = 0 to k = p^am of binom(p^am k) x^k, since we want the the coefficient of x^pa, we have to consider the term with k = p^a that is, the one whose coefficient is binom(p^am p^a). So binom(p^am p^a) ≡ m   mod p.   ■

4. Let G be a p-group. Let |G| = p^t. Since it's a double implication we must prove:

   1. "a finite group G is a p-group ⇒ every element of G has order a power of p."

   2. "a finite group G is a p-group ⇐ every element of G has order a power of p."

   (a). If H is a subgroup of G, By Lagrange's theorem we know that |H| | |G|. If h is an element of G, h generates a subgroup H = ⟨h⟩. Since |G| is a power of p so must be |H|, so the order of h is p^r with r ≤ t.

   (b) Conversely, let every element of G have order a power of p. Let the order of G be not a power of p. Then there exist a prime q ≠ p such that q | |G|. By Cauchy therem G contains at least an element of period q. Hence we get a contradiction and consequently, G is a p-group.   ■

5. We know that Z(G) ≠ {e} because the center of a p-group is non-trivial, and Z ≠ G since G is not abelian. In the case of |Z| = p², by Lagrange's theorem |G|/|Z(G)| = |G/Z(G)| = p and thus G/Z and by Corollary 7.9.3 it would be cyclic. But then G would be abelian (see Theorem 7.2.16). Thus Z has order p.  ■

6. As |S₃| = 3! = 3 ⋅ 2, the Sylow 2-subgroup of S₃ have order 2. Therefore n₂ = 3 and the Sylow 2-subgroups of S₃ are

   {1, (12)},   {1, (13)},   {1, (23)}

   while n₃ =1 since n₃ ≡ 1 mod 3; This Sylow 3-subgroup is

   {1, (123), (132)}

   Since |S₄| = 4! = 2³ ⋅ 3, we observe that the order of a Sylow 2-subgroup is 8 while that of a Sylow 3-subgroup is 3. The Sylow 3-subgroup of S₄ are n₃ = 4 since by Third SYlow theorem n₃ = 1 mod 3 and n₃| 24, that is by the 3-cycle of S₄. Each 3-Sylow having order 3 means it must be of the for <(a, b, c)> since a single 3-cycle is the only type of element of order 3 in S₄ (also in S₅, but not in S₆), this is because any group H, of prime order p must be cyclic which means H = ⟨h⟩ for some h in H, so h itself has order p. The Sylow 3-subgroups of S₄ which are all conjugated among themselves are

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

   For a subgroup of order 8, we might recall that the dihedral group D₄ (symmetries of the square) has order 8. The group D₄ generated by σ and τ is a Sylow 2-subgroup of S₄. Here's the group

   1,   σ = (1,2,3,4),   σ² = (1,3)(2,4),   σ³ = (4,3,2,1)
   τ = (2,4)   στ (1,2)(3,4)   σ²τ (1,3)   σ³τ = (2,3)(1,4)

   To find the others, we just have to conjugate it. Conjugating the permutations above by (1,2) we get the conjugate group (1,2)D₄(1,2)

   1,   (1,3,4,2),   (1,4)(2,3),   (2,4,3,1)
   (1,4)   (1,2)(3,4)   (2,3)   (1,3)(2,4)

   We need to find one more Sylow 2-subgroup. When we conjugate (1,2,3,4) by (1,2,3) we get (1,2,3)(1,2,3,4)(3,2,1) = (1,4,2,3), which is in neither of the preeceding Sylow 2-subgroup. Thus, we can be sure that the conjugate group (1,2,3)D₄(3,2,1) is the Sylow 2-subgroup we are looking for. Here it is

   1,   (1,4,2,3),   (1,2)(3,4),   (3,2,4,1)
   (3,4)   (1,4)(3,4)   (2,3)   (1,3)(2,4)

7. 65 = 13 ⋅ 5, the number of Sylow 5-subgroup must be a divisor of 13, that is either 13 or 1. Since it must also be conguent to 1 modulo 5, we conclude there is only a Sylow 5-subgroup which by Exercise 1 is unique. The same is true for the Sylow 13-subgorup which is also unique and normal.  ■

8. Since n_q must divide p, n_q = 1 or p; By Sylow counting theorem n_q ≡ 1 mod q. Since p < q, we conclude that n_q = 1; n_p can be either equal to q or 1, by the same argument. n_p being q means q ≡ 1 (mod p), that is p| (q − 1). In the hypothesis that p does not divide (q − 1) then n_p = 1, hence both Sylow p-subgroup and q-subgroup are normal. For the sake of simplicity let H be the Sylow p-subgroup and K the Sylow q-subgroup. By Lagrange's theorem the order of H ∩ K must divide both p and q, so is necessarily 1. Thus,

   H ∩ K = {e}.

   Now as p is prime, H must be cyclic, H ≃ Z_p, and the same for K ≃ Z_q. By Theorem 7.17.4

   G ≃ H x K ≃ Z_p x Z_q   ■

9. Let g ∈ G have order 2. Such an element exists necessarily by Cauchy's theorem. Let G act on itself by right multiplication and note that since g ≠ 1, the element g has no fixed points in this action: g * h = h ⇒ gg * h = gh ⇒ h = gh which cannot be since g is not e. Since g² = 1, the cycle structure of the permutation induced by g consists of only 2-cycles and 1-cycles, but since there are no fixed points, we have only 2-cycyles: if the group elements are e, x₁, x₂, ..., x_n and g is one of those, then the action of g on the group gives you a permutation sending e to ge, x₁ to gx₁, etc., which is equal to the permutation (e ge) (x₁ gx₁) (x₂ gx₂) (x_n gx_n); because there are no fixed points it can't be (e ge) (x₁ gx₁) (x₂ gx₂) (x_n) which would imply gx_n x_n. It follows that this permutation is a product of m disjoint transpositions and hence is odd.

   Considering the regular representation of φ: G ⟶ S_n, since G contains an odd permutation, by 7.16.15 Lemma, it has a normal subgroup of index 2. ■

10. Let G be a group of order 33 = 11 ⋅ 3. Let H, the Sylow 11-subgroup of G, the number of 11-subgroups could be either 1 or 3. We have n11 = 1 hence normal (since 3 is not equal to 1 modulo 11) by 7.17.10 Third Sylow Theorem, similarly letting K the Sylow 3-subgroup we have it is also unique (since 11 is not equal to 1 modulo 3) and hence normal. By Lagrange's theorem the order of H ∩ K must divide both p and q, so is necessarily 1. Thus,

    H ∩ K = {e}.

    Now as 11 is prime, H must be cyclic, H ≃ Z₁₁, and the same for K ≃ Z₃. By Theorem 7.17.4

    G ≃ H x K ≃ Z₁₁ x Z₃   ■