Symmetric Groups Spring 2008 Problem 2. for N ⩾ 3, the Center Of

Symmetric Groups

Spring 2008 Problem 2.

For n > 3, the center of the symmetric group Sn is trivial.

Proof. Let h be in the center. Then gh = hg for all g ∈ Sn. Let g ∈ Gx, the stabilizer of x ∈ X (we realize Sn as the group of permutations on a finite set X with n elements). Then g · hx = hx for all g ∈ Gx. So g ∈ Ghx, as well. This means that Gx 6 Ghx for all x ∈ X. Suppose hx 6= x for some x ∈ X. Then every permutation of X that fixes x must also fix hx. Since n > 3, let x, hx, y ∈ X, all distinct. Define σ : X → X by the following:

x 7→ x hx 7→ y y 7→ hx, and σ fixes everything else. Then σ is a permutation of X that fixes x but does not fix hx. This is a contradiction. Hence it must not be the case that hx 6= x for some x ∈ X, that is, hx = x for all x ∈ X. So the center is trivial.

Autumn 2010 Problem 2. Let G be a transitive subgroup of the symmetric group on the set X. Let N be a normal non-trivial subgroup of G. Then N has no ﬁxed points in X.

Proof. Let N be a normal subgroup of G. For all g ∈ G, g−1ng = n0 for some n0 ∈ N (depending, of course, on g). Let x ∈ X be ﬁxed by every element of N. Then for all g ∈ G n · (g · x) = ng · x = gn0 · x = g · x. But G is transitive on X, so n · x = x for all x ∈ X. Hence n must be the identity element of N. This proves the claim.

Autumn 2009 Problem 2. Let G be an abelian subgroup of the symmetric group on a set X.

Suppose G acts transitively on X. Then the stabilizer Gx is trivial for every x ∈ X.

Proof. Fix x ∈ X. Let h ∈ Gx. Since G is abelian, for all g ∈ G, we have

h · (g · x) = (hg) · x = (gh) · x = g · x.

Since G is transitive on X, it follows that h · y = y for all y ∈ X. Hence h is the identity element. 1 2

Autumn 2002 Problem 2.

The centralizer of (12345) in S5 is the cyclic group generated by (12345), and (12345) and (13524) are not conjugate in A5.

Proof. Let σ ∈ S5 such that σa = aσ. Note that a(n) = n + 1 for n ∈ {1, 2, 3, 4}, so ak+1(n) = ak(n). Note that

a2 = (13524) a3 = (14253) a4 = (15432).

So let k ∈ {1, 2, 3, 4} such that ak(1) = σ(1). Then σa(1) = aσ(1), so σ(2) = ak+1(1) = ak(2). Suppose σ(n − 1) = ak(n − 1) for some n ∈ {2, 3, 4, 5}. Then σa(n − 1) = aσ(n − 1), so σ(n) = ak+1(n − 1) = ak(n). Hence, by induction, σ = ak. Now, note that b = a2. Suppose, for contradiction, that a and b are conjugate −1 2 in A5. Let σ ∈ A5 such that σaσ = b = a . Then σ ∈ N(hai), the normalizer of hai, but σ 6∈ hai. Note that

σ2aσ−2 = a4 σ3aσ−3 = a3 σ4aσ−4 = a.

Hence, by part (a), σ4 ∈ hai. Let σ4 = ak. Then σ5 = σak = akσ. Hence

ak = σakσ−1 = (σaσ−1)k = a2k.

k 4 So a = σ = 1. Hence the order of σ in A5 is 4. So hσi generates a Sylow-2 subgroup of A5. But P = {e, (12)(34), (13)(24), (14)(23)} is another Sylow-2 subgroup of A5. Since Sylow-2 subgroups are conjugate, hence isomorphic, we have that hσi ' P. But this is impossible since P is not cyclic. It must be that a and b are not conjugate in A5. Autumn 2005 Problem 4(a).

S5 is generated by (12) and (12345).

Proof. Let n 6= m be non-identity integers. Then

(1n)(1m)(1n) = (1)(nm).

Since every element of S5 decomposes into a product of transpositions, we have that S5 is generated by (12), (13), (14), and (15). Now, note that (12)(12345)(12) = (21345). 3

Now, (21345)(12)(54312) = (13), and (13)(12345)(13) = (32145). Now, (32145)(12)(54123) = (14), and (14)(12345)(14) = (42315). So (42315)(13)(51324) = (15). Since the subgroup generated by (12) and (12345) contains all transpositions of the form (1n), it must, in fact, generate all of S5. Spring 2008 Problem 1.

The symmetric group S5 has six Sylow-5 subgroups. This implies that

S6 contains two copies of S5 that are not conjugate to each other.

Proof. Let n5 denote the number of Sylow-5 subgroups of S5. Then

n5 ≡ 1 mod 5 n5 | 24.

So n5 = 1 or n5 = 6. Suppose n5 = 1. If P is a Sylow-5 subgroup of S5, then P is normal in S5. But this is impossible since A5 is the only normal subgroup of S5.

So it must be that n5 = 6.

Now, let S5 act on its six Sylow-5 subgroups by conjugation. This induces a map ϕ from S5 −→ S6. Each Sylow-5 subgroup is conjugate to each other, so im ϕ > 6, so ker ϕ 6 20. Since ker ϕ is normal in S5, this implies that ker ϕ is trivial. So ϕ is an injection.

We realize S6 as the group of permutations on the set {1, 2, 3, 4, 5, 6}. Consider the subgroup H = {σ ∈ S6 : σ(6) = 6}. Then H is naturally isomorphic to S5. Suppose there exists σ ∈ S6 such that σH = im ϕσ. Since H ﬁxes 6, this implies that

σ(6) = τσ(6) for all τ ∈ im ϕ. But this means that S5 acts trivially by conjugation on some Sylow-5 subgroup, that is, that there is a normal Sylow-5 subgroup. This is impossible since it implies there is only one Sylow-5 subgroup.

So H and im ϕ are subgroups of S6, isomorphic to S5, and not conjugate.