Math 210A: Algebra, Homework 4

Ian Coley

October 30, 2013

Problem 1.Determine all conjugacy class in Sn for n ≤ 4.

Solution.Henceforth, for the sake of notation when we must use variables, let ki denote the elementi+ k in the symmetric group on n elements, where i+ k ≤ n.

S1 is the trivial group, so has one conjugacy class. S2 is abelian, so it has two conjugacyclasses of one element each. We proved in class that cycle type is invariant under conjugation,so every conjugacy class may contain only one cycle type. We use this for the next two cases.

The conjugacy classes for S3 are {e}, {(1 2), (1 3), (2 3)}, and {(1 2 3), (1 3 2)}. We seethis since (1 2)(1 3)(1 2) = (2 3) and (2 3)(1 3)(2 3) = (1 2), and (2 3)(1 2 3)(2 3) = (1 3 2).Therefore each conjugacy class contains all elements of a given cycle type.

The conjugacy classes for S4 are {e}, the transpositions, the 3-cycles, the 4-cycles, and the2-2-cycles, i.e. {(1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}. We can see the 2-2-cycles are all conjugate byconjugating the first listed element by (2 3) and (2 4). The transpositions are all conjugatesince, for any (i j) and (k l), we have (i k j l)(i j)(i j k l)−1 = (k l). The 3-cycles are allconjugate since, for any (i j k) and (l mn), we must have at most one element can differbetween the two. Without loss of generality, we have (i j k) and (i j l) (the other cases beinganalogous). Then (k l)(i j k)(k l) = (i j l), so all 3-cycles are conjugate. Finally, given two 4-cycles (i j k l) and (i k l j) (the other cases being analogous), we have (j k l)(i j k l)(j k l)−1 =(i k l j) as required. This completes the classification.

Problem 2.Determine all subgroups of A4. Show that A4 has no subgroups of order 6.

Solution.Recall that A4 is composed of the identity, the three 2-2-cycles, and the eight 3-cycles ofS4. Then there are eight cyclic subgroups of A4, one each for the identity and the 2-2-cyclesand one for each pair σ, σ2 = σ−1 of the 3-cycles. As discussed in class, A4 has a normalsubgroup comprised of the identity and the 2-2-cycles isomorphic to Z/2Z × Z/2Z. Since|A4| = 12, any other subgroups would have to have order 6.

We claim A4 has no such subgroup. Suppose N is a subgroup of A4 of order 6. ThenN /A4, since it has index 2, and hence N is composed entirely of conjugacy classes. However,from in-discussion exercise and easy verification, the orders of the conjugacy classes of A4

are 1,3, and 4 (the 3-cycles are split in half). No combination of these numbers sums to 6,so no such N exists. This completes the classification.

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Problem 3.(a) Prove that Sn is generated by (1 2), (1 3), . . . , (1n).

(b) Prove that Sn is generated by (1 2) and (1 2 · · ·n).

Solution.(a) By in-class theorem, Sn is generated by transpositions. We proceed by induction.

The case where n = 1 or n = 2 is clear. Now suppose that Sn can be generated bytranspositions of the form (1 i). Then each of the transpositions (i j) for i, j ≤ n arecontained in the spam of (1 2), . . . , (1n+1), so we need only show that we can generate(i n+ 1) for each 1 < i < n+ 1. But we have

(1 i)(1 1n)(1 i) = (i 1n)

for any i ∈ {2, . . . , n}. This means we can generate all transpositions from the given setof transpositions, so Sn+1 is generated by (1 2), . . . , (1 1n). Having shown the inductivestep, we are done.

(b) We proceed by induction. We claim that we can reduce to the case of (a). Let σ beour n-cycle. Then we may generate the transpositions (i 1i) by taking

σi−1(1 2)σ1−i = (σi−1(1)σi−1(2)) = (i 1i).

Now, we can generate transpositions of the form (1 i) by taking

(−1i i)(−2i −1i) · · · (2 3)(1 2)(2 3) · · · (−1i i)(−2i −1i)= (−1i i)(−2i −1i) · · · (13) · · · (−1i i)(−2i −1i) = (1 i).

Therefore we have reduced to (a), so these two cycles generate Sn.

Problem 4.Show that An (n ≥ 4) and Sn (n ≥ 3) have a trivial centre.

Solution.By considerations in class, An is simple for all n 6= 4. Since the centre of a group is a normalsubgroup, An has a trivial centre for n > 4. For n = 4, we established above that A4 has onlyone singleton conjugacy class, and since the union of singleton conjugacy classes is preciselythe centre, A4 also has a trivial centre.

Now consider Sn. Let σ ∈ Sn be a non-identity element, and suppose π(i) = j, j 6= i.Then since n ≥ 3, there exists k 6= i, j. Let τ = (j k). Then

τσ(i) = τ(j) = k 6= j = σ(i) = στ(i).

Hence for every non-identity permutation in Sn, there exists some element not commutingwith it. Therefore Z(Sn) must be trivial.

Problem 5.

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(a) Show that the centraliser of An in Sn is trivial if n ≥ 4.

(b) Let g ∈ Sn be an odd permutation. Show that the map f : An → An given byf(x) = gxg−1 is an automorphism. Prove that f is not an inner automorphism ifn ≥ 3.

Solution.

(a) Let N be the centraliser of An in Sn. Then N is normal in An, since σρσ−1 = ρ for allρ ∈ N, σ ∈ An. By earlier considerations, An is simple for n > 4. Hence since An isitself not abelian, we must have N = {e} for n > 4 If n = 4, we know that the onlynontrivial normal subgroup is N = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}. However it isclear that

(1 2 3)(1 2)(3 4) = (1 3 4) 6= (2 4 3) = (1 2)(3 4)(1 2 3).

Therefore A4 also has a trivial centre.

(b) Since An is a normal subgroup of Sn, conjugation by any element g ∈ Sn preserves An.Hence f is an automorphism of An. Assume that f is an inner automorphism. Thenfor an (even) σ ∈ An, we have for all ρ ∈ An,

σρσ−1 = gρg−1.

In particular, g−1σρ = ρg−1σ. Therefore g−1σ ∈ Z(Sn). If n ≥ 4, by (a) we haveg−1σ = e, which is a contradiction since g was odd and σ was even. If n = 3, thensince A3 is abelian, it has no inner automorphisms. But AutS3

∼= S3 (by the nextproblem) and every automorphism is given by conjugation by some g ∈ S3, so anyf : A3 → A3 given by σ 7→ gσg−1 is a nontrivial automorphism which cannot be aninner automorphism of A3. This completes the proof.guild

Problem 6.Prove that every automorphism of S3 is inner and AutS3 is isomorphic to S3.

Solution.First, as was proved in class, InnS3

∼= S3/Z(S3). By 4, Z(G) = {e} so InnS3∼= S3.

Therefore we have an injection S3 → AutS3.Conversely, let ϕ ∈ AutS3. Let τ be a transposition. Then ϕ(τ)2 = ϕ(τ 2) = e, so

ϕ(τ) has order 2, so must also be a transposition. Therefore AutS3 acts on the set X oftranspositions in S3, which has order 3. This action induces a homomorphism ψ : AutS3 →S(X) ∼= S3. Since every automorphism permutes S3 uniquely and the transpositions generateS3, each ϕ acts uniquely on X. Therefore ψ is also an injection. Therefore since we havetwo injections, we must have |AutS3| = |S3| and so AutS3

∼= S3.

Problem 7.Describe all Sylow subgroups in S5.

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Solution.We proceed prime by prime. Let np denote the number of Sylow p-subgroups. We note that,since S5 has a unique normal subgroup of index 2, we cannot have np = 1 for any prime.Further, since all Sylow p-subgroups are conjugate, they are pairwise isomorphic, so we needonly find the structure and number of each Sylow p-subgroup.

We have n5 ≡ 1 (mod 5) and n2 | 24. As stated above, we cannot have n5 = 1, so n5 = 6.Further, there is only one group of order 5 up to isomorphism, namely Z/5Z. This completesthe case of 5.

We have n3 ≡ 1 (mod 3) and n3 | 40. We have n3 = 4 or 10 as choices. Note that thestructure of a Sylow 3-subgroup must be cyclic of order 3. Therefore for each pair of 3-cyclesσ and σ−1 there is associated a cyclic subgroup generated by σ. Further, there are

(5

3=20

)3-cycles, hence 10 pairs of inverse 3-cycles, so at least 10 Sylow 3-subgroups. Since there areno more than 10 by the above reasoning, we conclude that there are 10 Sylow 3-subgroupsisomorphic to Z/3Z.

We have n2 ≡ 1 (mod 2) and n2 | 15. Therefore n2 = 3, 5 as choices. The order of thesesubgroups is 8, and we claim they have the structure of D8, the dihedral group of the square.D8 may be viewed naturally as a subgroup of S4 (generated by (1 2 3 4) and (1 2)(3 4), forexample), and since S4 injects into S5, there is a subgroup of S5 isomorphic to D8. Thereforesince all Sylow 2-subgroups must have the same structure, they are isomorphic to D8. Assuggested by the given generators of the subgroup in S4, we see that any subgroup generatedby (i j k l) and (i j)(k l) is isomorphic to D8. Since there are at least 5 unique choices fori, j, k, l (choosing one of 1,2,3,4,5 to exclude), n2 ≥ 5. Therefore n2 = 5 and we are done.

Problem 8.Show that every subgroup of Sn of index n is isomorphic to Sn−1.

Solution.Let H be a subgroup of index n in Sn. Consider the action of Sn on the cosets X = Sn/Hby left translation. This induces a group homomorphism ϕ : Sn → S(X). Then if σ ∈ kerϕ,we must have σρH = ρH for every ρH ∈ X, i.e. ρ−1σρ ∈ H for every ρH ∈ X. Since wemay vary the representative ρ over all of Sn, this implies that

kerϕ =⋂ρ∈Sn

ρHρ−1.

Therefore kerϕ is a normal subgroup of Sn contained in H. Suppose that n 6= 4. Thenthe only nontrivial normal subgroup of Sn is An, and since |Sn : H| = n > 2 = |Sn : An|,we cannot have An ⊂ H (excepting the easy cases of n = 1, 2). Therefore we must havekerϕ = {e}. Hence ϕ : Sn → S(X) is a injective homomorphism. Restricting to H, we seethat the action of H on X is also injective. Since the orbit of the coset eH ∈ X is a singleton(namely it is stable under multiplication by H), the action of H on X \ eH is still injective.Therefore there is an injective homomorphism H → S(X \ eH) = Sn−1, and since the ordersof the domain and codomain are equal, we have an isomorphism.

In the case of n = 4, we have a nontrivial normal N subgroup of order 4 in S4. However,since H has order 6, it must be isomorphic to Z/6Z or S3 by earlier considerations, and

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neither of these contain a subgroup of order 4. Therefore we are back in the previous casewhere kerϕ is trivial, so we are done.

Problem 9.

(a) Show that for n 6= 4, every proper subgroup in An has index ≥ n.

(b) Prove that there are no injective homomorphisms Sn → An+1 for n ≥ 2.

Solution.

(a) Let H be a proper subgroup of An and consider the action of An on X = An/H byleft translation. This induces a homomorphism ϕ : An → S(X). The kernel of ϕ is anormal subgroup of An. Further, as proved in class, An is simple for n 6= 4. Thereforekerϕ = An or {e}. Additionally, we cannot have kerϕ = An since the action is nottrivial, namely xH 6= H for x ∈ An \H 6= ∅.Therefore |X| = |An : H| must be sufficiently large such that ϕ : An → S(X) is aninjection. If |X| < n, then we have |S(X)| = (n − 1)! < n!/2 = |An| for n > 2.Therefore in this case we must have |An : H| ≥ n. In the case of n = 1 and 2, then thesole proper subgroup must have index 1 and 2 respectively. This completes the proof.

(b) Let ϕ : Sn → An+1 is an injective homomorphism. Then ϕ(Sn) is a subgroup of An+1

of index (n+1)!2·n! = n+1

2. For n ≥ 2, this is strictly less than n, which contradicts (a).

Hence no such ϕ exists.

Problem 10.

(a) Show that for any n ≥ 1, there is an injective homomorphism Sn → An+2.

(b) Prove that every finite group is isomorphic to a subgroup of a finite simple group.

Solution.

(a) We construct a homomorphism ϕ : Sn → An+2 in the following way: let i, j be twosymbols appearing in An+2 but not in Sn, and let τ = (i j). Then for σ ∈ Sn,

ϕ(σ) =

{σ sgnσ = 1

τσ sgnσ = −1

sgnϕ(σ) = 1 for all σ ∈ Sn, so ϕ maps Sn into An+2. Further, ϕ is a group homomor-phism since multiplication in An+2 and Sn is consistent and τ commutes with everyσ ∈ Sn, e.g. if σ is even and ρ is odd,

ϕ(σρ) = τσρ = στρ = ϕ(σ)ϕ(ρ).

An analogous result holds if σ and ρ have the same sign. Further, ϕ is injective since, ifϕ(σ) = ϕ(ρ), then either σ = ρ or τσ = τρ, which implies σ = ρ after left multiplicationby τ . This is our homomorphism, and we are done.

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(b) Let |G| = n. Then there exists an injective homomorphism ψ : G → Sn induced bythe group action of G on itself by left multiplication. By (a), there exists an injectivehomomorphism ϕ : Sn → An+2. Since the composition of injections is an injection, wehave an injective homomorphism ϕ ◦ ψ : G → An+2. By earlier considerations, An issimple for n 6= 4. Therefore if G is a finite group of order 6= 2, ϕ ◦ ψ embeds G into afinite simple group. If |G| = 2, then G itself is simple, so it is a subgroup of a finitesimple group. This completes the proof.

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