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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 k i denote the element
i + 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 conjugacy
classes 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 see
this 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 the
2-2-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 by
conjugating the first listed element by (2 3) and (2 4). The transpositions are all conjugate
since, 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 all
conjugate since, for any (i j k) and (l m n), we must have at most one element can differ
between the two. Without loss of generality, we have (i j k) and (i j l) (the other cases being
analogous). Then (k l)(i j k)(k l) = (i j l), so all 3-cycles are conjugate. Finally, given two 4cycles (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 of
S4 . Then there are eight cyclic subgroups of A4 , one each for the identity and the 2-2-cycles
and one for each pair σ, σ 2 = σ −1 of the 3-cycles. As discussed in class, A4 has a normal
subgroup 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. Then
N /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), . . . , (1 n).
(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 by
transpositions of the form (1 i). Then each of the transpositions (i j) for i, j ≤ n are
contained in the spam of (1 2), . . . , (1 n+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 1 n)(1 i) = (i 1 n)
for any i ∈ {2, . . . , n}. This means we can generate all transpositions from the given set
of transpositions, so Sn+1 is generated by (1 2), . . . , (1 1 n). Having shown the inductive
step, we are done.
(b) We proceed by induction. We claim that we can reduce to the case of (a). Let σ be
our n-cycle. Then we may generate the transpositions (i 1 i) by taking
σ i−1 (1 2)σ 1−i = (σ i−1 (1) σ i−1 (2)) = (i 1 i).
Now, we can generate transpositions of the form (1 i) by taking
(−1 i i)(−2 i −1 i) · · · (2 3)(1 2)(2 3) · · · (−1 i i)(−2 i −1 i)
= (−1 i i)(−2 i −1 i) · · · (13) · · · (−1 i i)(−2 i −1 i) = (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 normal
subgroup, An has a trivial centre for n > 4. For n = 4, we established above that A4 has only
one singleton conjugacy class, and since the union of singleton conjugacy classes is precisely
the 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
τ σ(i) = τ (j) = k 6= j = σ(i) = στ (i).
Hence for every non-identity permutation in Sn , there exists some element not commuting
with 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 by
f (x) = gxg −1 is an automorphism. Prove that f is not an inner automorphism if
n ≥ 3.
Solution.
(a) Let N be the centraliser of An in Sn . Then N is normal in An , since σρσ −1 = ρ for all
ρ ∈ N, σ ∈ An . By earlier considerations, An is simple for n > 4. Hence since An is
itself not abelian, we must have N = {e} for n > 4 If n = 4, we know that the only
nontrivial normal subgroup is N = {e, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)}. However it is
clear 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. Then
for an (even) σ ∈ An , we have for all ρ ∈ An ,
σρσ −1 = gρg −1 .
In particular, g −1 σρ = ρg −1 σ. Therefore g −1 σ ∈ Z(Sn ). If n ≥ 4, by (a) we have
g −1 σ = e, which is a contradiction since g was odd and σ was even. If n = 3, then
since A3 is abelian, it has no inner automorphisms. But Aut S3 ∼
= S3 (by the next
problem) and every automorphism is given by conjugation by some g ∈ S3 , so any
f : A3 → A3 given by σ 7→ gσg −1 is a nontrivial automorphism which cannot be an
inner automorphism of A3 . This completes the proof.guild
Problem 6.
Prove that every automorphism of S3 is inner and Aut S3 is isomorphic to S3 .
Solution.
First, as was proved in class, Inn S3 ∼
= S3 /Z(S3 ). By 4, Z(G) = {e} so Inn S3 ∼
= S3 .
Therefore we have an injection S3 → Aut S3 .
Conversely, let ϕ ∈ Aut S3 . 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 Aut S3 acts on the set X of
transpositions in S3 , which has order 3. This action induces a homomorphism ψ : Aut S3 →
S(X) ∼
= S3 . Since every automorphism permutes S3 uniquely and the transpositions generate
S3 , each ϕ acts uniquely on X. Therefore ψ is also an injection. Therefore since we have
two injections, we must have | Aut S3 | = |S3 | and so Aut S3 ∼
= 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 need
only 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 completes
the case of 5.
We have n3 ≡ 1 (mod 3) and n3 | 40. We have n3 = 4 or 10 as choices. Note that the
structure of a Sylow 3-subgroup must be cyclic of order 3. Therefore for each pair of 3-cycles
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σ and σ −1 there is associated a cyclic subgroup generated by σ. Further, there are 3=20
3-cycles, hence 10 pairs of inverse 3-cycles, so at least 10 Sylow 3-subgroups. Since there are
no more than 10 by the above reasoning, we conclude that there are 10 Sylow 3-subgroups
isomorphic to Z/3Z.
We have n2 ≡ 1 (mod 2) and n2 | 15. Therefore n2 = 3, 5 as choices. The order of these
subgroups 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), for
example), and since S4 injects into S5 , there is a subgroup of S5 isomorphic to D8 . Therefore
since all Sylow 2-subgroups must have the same structure, they are isomorphic to D8 . As
suggested by the given generators of the subgroup in S4 , we see that any subgroup generated
by (i j k l) and (i j)(k l) is isomorphic to D8 . Since there are at least 5 unique choices for
i, 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 /H
by left translation. This induces a group homomorphism ϕ : Sn → S(X). Then if σ ∈ ker ϕ,
we must have σρH = ρH for every ρH ∈ X, i.e. ρ−1 σρ ∈ H for every ρH ∈ X. Since we
may vary the representative ρ over all of Sn , this implies that
\
ker ϕ =
ρHρ−1 .
ρ∈Sn
Therefore ker ϕ is a normal subgroup of Sn contained in H. Suppose that n 6= 4. Then
the 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 have
ker ϕ = {e}. Hence ϕ : Sn → S(X) is a injective homomorphism. Restricting to H, we see
that 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 orders
of 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 case
where ker ϕ 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 by
left translation. This induces a homomorphism ϕ : An → S(X). The kernel of ϕ is a
normal subgroup of An . Further, as proved in class, An is simple for n 6= 4. Therefore
ker ϕ = An or {e}. Additionally, we cannot have ker ϕ = An since the action is not
trivial, namely xH 6= H for x ∈ An \ H 6= ∅.
Therefore |X| = |An : H| must be sufficiently large such that ϕ : An → S(X) is an
injection. 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 the
sole 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
= n+1
. For n ≥ 2, this is strictly less than n, which contradicts (a).
of index (n+1)!
2·n!
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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 two
symbols appearing in An+2 but not in Sn , and let τ = (i j). Then for σ ∈ Sn ,
(
σ
sgn σ = 1
ϕ(σ) =
τ σ sgn σ = −1
sgn ϕ(σ) = 1 for all σ ∈ Sn , so ϕ maps Sn into An+2 . Further, ϕ is a group homomorphism since multiplication in An+2 and Sn is consistent and τ commutes with every
σ ∈ Sn , e.g. if σ is even and ρ is odd,
ϕ(σρ) = τ σρ = στ ρ = ϕ(σ)ϕ(ρ).
An analogous result holds if σ and ρ have the same sign. Further, ϕ is injective since, if
ϕ(σ) = ϕ(ρ), then either σ = ρ or τ σ = τ ρ, which implies σ = ρ after left multiplication
by 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 by
the group action of G on itself by left multiplication. By (a), there exists an injective
homomorphism ϕ : Sn → An+2 . Since the composition of injections is an injection, we
have an injective homomorphism ϕ ◦ ψ : G → An+2 . By earlier considerations, An is
simple for n 6= 4. Therefore if G is a finite group of order 6= 2, ϕ ◦ ψ embeds G into a
finite simple group. If |G| = 2, then G itself is simple, so it is a subgroup of a finite
simple group. This completes the proof.
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