Fifth Homework Solutions

Math 4124 Wednesday, March 16
Fifth Homework Solutions
1. Let G be a group of order 10 with a normal subgroup H of order 2.
(a) Prove that H is contained in the center of G (hint: If 1 �= x ∈ H and g ∈ G, then
1 �= gxg −1 ∈ H).
(b) Prove that G is abelian.
(c) Let a ∈ G \ H. Prove that aH has order 5 in G/H.
(d) Prove that G has an element y of order 5.
(e) Let 1 �= x ∈ H. Prove that the order of xy is neither 2 nor 5.
(f) Prove that G is cyclic.
(g) Prove that G ∼ = Z/10Z.
(a) Since |H| = 2, it has two elements, one of which is the identity 1; we’ll call the
other element x. Let g ∈ G. The g1g −1 = 1. Also gxg −1 ∈ H \1 (because H ✁G),
hence gxg −1 = x. It follows that g commutes with every element of H and we
deduce that H is contained in the center of G.
(b) Note that G/H is cyclic because it has order 5, which is a prime, and all groups
of prime order are cyclic. Since H is contained in the center of G, we conclude
that G is abelian.
(c) By Lagrange’s theorem, |aH| divides |G/H| = 5. Since aH �= 1 because a /∈ H,
we see that |aH| = 5.
(d) Since |aH| = 5, we see that |a| �= 1 or 2. Also |a| divides 10 by Lagrange’s
theorem. Therefore |a| = 5 or 10. If |a| = 5, we’re finished. On the other hand if
|a| = 10, then |a 2 | = 5 as required.
(e) Note that |x| = 2. Since G is abelian, (xy) 2 = x 2 y 2 = y 2 �= 1 and (xy) 5 = x 5 y 5 =
x �= 1. Therefore |xy| �= 2 or 5.
(f) Note that xy �= 1. Therefore |xy| �= 1, 2 or 5. It follows from Lagrange’s theorem
that |xy| = 10 and we conclude that G is cyclic.
(g) All cyclic groups of order 10 are isomorphic to Z/10Z. The result follows, because
G is cyclic of order 10.
2. 3.3.8 on page 101. Let p be a prime and let G be the group of p-power roots of 1
in C. Prove that the map z ↦→ z p is a surjective homomorphism. Deduce that G is
isomorphic to a proper quotient of itself.
Define θ : G → G by θg = g p . Then for x,y ∈ G, we have (xy) p = x p y p because G is
abelian, consequently
θ(xy) = (xy) p = x p y p = θxθy,

so θ is a homomorphism. Suppose g ∈ G, then g = e2kπi/pn for some k,n ∈ Z. Set h =
e2kπi/pn+1. Then h ∈ G and θh = g. Therefore θ is surjective. Finally e2πi/p ∈ kerθ,
so kerθ �= 1. The result now follows from the first isomorphism theorem.
3. Let G be a group, and suppose there is a homomorphism of G onto S3 (the symmetric
group of degree 3) with kernel K. Determine the number of subgroups of G which
contain K, and show that exactly three of these subgroups are normal.
By the subgroup correspondence theorem, the subgroups containing K are in a oneto-one
correspondence with the subgroups of G/K, and this correspondence preserves
normality. This means we have to determine the subgroups of S3. Any subgroup
of S3 has order dividing 6, so any subgroup not equal to 1 or S3 has prime order
and is therefore cyclic. It follows easily that the subgroups of S3 are 1, S3, 〈(12)〉,
〈(23)〉, 〈(31)〉, 〈(123)〉, so there are six subgroups of G which contain K. Finally the
normal subgroups of S3 are 1, S3 and 〈(123)〉. We conclude that the number of normal
subgroups containing K is 3, as required.
4. 3.5.2 on page 111. Prove that σ 2 is an even permutation for every permutation σ.
We may write σ as a product of n transpositions for some positive integer n. Then
we may write σ 2 as a product of 2n transpositions, which shows that σ is an even
permutation.
5. 4.1.5(a) on page 116. Let S3 act on the set A of 27 ordered triples {(i, j,k) | 1 ≤ i, j,k ≤
3} by σ(i, j,k) = (σi,σ j,σk). Find the orbits of S3 on A. For each σ ∈ S3, find the
cycle decomposition of σ under this action. For each orbit O of S3 acting on these 27
points, pick some a ∈ O and find the stabilizer of a in S3.
The size of an orbit divides the order of the acting group, which is here S3, so we are
looking for orbits of size 1,2,3 or 6. By considering just the first coordinate, it is easy
to see that all the orbits must have size at least 3, so each orbit has size 3 or 6. We will
consider S3 as {1, (1 2), (1 3), (2 3), (1 2 3), (1 3 2)}. Then the orbits are
{(1,1,1), (2,2,2), (3,3,3)}
{(1,1,2), (2,2,1), (3,3,2), (1,1,3), (2,2,3), (3,3,1)}
{(1,2,1), (2,1,2), (3,2,3), (1,3,1), (2,3,2), (3,1,3)}
{(1,2,3), (2,1,3), (3,2,1), (1,3,2), (2,3,1), (3,1,2)}
{(1,2,2), (2,1,1), (3,2,2), (1,3,3), (2,3,3), (3,1,1)}
The stabilizer of (1,1,1) in S6 is {1, (2 3)}, while the stabilizer of all elements in any
of the other four orbits is 1.
We will let (a,b,c) correspond to 9(a − 1) + 3(b − 1) + c. This means the elements of
S are

(1,1,1)=1, (1,1,2)=2, (1,1,3)=3, (1,2,1)=4, (1,2,2)=5, (1,2,3)=6, (1,3,1)=7, (1,3,2)=8,
(1,3,3)=9, (2,1,1)=10, (2,1,2)=11, (2,1,3)=12, (2,2,1)=13, (2,2,2)=14, (2,2,3)=15,
(2,3,1)=16, (2,3,2)=17, (2,3,3)=18, (3,1,1)=19, (3,1,2)=20, (3,1,3)=21, (3,2,1)=22,
(3,2,2)=23, (3,2,3)=24, (3,3,1)=25, (3,3,2)=26, (3,3,3)=27
Element Cycle Decomposition
1 (1)
(1 2) (1 14)(2 13)(3 15)(4 11)(5 10)(6 12)(7 17)(8 16)(9 18)(19 23)(20 22)(21 24)(25 26)
(1 3) (1 27)(2 26)(3 25)(4 24)(5 23)(6 22)(7 21)(8 20)(9 19)(10 18)(11 17)(12 16)(13 15)
(2 3) (2 3)(4 7)(5 9)(6 8)(10 19)(11 21)(12 20)(13 25)(14 27)(15 26)(16 22)(17 24)(18 23)
(1 2 3) (1 14 27)(2 15 25)(3 13 26)(4 17 21)(5 18 19)(6 16 20)(7 11 24)(8 12 22)(9 10 23)
(1 3 2) (1 20 16)(2 25 15)(3 26 13)(4 21 17)(5 19 18)(6 20 16)(7 24 11)(8 22 12)(9 23 10)