Fifth Homework Solutions

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