Graded Homework 4 - Solution
Math 303, Fall 2013 (Brown)
Due: Nov 1
1. Suppose that K is a proper subgroup of H and H is a proper subgroup of G. If |K| = 42
and |G| = 420, what are the possible orders of H?
Since K is a proper subgroup of H, Lagrange’s Theorem implies that |H| = 42j, where
j ≥ 2 is an integer. Since H is a proper subgroup of G, Lagrange’s Theorem also implies
420
that |H| divides 420; that is, 420 = ` · |H|, or |H| =
, for some integer `. Then,
`
42j = |H| =
420
10
=⇒ j =
∈ Z.
`
`
Thus, j = 2 or 5. Hence, |H| = 84 or 210.
2. Prove that every subgroup of Dn of odd order is cyclic.
Suppose that H is a subgroup of Dn of odd order. Dn consists of n rotations and n
reflections. If H contained a reflection ρ, then |H| would be even because ρ has order 2
and the order of an element of a group must divide the order of the group. So, H is a
subgroup of the full rotation subgroup of Dn . Since the rotations form a cyclic subgroup
of Dn , we conclude that H must also be cyclic.
3. Let G be a group of order 49. Prove that G is cyclic or g 7 = e for all g ∈ G.
Suppose that G is a group of order 49. By Lagrange’s Theorem, the orders of elements
in G must divide 49. Thus, non-identity elements in G can have order 7 or 49. If an
element has order 49, then G is cyclic; for example G = Z49 .
Now, suppose that G is not cyclic. Then, g ∈ G must have order 7 (or 1 if g = e is the
identity.) In any case, we then have g 7 = e for all g ∈ G.
4. Let G be a group of permutations acting on a set S. For i ∈ S, prove that stabG (i) is a
subgroup of G.
1
Let i ∈ S. First, stabG (i) is not empty since e ∈ stabG (i). Assume that a, b ∈ stabG (i).
This means that a(i) = i and b(i) = i. So,
(ab−1 )(i) = (ab−1 )(b(i)) = (ab−1 b)(i)) = a(i) = i.
Thus, ab−1 ∈ stabG (i), proving that stabG (i) is a subgroup of G (using the One-step
subgroup test.)
5. Determine the number of different 11-bead necklaces that can be made with two colors of
beads.
We’ll use Burnside’s Lemma to solve this problem. First, we identify the group G and
the set S on which G acts. Let S be the set of all possible beads arrangements. Note,
that S = 211 (two choices for each of 11 spots.)
G is the group of all transformations of a necklace that preserves it (in the same orientation.) Viewing the beads as vertices of a regular 11-gon, we see that G ∼
= D11 .
By Burnside’s Lemma, we have
the number of distinct necklaces =
1 X
|fix(φ)|.
|G|
φ∈G
To help us organize our necklaces, suppose the two colors are black and white. Then, we
denote a (b, w)-necklace as a necklace strung with b black beads and w white beads.
• The identity fixes all 211 configurations;
• The other ten rotations each fix the one (11, 0)-necklace and the one (0, 11)-necklace;
and,
• Each of the 11 reflections fix
–
–
–
–
–
–
The one (11, 0)-necklace and the one (0, 11)-necklace;
One (10, 1)-necklace and one (1, 10)-necklace;
Five (9, 2)-necklaces and five (2, 9)-necklaces;
Five (8, 3)-necklaces and five (3, 8)-necklaces;
Ten (7, 4)-necklaces and ten (4, 7)-necklaces;
Ten (6, 5)-necklaces and ten (5, 6)-necklaces.
Putting all of this together, we find that the number of distinct necklaces is
1 11
2 + 10 · 2 + 11 · 2 · (1 + 1 + 5 + 5 + 10 + 10) = 126.
22
6. Determine the number of ways that in which the faces of a cube can be colored with three
colors.
2
Let S be the set of the 36 possible colorings of the faces of a cube with 3 colors. Let G
be the group of rotations of the cube. Note that |G| = 24.
• The identity fixes all 36 colorings;
• The six face rotations of π/2 fix 33 colorings;
• The three face rotations of π fix 34 colorings;
• The eight vertex rotations of 2π/3 fix 32 colorings; and,
• The six edge rotations of π fix 33 colorings.
1
1 · 36 + 6 · 33 + 3 · 34 + 8 · 32 + 6 · 33 = 57 rotationally different colThus, there are
24
orings.
3