Homework 10
5.1.4. Claim: Let G be a group and H a subgroup of G. Then G acts on G/H, the set
of left cosets of H, by left multiplication, and this action is transitive.
Proof: First, if e ∈ G is the identity, and aH ∈ G/H, then e(AH) = (ea)H = aH,
so e fixes elements of G/H. Next, if x, yinG, and aH ∈ G/H, then (xy)aH = (xya)H.
And x(y(aH)) = x(yaH) = (xya)H. Hence (xy)(aH) = x(y(aH)) for all aH ∈ G/H.
This proves left multiplication is a group action on G/H. Now pick a, b ∈ G. Then
aH = (ab−1 )bH, so the action is also transitive.
5.1.6. Claim: Suppose G acts on X and x ∈ X. Then Stab(x) ≤ G. Further, if y ∈ O(x)
then Stab(x) and Stab(y) are conjugate subgroups of X.
Proof: Let x ∈ X. Then Stab(x) = {g ∈ G | gx = x}. We apply the subgroup test
to Stab(x). First, Stab(x) 6= ∅ since ex = x. Next, suppose g, h ∈ Stab(x). Then
(gh)x = g(hx) = gx = x, so gh ∈ Stab(x). Next, if gx = x then x = ex = (g −1 g)x =
g −1 (gx) = g −1 x. So g −1 ∈ Stab(x). This proves Stab(x) is a subgroup of G.
Now suppose y ∈ O(x). So there exists g ∈ G so that gx = y. We claim that
gStab(x)g −1 = Stab(y). First, if a ∈ Stab(x), then (gag −1 )y = (ga)(g −1 y) = (ga)x. And
(ga)x = g(ax) = gx = y. Therefore gag −1 ∈ Stab(y). Now suppose b ∈ Stab(y). Then
(g 1 bg)x = g −1 by = g −1 y = x. So g −1 bg ∈ Stab(x) and b = g(g −1 bg)g −1 ∈ gStab(x)g −1 .
This proves that gStab(x)g −1 = Stab(y), so Stab(x) and Stab(y) are conjugate subgroups.
5.2.2. There are necklaces with two red beads, two green beads, and two violet beads.
Proof: There are 6 beads to build necklaces out of, so the group D6 acts on the set of
all possible necklaces where the beads lie at vertices of the hexagon, and the number of
different necklaces isP
the number of orbits of this action. By Burnside’s Lemma, the number
1
2 3 4 5
of these orbits is |G|
g∈G |Fix(g)|. The elements of D6 are {e, r, r , r , r , r , a, b, c, d, f, g}
where r is the rotation, a, f, g are flips over the centers of opposite sides of the hexagon,
and b, c, d are flips over diagonals of the hexagon.
First, all necklaces are fixed by e, so
|Fix(e)| =
6!
= 90.
2!2!2!
Next, since a, f, g are flips over axes which divide sides of the hexagon, they swap three
vertices of the hexagon for the three opposite vertices. This means there must be one bead
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of each color on one side of the axis, and one of each color on the opposite side, in the
same order. Therefore the number of necklaces fixes by a, f, g is the number of orders for
the three colors of bead, which is 3! = 6. Therefore
|Fix(a)| = |Fix(f )| = |Fix(g)| = 6.
Next, consider the elements b, c, d which flip the hexagon over a diagonal, then the two
vertices on one side of the diagonal must be different colors, and the vertices opposite these
must match. This leaves one color of bead for each vertex on the diagonal. Hence the fixed
necklaces are again in one-to-one correspondence with the number of ways to order the
three colors, which is 6. So
|Fix(b)| = |Fix(c)| = |Fix(d)| = 6.
Now we consider the necklaces fixed by rotations of the hexagon. First, the rotation r
which moves each vertex one to the right, fixes only necklaces all of whose beads are the
same color. Since we have only 2 beads of each color, this is not possible. Similarly, the
rotation r5 moves each vertex one to the left, so it can fix no necklaces that have multiple
colors of bead. Therefore
|Fix(r)| = |Fix(r5 )| = 0.
If a necklace is fixed by r2 or r4 , then every other vertex must have the same color. But
there are only 2 beads of each color, so
|Fix(r2 )| = |Fix(r4 )| = 0.
Finally, if r3 fixes a necklace, then opposite vertices must have the same color. The number
of fixed necklaces is the number of orderings of the first three beads, so
|Fix(r3 )| = 6.
Now we apply Burnside’s Lemma to find the number of necklaces is
1
(90 + 7(6)) = 11.
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5.3.7.
(a) Claim: The automorphisms of Z2 × Z2 are determined by elements of the group of
invertible 2 × 2 matrices with entries in Z2 , which is isomorphic to S6 .
Proof: Let G = GL(2, Z2 ) be the group of 2 × 2 matrices with entries in Z2 . This
group has 6 elements,
1 0
M1 =
,
0 1
2
M2 =
M3 =
M4 =
M5 =
M6 =
1 1
0 1
1 1
1 0
1 0
1 1
0 1
1 1
0 1
1 0
,
,
,
,
.
From each Mi we get an automorphism φi of Z2 × Z2 which send (a, b) to the transpose
of Mi (a, b)T . That is,
m00 m10
a
φi (a, b) =
.
m01 m11
b
First, φi is an automorphism of Z2 × Z2 . If (a, b), (c, d) ∈ Z2 × Z2 , then Mi ((a, b) +
(c, d))T = Mi ((a+c, b+d)T ) = Mi (a, b)T +Mi (c, d)T . Further, if Mi−1 = Mj , φ−1
i = φj .
Next, every automorphism of Z2 × Z2 can be written in this way. Suppose φ is
an automorphism of Z2 × Z2 , and φ((1, 0)) = (a, b), and φ((0, 1)) = (c, d). Then
φ((0, 0)) = (0, 0) and φ((1, 1)) = (a + c, b + d) because φ is a homorphism. Since φ is
an isomorphism,
a c
M=
b d
is in G. Further M (0, 0)T = (0, 0)T , M (1, 0)T = (a, b)T , M (0, 1)T = (c, d)T , and
M (1, 1)T = (a + c, b + d)T .
Finally, G is isomorphic to S3 . below we give the multiplication tables of each, ordered
to show that there is a multiplication-preserving bijection between them.
3
S3
e
(12)
(23)
(13)
(123)
(132)
e
e
(12)
(23)
(13)
(123)
(132)
(12)
(12)
e
(123)
(132)
(23)
(13)
(23)
(23)
(132)
e
(123)
(13)
(12)
(13)
(13)
(123)
(132)
e
(12)
(23)
(123)
(123)
(13)
(12)
(23)
(132)
e
(132)
(132)
(23)
(13)
(12)
e
(123)
1 0
0 1
1 1
0 1
1 0
1 1
0 1
1 0
0 1
1 1
1 1
1 0
GL(2, Z2 )
1 0
0 1
1 0
0 1
1 1
0 1
1 0
1 1
0 1
1 0
0 1
1 1
1 1
1 0
1 1
0 1
1 1
0 1
1 0
0 1
0 1
1 1
1 1
1 0
1 0
1 1
0 1
1 0
1 0
1 1
1 0
1 1
1 1
1 0
1 0
0 1
0 1
1 1
0 1
1 0
1 1
0 1
0 1
1 0
0 1
1 0
0 1
1 1
1 1
1 0
1 0
0 1
1 1
0 1
1 0
1 1
0 1
1 1
0 1
1 1
0 1
1 0
1 1
0 1
1 0
1 1
1 1
1 0
1 0
0 1
1 1
1 0
1 1
1 0
1 0
1 1
0 1
1 0
1 1
0 1
1 0
0 1
0 1
1 1
(b) Claim: The automorphisms of Z2 × Z2 are in one-to-one correspondence with the
permutations of the nonidentity elements of the group.
Proof: The elements of Z2 × Z2 are (0, 0), (1, 0), (0, 1) and (1, 1). Let φ be an
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automorphism of Z2 × Z2 . Then φ((0, 0)) = (0, 0) since (0, 0) is the identity. Let
φ((1, 0)) = a. There are three choices for a. Then let φ((0, 10)) = b. There are two
choices left for b, since φ((1, 0)) and φ((0, 1)) must be different. Finally, φ((1, 1)) =
φ((1, 0) + (0, 1)) = a + b, so there is only one choice for φ((1, 1)). Hence there are 6
automorphisms of Z2 × Z2 .
Let σ ∈ S3 . Then σ defines an automorphism φσ of Z2 × Z2 as follows. Set g1 = (1, 0),
g2 = (0, 1) and g3 = (1, 1). Then φσ (g1 ) = gσ(1) , φσ (g2 ) = gσ(2) , and φσ (g3 ) = gσ(3) .
This gives a one-to-one correspondence between automorphisms of Z2 × Z2 with S3 .
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