Solutions for Exercises for TATA55, batch 2
December 14, 2016
1. Let G ⊆ S6 consist of all permutations σ such that i ≤ 3 ⇐⇒ σ(i) ≤ 3. Find the
cardinality of G and show that it is a subgroup. Then find two proper nontrivial normal
subgroups H, K / G such that H ∩ K = {()}, HK = G, and conclude that G ' H × K.
Solution: Let A = {1, 2, 3} and B = {4, 5, 6}. Then G consists of those σ for which
σ(A) = A, σ(B) = B. Let H be those σ ∈ G that, in addition, satisfy σ(b) = b for
all b ∈ B, and similarly, let K be those that fix all elements in A. Then H ' K ' S3
are both subgroups of S6 . Furthermore, since A and B are disjoint, everything in H
commutes with everything in K, H ∩ K = {()}, and HK = G. Thus H, K / G, and
G ' H × K. It follows that |G| = |H||K| = 62 = 36.
2. (a) If G is a group, then Aut(G) is the set of all automorphisms on G, i.e. all isomorphisms from G to itself. Show that Aut(G) is a group under composition of
functions.
(b) If g ∈ G, denote by cg the map cg (x) = gxg −1 . Show that cg ∈ Aut(G).
(c) Automorphisms of the previous type are called inner, and the set of those are
denoted by Inn(G). Non-inner automorphisms are called outer automorphisms.
Show that Inn(G) ≤ Aut(G). Is it always a normal subgroup?
(d) What are the inner automorphisms of an abelian group?
(e) If G is abelian, show that g 7→ g −1 is an outer automorphism.
(f) Describe the automorphism group of a finite cyclic group. Describe the automorphism group of an infinite cyclic group.
Solution: (a) and (b) are easily checked, as is the first part of (c). Since φ(cg (φ−1 (h))) =
cφ(g) (h), Inn(G) is indeed always a normal subgroup of Aut(G). For an abelian group,
all cg (h) = h, thus Inn(G) is trivial. In the abelian case, (gh)−1 = h−1 g −1 = g −1 h−1 ,
so inversion is a homomorphism.
If G = hgi, then any endomorphism φ is determined by its action on the generator g,
hence is of the form φ(g s ) = (g s )k = g sk for some integer k. For this to be injective, the
generator must be mapped to another generator, thus in the finite case we must have
that gcd(k, n) = 1, where n is the order of the group, and in the infinite case we must
have k ∈ {−1, 1}. These conditions are also sufficient for surjectivity.
So, in the infinite case the automorphism group is the cyclic group with two elements.
In the finite case, we have that
(g k1 s )k2 = g k1 k2 s
and furthermore g k1 and g k2 represent the same map iff k1 ≡ k2 mod n. Thus, the
automorphism group is isomorphic to Z∗n .
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3. (a) Let G be a group, N / G, K ≤ G, N ∩ K = {1}, N K = G. G is then called the
interior semidirect product of N and K. Show that S3 is the semidirect product
of h(123)i and h(12)i.
(b) Show that φ : K 3 g 7→ cg ∈ Aut(K) is a group automorphism.
(c) Let M = N × K as a set, and introduce a multiplication by
(n1 , k1 ) ∗ (n2 , k2 ) = (n1 φ(k1 )(n2 ), k1 k2 ).
(1)
Show that this turns M into a group isomorphic to G.
(d) Conversely, Let N, K be groups, and let φ : K → Aut(K) be a homomorphism.
Define the exterior semidirect product of N and K to be the set N × K, with
the multiplication (1). Show that this is the interior semidirect product of two
subgroups Ñ and K̃ isomorphic to N and K, respectively.
(e) Show that the dihedral group Dn is the semidirect product of hri, the subgroup of
rotations and the order two subgroup hsi, generated by the reflection in the x-axis.
(Hint: srs = r−1 so cs is inversion.)
Solution: : (a) is immeditate, and (b) is equivalent to showing that cg ((ch (x)) = cgh (x),
which is straightforward. For (c), note that the conditions imposed implies that every
g ∈ G can be uniquely written as g = nk, with n ∈ N , k ∈ K. The multiplication on
M can be written as
(n1 , k1 ) ∗ (n2 , k2 ) = (n1 k1 n2 k1−1 , k1 k2 ),
where the first factor belongs to N since k1 n2 k1−1 ∈ N since N is normal. The product
−1
is checked to be associative, and the inverse of (n1 , k1 ) is (k1−1 n−1
1 k1 , k1 ). The map
φ : M → G given by
φ((n, k)) = nk,
is obviously surjective, and injective for the reason stated above. We have that
φ((n1 , k1 )∗(n2 , k2 )) = φ((n1 k1 n2 k1−1 , k1 k2 )) = n1 k1 n2 k1−1 k1 k2 = n1 k1 n2 k2 = φ((n1 , k1 ))φ((n2 , k2 )).
For (e) we take N = hri and K = hsi. Since every element in Dn can be uniquely
written as either rk or rk s, and hri is normal in Dn , the assertion follows.
4. The rigid symmetry group of the cube (considered in batch 1) acts on the set of colorings
of the faces of the cube using k colors. Compute the number of such colorings. Compute
the number of such colorings that color the top face using the first color.
Solution: Of course what is meant is the number of non-equivalent colorings. Burnside’s lemma gives that this number is
1
1
k 6 + 6k 3 + 3k 4 + 6k 3 + 8k 2 =
k 6 + 3k 4 + 12k 3 + 8k 2 ,
24
24
since, when acting on the six faces of the cubes, rotations around an axis through the
midpoints of opposite faces have cycle type 4 + 1 + 1 or 2 + 2 + 1 + 1, rotations around
an axis joining opposite midpoints of edges have cycle type (2 + 2 + 2, and rotations
through opposite vertices have cycle type 3 + 3. The proportion of those colorings that
color the top face using color 1 is k1 .
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5. Write down the class equation for A3 and A4 . Solution: Since A3 is abelian, the class
equation is 3 = 1+1+1. In A4 , the permutations with cycle type 2+2 are all conjugate.
The 3-cycles, however, which form one cycle type in S3 , split into two conjugacy classes
in A4 . Namely, by laborious calculations one can show that no 3-cycle is conjugate to
its inverse, so that the 3-cycles form the two classes
{(1, 2, 3), (1, 3, 4), (1, 4, 2), (2, 4, 3)} and {(1, 3, 2), (1, 4, 3), (1, 2, 4), (2, 3, 4)}.
Note that
{[1, 2, 3, 4], [1, 3, 4, 2], [1, 4, 2, 3], [2, 4, 3, 1]}
are even, and
{[1, 3, 2, 4]), [1, 4, 3, 2], [1, 2, 4, 3], [2, 3, 4, 1]}
are odd.
Hence , the class equation is 12 = 1 + 4 + 4 + 3.
6. Let G be a group. Is it always true that for any g ∈ G, g and g −1 are conjugate? Is it
always true that the subgroup generated by the conjugacy class of g is normal in G?
Solution: In an abelian group, no two elements are conjugate, so the first assertion is
false. The second question has an affirmative answer, however.
Let g G = xgx−1 x ∈ G be the conjugacy class containing g, and denote by N the
subgroup of G that it generates. Then any element in N can be written
ar −1
a2 −1
m = x1 g a1 x−1
1 x2 g x2 · · · xr g xr ,
ai ∈ {−1, +1},
xi ∈ G.
Conjugating with an arbitrary y ∈ G we get
ar −1 −1
a2 −1
ymy −1 = yx1 g a1 x−1
1 x2 g x2 · · · xr g xr y
−1
ar −1 −1
a2 −1 −1
= yx1 g a1 (y −1 y)x−1
1 x2 g x2 (y y) · · · (y y)xr g xr y
ar −1 −1
−1
a2 −1 −1
= (yx1 g a1 x−1
1 y )(yx2 g x2 y ) · · · (yxr g xr y )
= (yx1 )g a1 (yx1 )−1 (yx2 )g a2 (yx2 )−1 · · · (yxr )g ar (yxr )−1 ,
which again lies in N .
7. An ancient device of mysterious origin allows enlightened disciples to mentally communicate any non-negative integer strictly less than 64. Having received the number,
the device shivers perceptibly, and some of the originally inert embedded gemstones,
arranged symmetrically in an equidistant fashion around the rim of the device, lit up,
whereas the remainder of the six gemstones remain dim. It has been known since the
dawn of time that the pattern of lit and unlit gemstones represent the binary expansion
of the communicated integer (with lit gemstones indicating ones). However, from which
gemstone to start the decoding, and whether to proceed in a clockwise or counterclockwise fashion — this is something which varies from occasion to occasion, depending on
the sacred constellations of the seven heavens!
Learned maestros have taken to calling two integers which may, given fortuitious celestal
circumstances, produce the same divine lit-unlit gemstone configuration, mystically connected. The first task any novice in the order of devotees of the ancient device is given,
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is to calculate the largest number of pairwise non-mystically connected integers in the
prescribed range. Do so!
Solution: Stated less floridly, we should color the vertices of a regular hexagon using
zero or one. There is rotational symmetry, and we are also allowed to read clockwise
or counter-clockwise. Thus, if we number the vertices counter-clockwise, 1,2,3,4,5, and
6, and proceed to read the number of the vertices clockwise, still starting at 1, we get
1,6,5,4,3,2. But the permutation [1, 6, 5, 4, 3, 2] is obtained by reflecting in the x-axis,
hence we conclude that we are to count inequivalent colorings under dihedral symmetry.
The rotations r0 , r1 , r2 , r3 , r4 , r5 induce the following permutations on the vertex set:
(1)(2)(3)(4)(5)(6), (123456), (135)(246), (14)(25)(36), (153)(264), (1654321).
The 3 reflections through lines throug vertices induce
(1)(4)(26)(35), (2)(5)(13)(46), (3)(6)(15)(24),
and the 3 reflections through lines through midpoints induce
(12)(36)(45), (14)(23)(56), (16)(25)(34).
Thus the number of inequvalent k-colorings are
1
1
k 6 + 2 ∗ k 1 + 2 ∗ k 2 + k 3 + 3k 4 + 3 ∗ k 3 =
k 6 + 3k 4 + 4k 3 + 2k 2 + 2k
12
12
so the number of inequivalent 2-colorings is
1
1 6
2 + 3 ∗ 24 + 4 ∗ 23 + 2 ∗ 22 + 2 ∗ 2 =
(64 + 48 + 32 + 8 + 4) = 13.
12
12
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