Solutions for Exercises for TATA55, batch 1, 2016
October 11, 2016
1. Let G be a group. Show that ∅ 6= H ⊆ G is a subgroup iff xy −1 ∈ H for
all x, y ∈ H.
Solution: Take x ∈ H \ {1}. Then xx−1 = 1 ∈ H. Hence, 1x−1 = x−1 ∈
H. Finally, x(y −1 )−1 = xy ∈ H for all x, y ∈ H.
2. If σ = [σ(1), . . . , σ(n)] is a permutation, then its set of inversions is
inv(σ) = { (i, j) i < j, and σ(i) > σ(j) }.
(a) Find inv([2, 3, 1, 5, 4]).
(b) An adjacent transposition τ is a transposition that transposes two
adjacent integers, i.e., τ = (k, k + 1) for some k. What is inv(τ )?
(c) Consider the transposition γ = (k, k + r). What is inv(γ)? Show
that γ can be written as an odd number of adjacent transpositions.
Solution: inv([2, 3, 1, 5, 4]) = {(1, 3), (2, 3), (4, 5)}.
inv((k, k+1)) = {(k, k + 1)}, and inv((k, k+r)) is the union of {(k, k + 1), (k, k + 2), . . . , (k, k + r)}
and {(k + 1, k + r), (k + 2, k + r), . . . , (k + r − 1, k + r)}.
We illustrate that (1, 4) = (1, 2)(2, 3)(3, 4)(2, 3)(1, 2), this can be modified
for any transposition.
3. Describe the subgroup of Sn generated by all n-cycles. Solution: Let
r = (1, 2, . . . , n), s = (n, n − 1, . . . , 3, 1, 2). Then sr = (1, 3, 2) and srsr =
(1, 2, 3). Modifying this construction, we can get any 3-cycle as a product
of n-cycles.
If n is odd, then all n-cycles are even permutations. Since the 3-cycles
generate the alternating group An , it follows that the group generated by
the n-cycles is precisely An .
If n is even, then n-cycles are odd permutations. In this case, the subgroup
generated by the n-cycles contains all 3-cycles, hence all of An , as well
as some odd permutations. Since An is a maximal subgroup of Sn , the
subgroup generated by the n-cycles is the whole of Sn .
4. Let C∗ be the set of all non-zero complex numbers.
(a) Show that C∗ is a group under multiplication.
(b) Call a subroup H ⊆ C ∗ bounded if H is contained in some open disc
of finite radius. Show that the circle group T , consisting of complex
number of unit modulus, is bounded, and contains all such bounded
subgroups.
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(c) A subgroup K ⊂ C ∗ is called discrete if every open disc of finite
radius intersection K in a finite number of points. Show that every
such discrete subgroup is in fact bounded. Must it also be cyclic?
(d) Regard R2 as an abelian group, with the natural addition. Define
the notions of discrete and bounded subgroups as before, mutatis
mutandis. Show that the subgroup of R2 generated by two non-zero,
non-parallell vectors yields an example of a discrete, non-bounded
subgroup of R2 .
(e) Show that the cyclic subgroup C generated by a non-zero vector is
another such example. Show that the quotient R2 /C is isomorphic
to the cylinder T × R.
Solution: T is obviously bounded. A bounded subgroup can not contain
any z with norm > 1, since otherwise the sequence z k tends to infinity.
Nor can it contain any z with norm < 1, since otherwise z −1 has norm
> 1. It follows that such a bounded subgroup must be contained in T .
A non-bounded subgroup contains a sequence zk with |zk | → ∞, hence
zk−1 → 0 ∈ C, whence any open disc containing the origin contains infinitely many elements from the subgroup, contradicting discreteness.
Hence, a discrete subgroup D must be a subgroup of T . Furthermore,
since T has finite length, it follows that D must in fact be finite. In
particular, all elements must have finite order. Since the elements of order
n in T are precisely e2πik/n , gcd(n, k) = 1, there is some N such that the
subgroup D ≤ he2πi/N i. We know that ever subgroup of a cyclic group is
cyclic, hence D is cyclic.
Let (a1 , a2 ) and (b1 , b2 ) be two linearly independent vectors in R2 . Then
the subgroup they generate is { (ra1 + sb1 , ra2 + sb2 ) r, s ∈ Z }, which is
clearly bounded and discrete.
We can, without loss of generality, assume that the vector is (1, 0), so that
C = { (n, 0) n ∈ Z } .
The map
R2 3 (x, y) 7→ (e2πix , y) ∈ T × R
is surjective homomorphism with kernel C, hence the first isomorphism
theorem yields the desired result.
5. Let p be an odd prime number. Show

1 a −a



0 1 0
G= 
0 0 1



0 0 0
that the set of matrices


b



b
 a, b ∈ Zp

b



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is (under multiplication) a finite abelian group.
Solution: The exercise got truncated; the intended question was to classify which finite abelian group this is.
6. In R3 , the linear isometries are the rotations (around a line through the
origin) and the reflections (in a plane through the origin). Let K be
the cube with vertices in { (x, y, z) x, y, z ∈ {−1, 1} } . Let R be the set of
rotations that preserve K.
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(a) Show that R is a group with 24 elements.
(b) Show that there are precisely 8 elements of order 3, corresponding to
rotations by 2π/3 radians around an axis through two diametrically
opposed vertices, and all these rotations are conjugate, but need not
commute.
(c) What are the elements of order 4? Are they all conjugate?
(d) What about the elements of order 2?
(e) What other order do the elements in R have?
(f) List all normal subgroups of R.
Solution: Suppose that a plane with a unit square is given; then a rotation of the cube is specified by placing the cube on the square. Choosing
which face is to touch the plane can be done in 6 ways, and once this is
done, the cube can be rotated in 4 ways, hence the number of placements
are 24.
Now, label the vertices on the bottom face of the cube 1,2,3,4, and label the dimaterically oppsed vertex to a by â. The pair {a, â} can be
identified with the line segment joining these vertices. The group acts
by permuting these four line segments, and different symmetries induce
different permutations. It follows that the group is isomorphic to S4 . In
this correspondence, conjugacy classes maps to conjugacy classes, hence
we have
(a) The identity (),
(b) Rotations by 180 degrees around an axis connecting midpoints of
opposing edges, {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}, order 2,
(c) Rotations by 120 degrees around an axis connecting opposite vertices,
all 3-cycles, 8 of them, order 3,
(d) rotation by 90 degrees around an axis connecting midpoints of opposite faces, all 4-cycles, 6 of them, order 4,
(e) rotation by 180 degrees around an axis connecting midpoints of opposite faces, {(1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)}, order 2.
Since a normal subgroup is a union of conjugace classes, the only proper,
non-trivial ones are A4 and {(), (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)}.
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