Download INTRO GROUP THEORY 2015: SOLUTIONS [B] denotes bookwork

INTRO GROUP THEORY 2015: SOLUTIONS
[B] denotes bookwork. [H] denotes a problem similar to homework. [U] means “unseen”.
A1. All [B].
(a) A group (G, ∗) consists of a set G and a binary operation ∗ : G × G → G such that
• (x ∗ y) ∗ z = x ∗ (y ∗ z) for all x, y, z ∈ G.
• There is e ∈ G such that e ∗ x = x = x ∗ e for all x ∈ G.
• For every x ∈ G there is x−1 ∈ G such that x ∗ x−1 = e = x−1 ∗ x.
(b) (G, ∗) is abelian if x ∗ y = y ∗ x for every x, y ∈ G.
(c)
(i) xy = x ⇒ x−1 (xy) = x−1 x ⇒ (x−1 x)y = x−1 x ⇒ ey = e ⇒ y = e.
(ii) xy = e ⇒ x−1 (xy) = x−1 e ⇒ (x−1 x)y = x−1 ⇒ ey = x−1 ⇒ y = x−1 .
(iii) (xy)(y −1 x−1 ) = ((xy)y −1 )x−1 = (x(yy −1 ))x−1 = (xe)x−1 = xx−1 = e. and so
by (ii) y −1 x−1 = (xy)−1 .
A2. (a),(b) [B], (c),(d) [U].
(a) If x ∈ G is an element of a multiplicatively written group, then if there is no positive
integer n with xn = e, then the order of x is infinite. Otherwise ord(x) is the smallest
positive integer n such that xn = e.
(b) A subset H of a group G is a subgroup if it is a group with the same operation as
G. [Or alternatively, if it is closed under the group operation, contains the identity
element and is closed under taking inverses.]
(c) Let x, y ∈ H. Then
• (xy)n = xn y n = e since H is abelian, and so xy ∈ H.
• en = e, and so e ∈ H.
• If x ∈ H, then (x−1 )n = (xn )−1 = e−1 = e, and so x−1 ∈ H.
(d) Let G be the dihedral group of order 6, and n = 2. Then the reflections ab and b are
in H, but (ab)b = a is not, and so H is not closed. [a and b are standard notation for
a rotation and reflection in a dihedral group.]
A3. All [B] and [H]
(a) στ = (1, 3)(2, 4, 6, 5, 7).
(b)
(i) ord(σ) = 12, ord(τ ) = 7 and ord(στ ) = 10, since the order of a permutation
written as a product of disjoint cycles is the lowest common multiple of the cycle
lengths.
(ii) σ and στ are odd, and τ is even, since a cycle is even iff the cycle length is odd,
and the product of an odd and even permutation is odd.
(iii) σ = (1, 2)(2, 3)(3, 4)(5, 6)(6, 7), τ = (1, 2)(2, 3)(3, 4)(4, 5)(5, 6)(6, 7), and στ =
(1, 3)(2, 4)(4, 6)(6, 5)(5, 7).
A4. (a) and (b) [B], (c) [U].
(a) If G and H are groups, a function ϕ : G → H is a group homomorphism if ϕ(xy) =
ϕ(x)ϕ(y) for all x, y ∈ G. The kernel of ϕ is {x ∈ G : ϕ(x) = e}.
(b) A subgroup N ≤ G is normal if, for all g ∈ G and n ∈ N , gng −1 ∈ N . [Alternatively,
if gN = N g for every g ∈ G.] Let n ∈ ker(ϕ) and g ∈ G. Then
ϕ(gng −1 ) = ϕ(g)ϕ(n)ϕ(g)−1 = ϕ(g)eϕ(g)−1 = e,
and so gng −1 ∈ ker(ϕ).
(c) Let x, y ∈ H. Since ϕ is surjective, x = ϕ(a) and y = ϕ(b) for some a, b ∈ G. Then
xy = ϕ(a)ϕ(b) = ϕ(ab) = ϕ(ba) = ϕ(b)ϕ(a) = yx,
and so H is abelian.
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A5. All [B].
(a) An action of G on X is a function G × X → X (denoted by (g, x) 7→ g · x) such that
• e · x = x for every x ∈ X.
• (gh) · x = g · (h · x) for all g, h ∈ G and x ∈ X.
(b) G · x = {g · x : g ∈ G} and Gx = {g ∈ G : g · x = x}.
Let g, h ∈ Gx . Then
(gh) · x = g · (h · x) = g · x = x,
and so gh ∈ Gx .
Since e · x = x, e ∈ Gx .
Let g ∈ Gx . Then
x = e · x = (g −1 g) · x = g −1 · (g · x) = g −1 · x,
and so g −1 ∈ Gx .
(c) If a finite group G acts on a set X, and x ∈ X, then
|G| = |G · x||Gx |.
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B1 (a) [B]
(i) An isomorphism ϕ : G → H is a bijective function such that ϕ(xy) =
ϕ(x)ϕ(y) for all x, y ∈ G. G and H are isomorphic if there is an isomorphism between them.
(ii) Since ϕ(eG )ϕ(eG ) = ϕ(e2G ) = ϕ(eG ), uniqueness of the identity implies that
ϕ(eG ) = eH .
(iii) Since ϕ(x)ϕ(x−1 ) = ϕ(xx−1 ) = ϕ(eG ) = eH , uniqueness of inverses implies
that ϕ(x−1 ) = ϕ(x)−1 .
(iv) Let ϕ : G → H be an isomorphism. If x ∈ G with ord(x) = n, then
ϕ(xi ) = ϕ(x)i
and so
ϕ(x)i = eH ⇔ ϕ(xi ) = eH ⇔ xi = eG ,
since ϕ is bijective and ϕ(eG ) = eH . So ord(x) = n ⇔ ord (ϕ(x)) = n, and ϕ
gives a bijection between the sets of elements of order n in G and H.
(b) [B] except (iii) [H]
(i) If K and L are groups, then the direct product is the Cartesian product K ×L
with group operation
(x, y)(x0 , y 0 ) = (xx0 , yy 0 )
for x, x0 ∈ K, y, y 0 ∈ L.
(x, y)(x0 , y 0 ) (x”, y”) = (xx0 , yy 0 )(x”, y”) = (xx0 x”, yy 0 y”)
which is the same as
(x, y) (x0 , y 0 )(x”, y”) = (x, y)(x0 x”, y 0 y”) = (xx0 x”, yy 0 y”),
so the operation is associative.
(eK , eL )(x, y) = (eK x, eL y) = (x, y) = (xeK , yeL ) = (x, y)(eK , eL ),
so (eK , eL ) is an identity element.
(x, y)(x−1 , y −1 ) = (xx−1 , yy −1 ) = (eK , eL ) = (x−1 x, y −1 y) = (x−1 , y −1 )(x, y),
so (x−1 , y −1 ) is an inverse of (x, y).
(ii)
(x, y)i = (eK , eL ) ⇔ (xi , y i ) = (eK , eL ) ⇔ xi = eK and y i = eL ,
which is true if and only if i is a multiple of ordK (x) and a multiple of ordL (y).
But ordK×L (x, y) is the smallest positive such i, which is lcm (ordK (x), ordL (y)).
(iii) D4n has a rotation of order 2n. But D2n × C2 has no elements of order 2n,
since every element of D2n or C2 has order dividing n (true for the reflections
and the generator of C2 since n is even). So by (ii), every element of D2n ×C2
has order dividing n. So D4n and D2n ×C2 have different numbers of elements
of order 2n, and so are not isomorphic, by (a)(iv).
(c) [U] First check H is a subgroup. Since a2i a2j = a2(i+j) , a2i (a2j b) = a2(i+j) b,
(a2i b)a2j = a2(i−j) b and (a2i b)(a2j b) = a2(i−j) , it is closed under multiplication. It
contains the identity. Since (a2i )−1 = a−2i and (a2i b)−1 = a2i b, it is closed under
taking inverses.
Let α and β be a rotation and a reflection in D2n with ord(α) = n.
Define ϕ : D2n → D4n by
ϕ(αi ) = a2i
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and
ϕ(αi β) = a2i b.
Then ϕ is a bijection from D2n to H, and
ϕ(αi )ϕ(αj ) = a2i a2j = a2(i+j) = ϕ(αi+j ),
ϕ(αi )ϕ(αj β) = a2i a2j b = a2(i+j) b = ϕ(αi+j β),
ϕ(αi β)ϕ(αj ) = a2i ba2j = a2(i−j) b = ϕ(αi−j β) = ϕ (αi β)αj ,
ϕ(αi β)ϕ(αj β) = a2i ba2j b = a2(i−j) = ϕ(αi−j ) = ϕ (αi β)(αj β) ,
so it is an isomorphism.
B2 (a) [B]
(i) Z/nZ is closed under addition since [a] + [b] = [a + b]. Addition (mod n) is
associative, since
([a] + [b]) + [c] = [a + b + c] = [a] + ([b] + [c]).
The identity element is [0] since
[a] + [0] = [a] = [0] + [a]
for any a ∈ Z. The inverse of [a] is [−a], since
[a] + [−a] = [0] = [−a] + [a].
So (Z/nZ, +) is a group.
(ii) Suppose [a], [b] ∈ Un , so hcf(a, n) = 1 = hcf(b, n). Then hcf(ab, n) = 1 and
so [ab] ∈ Un and Un is closed under multiplication.
Since
([a][b])[c] = [abc] = [a]([b][c])
multiplication on Un is associative.
[1] is an identity element, since [1][a] = [a] = [a][1] for any a.
If [a] ∈ Un , so hcf(a, n) = 1, then as+nt = 1 for integers s, t, and hcf(s, t) = 1.
So [s] ∈ Un is an inverse of [a].
So Un is a group under multiplication.
(b) [B]
(i) Let G be a finite group and H a subgroup. Then |H| divides |G|. If x ∈ G,
then hxi is a subgroup of G with order ord(x), so by Lagrange’s Theorem,
ord(x) divides |G|.
(ii) The order of Un is φ(n), and so the order of any element [a] ∈ Un divides
φ(n). So [a]φ(n) = [1] in Un , which means that aφ(n) ≡ 1 (mod n).
(c) (i),(ii) [H], (ii),(iv) [U]
(i) By Lagrange, the order divides pn and so must be pk for some k with 0 ≤
k ≤ n.
n−1
(ii) If xp
6= e, then ord(x) does not divide pn−1 . Since it does divide pn , the
only possibility is ord(x) = pn , so the cyclic subgroup hxi has order pn and
so must be the whole of G.
(iii) |U17 | = 16 = 24 , so (ii) applies, and if we can find an element [a] with [a]8 6= [1]
then it follows that U17 is cyclic. Trying [3], 32 = 9, 34 = 92 = 81 ≡ 13
(mod 17), 38 ≡ 132 ≡ 169 ≡ 16 (mod 17). So [3]8 6= [1] and so U17 is cyclic.
(iv) 51 = 3 × 17, so U51 has order 2 × 16 = 32. If hcf(a, 51) = 1, then a2 ≡ 1
(mod 3) and a16 ≡ 1 (mod 17), so a16 − 1 is divisible by both 3 and 17 and
hence by 51. So ord([a]) divides 16, and so U51 has no element of order 32
and so is not cyclic.
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