Download 201A Homework

Math 201A Homework
Edward Burkard
Chari Problems
9/23/10.
Exercise 1. Let X be a nonempty set. Chech that Aut(X) = {bijections from X to X} is a group with composition
of functions as the binary operation.
Proof. First let’s check if Aut(X) is closed under composition. Let f, g ∈ Aut(X). Then f and g are bijections and
hence f ◦ g is also a bijection. Thus Aut(X) is closed under composition.
Now the identity of Aut(X) is the identity map, id : X → X (clearly id ∈ Aut(X)) since given any f ∈ Aut(X)
we have id ◦ f = f = f ◦ id.
Lastly, since every bijection has a well defined inverse, we have that for any f ∈ Aut(X), there is a bijection f −1
such that f −1 ◦ f = id = f ◦ f −1 .
Thus Aut(X) is a group under composition.
Exercise 2. Show S2 is abelian, and Sn is never abelian for n > 2.
Proof. S2 is the permutation group on two letters, thus S2 = {e, (1, 2)}. Since ee = e, e(1, 2) = (1, 2), (1, 2)e = (1, 2),
and (1, 2)(1, 2) = e, we see that S2 is clearly an abelian group. For any Sn with n > 2 we have the permutations
(1, 2) and (2, 3) in the group. Since (1, 2)(2, 3) = (2, 3, 1) and (2, 3)(1, 2) = (1, 3, 2) which are not equal. Thus Sn is
never abelian for n > 2.
Exercise 3. Show that any subgroup of Z is of the form nZ for some n ∈ Z. (Use the division algorithm.)
Proof. Consider the set nZ = {nx | x ∈ Z}, where n ∈ Z. We want to show that nZ < Z.
Clearly we have that nZ ⊂ Z. Let a, b ∈ nZ, then a = nc and b = nd for some c, d ∈ Z. Since we have
a − b = nc − nd = n(c − d) ∈ nZ
we see that nZ is closed under addition. Thus nZ < Z.
Now suppose that H < Z. We want to show that H = nZ for some n ∈ Z. If H = h0i, then H = 0Z and we are
done; so suppose that H is nontrivial. Since H is a subgroup of Z it must have a smallest positive element, call it
n. (This smallest positive element must exist because H 6= h0i and H cannot contain only negative numbers since
it is a subgroup and must have inverses.) If n = 1 we get that H = Z since Z = h1i. In this case H = 1Z = Z, so
suppose that n > 1. We clearly have that nZ ⊂ H since n, −n ∈ H and anything in nZ can be written as a sum of
n0 s or −n0 s. Now suppose that h ∈ H. Then by the division algorithm, we can write h = nq + r where q, r ∈ Z and
0 ≤ r < n. Notice that qn ∈ H since nZ ⊂ H. Thus r = h − qn ∈ H, therefore, by the minimality of n, r = 0 and
h = qn. Thus H ⊂ nZ; and therefore H = nZ.
Exercise 4. Let X be a nonempty set. Let ∼ be an equivalence relation on X. Given any x ∈ X, the equivalence
class of x is [x] = {y ∈ X | y ∼ x}. Check that [x] is indeed an equivalence class.
Proof.
Exercise 5. Show Ker f C G1 where f : G1 → G2 is a group homomorphism.
Proof. Recall that Ker f = {g ∈ G1 | f (g) = e}. We would like to show that Ker f is a normal subgroup of G1 .
Clearly we have that Ker f ⊂ G1 . Let g, h ∈ Ker f . We would like to show gh−1 ∈ Ker f . Since:
f gh−1 = f (g)f h−1 = ef (h)−1 = ee = e
1
2
we have Ker f < G1 . Now let g ∈ G1 and k ∈ Ker f , we would like to show that g −1 kg ∈ Ker f . We have:
f g −1 kg = f g −1 f (k)f (g) = f (g)−1 ef (g) = f (g)−1 f (g) = e.
Therefore we have Ker f C G1 .
Exercise 6. Consider H = {e, (i, j)} ⊂ Sn . Show H is isomorphic to Z2 and that H is not normal in Sn .
Proof. First let’s show that H is isomorphic to Z2 . To do this, look at the multiplication table for H:
e
(i, j)
·
e
e
(i, j)
(i, j) (i, j)
e
and the addition table for Z2 :
+ 0
0 0
1 1
1
1
0
Thus the isomorphism is clear. Define f : H → Z2 by:
f (e) = 0 f ((i, j)) = 1.
Then we can see by the multiplication table that this is an isomorphism.
In order to do the next part it must be that n > 2 otherwise H 6⊂ S1 or H = S2 (with i = 1 and j = 2). So
assume that n > 2. Consider the element (i, k) ∈ Sn (recall that (i, k)−1 = (i, k)). Then consider (i, j) ∈ H. Since
we have:
(i, k)−1 (i, j)(i, k) = (i, k)(i, j)(i, k) = (j, k)
which is not an element of H. Thus H is not normal in Sn .
9/28/10.
Exercise 7. Let G be a group. Show that H C G iff for all g ∈ G, gH = Hg.
Proof.
(=⇒) Suppose H C G. Then for all g ∈ G and h ∈ H we have g −1 hg ∈ H. Thus g −1 Hg = H. Multiply on the
left by g to get Hg = gH.
(⇐=) Suppose for all g ∈ G, gH = Hg. This says that for all g ∈ G and h ∈ H that gh = hg which implies
h = g −1 hg ∈ H for all g ∈ G and h ∈ H. Thus H C G.
Exercise 8. Let π : G1 → G2 be a group homomorphism. Show:
(a) π(e1 ) =
e2
(b) π g −1 = π(g)−1
Proof.
(a) Since:
π(g) = π(ge1 ) = π(g)π(e1 )
by group cancellation we have:
e2 = π(e1 ).
(b) We have:
e2 = π(e1 ) = π gg −1 = π(g)π g −1 .
Multiplying on the left by (π(g))
−1
we have:
(π(g))
−1
= π g −1 .
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Exercise 9.
Proof.
Exercise 10.
Proof.
Exercise 11.
Proof.
Exercise 12.
Proof.
Exercise 13.
Proof.
Exercise 14.
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Exercise 15.
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Exercise 16.
Proof.
Exercise 17.
Proof.
Exercise 18.
Proof.
Exercise 19.
Proof.
Exercise 20.
Proof.
Exercise 21.
Proof.
Exercise 22.
Proof.
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Exercise 23.
Proof.
Exercise 24.
Proof.
Exercise 25.
Proof.
Exercise 26.
Proof.