Mathematics 581–Problem Set 4–Due Monday, April 29

1. Recall that the index of a subgroup H in a group G is the number of left cosets of H in G. Define a function f of the set of left cosets into the set of right cosets by

f (aH) = Ha¢ .

Prove

(a) f is well-defined,

(b) f is one-to-one,

(c) f is onto.

(This is without regard to whether G is finite or whether the number of left or right cosets is finite.)

2. (a) If s is an element of Sn and s = (a b)(b c) with a, b, c distinct, show that s can be written as a single 3-cycle.

(b) If s is an element of Sn and s = (a b)(c d) with a, b, c, d distinct, show that s is a product of two 3-cycles by inserting e = (b c)(b c) between (a b) and (c d).

(c) Prove that every even permutation in Sn can be written as a product of 3-cycles. (Each individual 3-cycle is even, so whether there is an even number of 3-cycles in the product or an odd number is irrelevant.)

3. Show that if H is a subgroup of G and the index (G: H) = 2, then H is a normal subgroup of G. (Hint. As a start you could show that for every element a not in G, aH = Ha, because each must be the set difference G - H.)

OVER

 

 

4. Let G and C be groups and let F be a homomorphism of G into C. Let a be any element of G and let k be any integer. Let e be the identity in G and f be the identity in C. We proved in class that F(e) = f. Since a0 = e and F(a)0 = f, we now know that F(a0) = F(a)0.

(a) Prove by induction that if k is positive then F(ak) = F(a)k.

(b) We proved in class that F(a¢ ) = F(a)¢ . Put this together with part (a) to deduce that if k is negative, then F(ak) = F(a)k.

[You've now shown formally that for any integer k, F(ak) = F(a)k.]

5. Let G = (g) be a cyclic group. Our main unresolved question about cyclic groups is to determine which elements gk other than g itself can be generators of G. Another way to phrase this is to observe that for every integer k (positive or negative) we get a subgroup (gk) of G. Then the question becomes for which values of k is (gk) = G?

(a) Show that if G is infinite then (gk) = G if and only if g = 1(trivial) or g = -1.

(b) Show that if G is finite and has n elements, then (gk) = G if and only if the greatest common factor (n, k) = 1. (You may use everything that we did in Math 580 with integers, factors, and greatest common factors.)