Handout #8 Math 545 Fall, 2010

1. Properties of Cosets: Let G be a group, a subgroup of G, and a, ∈G. Then (a) a ∈ aH. (b) aH = H ⇐⇒ a ∈ H. () aH = bH ⇐⇒ a ∈ bH. () aH = bH or aH ∩ bH = ∅. () |aH| = |bH|. () aH is a subgroup of G ⇐⇒ a ∈ H. (g) aH = bH ⇐⇒ a−1b ∈ H. (h) aH = Ha ⇐⇒ H = aHa−1. 2. Lagrange’ Theorem: If G is a finite group and H is a subgroup of G, then |H|||G|. |G| (a) Corollary 1: |G : H| = |H| . (b) Corollary 2: The order of an element of a finite group divides the order of the group. (c) Corollary 3: Let G be a finite group and a ∈ G. Then a|G| = e. (d) Corollary 4: A group of prime order is cyclic. (e) Corollary 5: For every integer a and every prime p, ap ≡ a (mod p).

Examples:

1. Find all left cosets of the subgroup < 8 > in the group Z20, of integers modulo 20 Let H =< 8 >= {0, 4, 8, 12, 16}. The left cosets of H in Z20 are: 0 + H = H = {0, 4, 8, 12, 16}, 1 + H = H = {1, 5, 9, 13, 17}, 2 + H = H = {2, 6, 10, 14, 18}, 3 + H = H = {3, 7, 11, 15, 19}. 2. Let G be a group and a ∈ G such that |a| = 25. Find all of the left cosets of < a5 > in < a >. < a >= {e, a, a2, . . . , a24}. The distinct left cosets of < a5 > in < a > are: < a5 >= {e, a5, a10, a15, a20}; a < a5 >= {a, a6, a11a16, a21}; a2 < a5 >= {a2, a7, a12, a17, a22}; a3 < a5 >= {a3, a8, a13, a18, a23}; a4 < a5 >= {a4, a9, a14, a19, a24}.

3. Suppose is a proper subgroup of H and H is a proper subgroup of the group G. If |K| = 50 and |G| = 600, what are the possible orders of H? K is a subgroup of H =⇒ |K|||H|; that is 50||H|. So |H| = 50k. Now H is a subgroup of the group G =⇒ |H|||G|; that is, 50k|600 =⇒ k|12. Then k = 1, 2, 3, 4, 6, 12. Thus |H| = 50k = 50, 100, 150, 200, 300, or 600. However, |K| < |H| < |G|; that is 50 < |H| < 600. Hence |H| = 100, 150, 200, or 300.

4. Suppose H and K are subgroups of a group G. If |H| = 12 and |K| = 35, then find |H ∩ K|. H ∩ K ≤ H and H ∩ K ≤ K =⇒ |H ∩ K|||H| and |H ∩ K|||K| =⇒ |H ∩ K||12 and |H ∩ K||35 =⇒ |H ∩ K|| gcd(12, 35) =⇒ |H ∩ K||1. So |H ∩ K| = 1.

5. Let G be a group such that |G| = 16. Show that G must have an element of order 2. Let g ∈ G. Then |g|||G| (The order of an element of a finite group divides the order of the group). Thus |g||16 =⇒ |g| = 1, 2, 4, 8, or 16. Case 1. If G has an element, say g, of order 16. Then g8 ∈ G has order 2. Case 2. Assume there is no element of order 16 in G and G has an element, say g, of order 8. Then g4 ∈ G has order 2. Cases 2. Assume there are no elements of order 8 or 16 in G and G has an element, say g, of order 4. Then g2 ∈ G has order 2. Case 3. Assume G has no element of orders 4, 8 or 16. Then all elements of G have orders 1 or 2. Since the identity element of G is the only element of order 1, G has 15 elements of order 2.

1 6. Let G be a group of odd order . For each a in G, show that the equation x2 = a has a solution in G. n is odd =⇒ 2 does not divide n =⇒ gcd(n,2) = 1 =⇒ ∃s, ∈ such that + 2t = 1. Thus ans+2t = 1 =⇒ (an)s(at)2 = a1 =⇒ (at)2 = a =⇒ (at)2 = a. Thus = at ∈ G (since a ∈ G and t ∈ Z) is a solution to x2 = a.

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