Math 3121 Abstract Algebra I Lecture 13 Midterm 2 back and over And Start Section 15

HW: Section 14 Don’t hand in Hand in (Tues Nov 25): Pages 142-143: 1, 3, 5, 9, 11, 25, 29, 31 Hand in (Tues Nov 25): Pages 142-143: 24, 37

Midterm 2 Midterm 2 is back and over

Section 15: Factor Groups Examples of factor groups When G/H has order 2 – H must be normal Falsity of converse to Lagrange’s Theorem – example A4 ℤn×ℤm/<(0,1)> G1×G2/i1 (G1) and G1×G2/i2 (G2) ℤ4×ℤ6/<(2,3)> Th: Factor group of a cyclic group is cyclic Th: Factor group of a finitely generated group is finitely generated. Def: Simple groups Alternating group An, for 5≤ n, is simple (exercise 39) Preservation of normality via homomorphisms Def: Maximal normal subgroup Th: M is a maximal normal subgroup of G iff G/M is simple Def: Center Def: Commutator subgroup

Examples of Factor Groups G/{e} isomorphic to G G/G isomorphic to {e}

Index 2 Subgroups are normal Theorem: If H is a subgroup of index 2 in a group finite group G, then H is normal. Proof: H and G-H partition the group G in halves. Thus G-H must be the left and the right coset of H in G other than H itself. Thus all left cosets are right cosets.

Falsity of the Converse of Lagrange’s Theorem Example: A4 Theorem: A4 has order 12, but none of its subgroups has order 6. Proof: The elements of G=A4 are (), (1 2 3), (1 3 2), (1 2 4), (1 4 2), (1 3 4), (1 4 3), (2 3 4), (2 4 3), (1 2)(3 4), (1 3)(2 4), (1 4)(2 3). Suppose H is a subgroup of order 6 of A4. Then G/H has two elements H and a H, for some a not in H. It is isomorphic to Z2. Its multiplication table has H H = H and (a H) (a H) = H. Thus x2 is in H for all x in G. Note that (1 2 3)2 = (1 3 2), (1 3 2)2 = (1 2 3), etc. Thus all three cycles are squares and hence are in H. However, there are 8 of them. So order H is larger that 6.