Math 3121Abstract Algebra I

Lecture 13Midterm 2 back and over

And Start Section 15

HW: Section 14

• Don’t hand inPages 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.