MATH 113 Midterm: MATH113 Midterm 2015 Fall Solution
MATH 113: SAMPLE MIDTERM SOLUTIONS AND HINTS
1. Just check by definition.
2. Ker(φ) is a subgroup of G. As |G|is prime, according to the theorem of
Lagrange, |Ker(φ)|=|G|or 1. In the first case, φis a trivial homomorphism. In the
second case, φis injective.
3. Assume φis a homomorphism. It is clear φis determined by φ((1,0)) and
φ((0,1)). It is clear that φ(a,0) =0 and φ(a,0) =amod 2 define homomorphisms
from Z10 to Z2. On the other hand, the homomorphism from Z3to Z2can only be
the trivial one: if φ((0,1)) =a∈Z2, 0 =φ((0,3)) =3a. Thus a=0.
Therefore, there are exactly two homomorphisms from Z10 ×Z3to Z2.
4. True or False questions:
(a)True.
(b) True.
(c)False.
5. We have done this in class.
1
Document Summary
Math 113: sample midterm solutions and hints: just check by de nition, ker( ) is a subgroup of g. as |g| is prime, according to the theorem of. In the rst case, is a trivial homomorphism. In the second case, is injective: assume is a homomorphism. It is clear is determined by ((1, 0)) and. It is clear that (a, 0) = 0 and (a, 0) = a mod 2 de ne homomorphisms from z10 to z2. On the other hand, the homomorphism from z3 to z2 can only be the trivial one: if ((0, 1)) = a z2, 0 = ((0, 3)) = 3a. Therefore, there are exactly two homomorphisms from z10 z3 to z2: true or false questions: (a)true. (b) true. (c)false, we have done this in class.
