Thread: Abstract Algebra homomorphism proof

Let ϕ: ℤ70 →ℤ5 be the group homomorphism that sends a class [a]70 to [a]5
(a) Determine the subgroup lattice of ℤ70
(b) Identify the subgroup of ℤ70 which is the kernel of ϕ
(c) Is the kernel a cyclic group?

(a) the order of a subgroup must _____ the order of the group
(b) 5 (mod 70) is an element. can yuo think of others? what is the general pattern?
(c) use the fact that φ is a homomorphism.

This is what I have so far


(a) Z₇₀ has lattice corresponding to the factors of 70:

(Dropping coset notation for simplicity,)
.........Z₇₀ = <1>
......./......|......\
....<2>..<5>..<7>
..../....X......X....\
<10>..<14>..<35>
........\.....|....../
......<70> = {0}

(b) ker ϕ = {[x]₇₀ in Z₇₀ : ϕ([x]₇₀) = [x]₅ = [0]₅} = <14>.

(c) Yes; subgroups of a cyclic group are cyclic.

14 = 4 (mod 5), not 0.

by the FIT, Z70/ker(φ) has order 5. but |Z70/ker(φ)| = 70/|ker(φ)| = 5, so |ker(φ)| = 70/5 = 14.

does |<14>| = 14?