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?

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- November 14th 2011, 12:06 PMThatPinkSockAbstract 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? - November 14th 2011, 06:54 PMDevenoRe: Abstract Algebra homomorphism proof
(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. - November 14th 2011, 07:36 PMThatPinkSockRe: Abstract Algebra homomorphism proof
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. - November 14th 2011, 08:07 PMDevenoRe: Abstract Algebra homomorphism proof
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?