So there are 4 homomorphisms onto Z10. Now, let’s examine homomorphisms to Z10. Then φ(1) must have an order that divides 10 and that divides 20.

As a conclusion, the answer is 2.

Because all homomorphisms must take identities to identities, there do not exist any more homomorphisms from Z to Z. Clearly, the identity map is the only surjective mapping. Thus there exists only one homomorphism from Z to Z which is onto.

six homomorphisms
Conclusion: There are only six homomorphisms from Z to Z6.

(5) The groups S3 and Z6 are not isomorphic. Answers: (1) FALSE. For example, both Z and Z × Z are countably infinite abelian groups where every nontrivial element has infinite order.

I have actually given you more information that you need, but to sum it up, there are no non-trivial homomorphisms from S3 to Z3. In a fancy way, they write this as Hom(S3,Z3)={e} where e:S3→Z3 is defined as e(σ)=ˉ0 for any σ∈S3.

four
# 26: Determine all homomorphisms from Z4 to Z2 ⊕ Z2. There are four such homomorphisms. The image of any such homomorphism can have order 1, 2 or 4. If it has order 1, then φ maps everything to the identity or φ(x) = (0,0.

So there are four homomorphisms, each determined by choosing the common image of a,b.

2 ring homomorphisms
ϕ(1) = 0 clearly defines a group homomorphism, while ϕ(1) = 3 defines a group homomorphism ϕ(n) = 3n ∈ Z6. Checking that it preserves multiplication: ϕ(mn) = 3mn, whereas ϕ(m)ϕ(n)=3m3n = 9mn = 3mn in Z6 since 3 is idempotent there. So there are 2 ring homomorphisms between Z4 → Z6. 11.

4 homomorphisms
Hence there are 4 homomorphisms to Z8. # 21: If φ is a homomorphism from Z30 onto a group of order 5, determine the kernel of φ.

The elements in S3 with order dividing 4 are just the identity and trans- positions. Thus the homomorphisms φ : Z4 → S3 are defined by: φ(n)=1 φ(n) = (12)n φ(n) = (13)n φ(n) = (23)n Problem 5: (a) Firstly, 6 – 4=2 ∈ H + N, so <2> C H + N.

Since |S3|=3! =6 and |Z3|=3 then any function φ:S3→Z3 will be not one-to-one. So, the kernel must be non-trivial. If φ:S3→Z3 is a homomorphism that doesn’t send everything to ˉ0∈Z3 then it must be surjective (According to what I said in 1).