There exists a homomorphism $\sigma$ from $[a]$ to $[b]$ such that $\sigma(a)=b^k$

Question

Assume that $[a]$ is a cyclic group of order $m$ and $[b]$ is a cyclic group of order $n$.
Prove that:
There exists a homomorphism $\sigma$ from $[a]$ to $[b]$ such that $\sigma(a)=b^k$, iff $n$ divides $mk$.

Also, Prove that if $mk=qn$, Then $\sigma$ is an automorphism iff $(m,q)=1$

The question is taken from the book 'Abstract Algebra' by Bhattacharya . (Chapter 5 - Section 2 - Question #6)

I don't know what to do. I should provide a homomorphism but how on earth should this come to my mind? There's no clue!

Answer 1

The first if exist homomorphism $$\sigma : [a] \to [b]$$ $$\sigma(a) = b^{k}$$ $$\sigma(a^{m}) = b^{km} = 0$$

So we must have $n$ is a divisor of $km$ . Now the converse, we have $n \mid km$

Define :

$$\sigma : [a] \to [b]$$ $$\sigma(a) = b^{k}$$

Now let me check it's well-defined

$$a^{u}=a^{v}$$

$$ u \equiv v (\mod m )$$

We need $\sigma(a^{u})=\sigma(a^{v})$ , it's equivalence with

$$ku \equiv kv (\mod n )$$

The remainder is left to you