$f(2_8)=(1234)$? an homomorphism?

Question

Proposition: Let $G,G´$ be groups, and let $G=\langle a\rangle$ be a cyclic group. If the order of $G$ is finite then, for any $b\in G´$, there is an homomorphism $f:G\to G´$ such that $f(a)=b$ iff $ord\: b|ord\: a$.

Problem:There is an unique homomorphism $f:\mathbb{Z}_8\to S_4$ such that $f(5_8)=(1234)$, given by $f(m.5_8)=(1234)^m$.

But there is no homomorphism $f:\mathbb{Z}_8\to\S_4$ such that $f(2_8)=(1234)$, despite $ord(1234)|ord(2_8)$.

I tried to check if this is an homomorphism but my attempt is wrong because I already assumed an homomorphism($f(2_8+2_8)=f((2_8)^2)=(1234)^2=(1234)(1234)=f(2_8)f(2_8)$).

Question:

How can it be there is no homomorphism between $f(2_8)=(1234)$? How can we know there is no homomorphism?

Thanks in advance!

Answer 1

The key point is that the proposition requires that the input which the homomorphism is defined at, $a$, is a generator of the group. $2$ is not a generator of $\mathbb{Z}_8$, so the proposition does not apply.

Essentially, the problem is the existence of $1$, which is half of $2$. Any homomorphism would require $f(1)$ to be a square root of $f(2)$ but no square root of $(1234)$ exists.

Answer 2

I think you should ask yourself

What happens if I define $f(2_8)=(1234),$ then where do $3_8, 5_8, 7_8$ go?

I think you'll find the answer to this to be the answer to your question.