Difficulty in understanding isomorphism proof between 2 groups $Z_4$ and $U_5$

Question

What I got is (please correct if anything is wrong)

Given 2 groups $\mathbb{Z}_4$ and $U_5$, proving isomorphism starts with proving there exist a bijection between 2 groups. Since there are same number of elements in them we can always construct a bijection.

But the mapping which preserves group structure should be bijective.

As both groups are cyclic, and $\mathbb{Z}_4$ is with addition operation and $U_5$ is with multiplication operation.

I get that if one generator is mapped to another which could be $1\rightarrow2$, mapping is complete.

Mapping function could be defined as $f(i) = a^i$

By this, required condition $f(a*b)=f(a)*f(b)$ is easily satisfied.

All this is understood intuitively. But how do I convey what generators are mapped and how do I prove there is bijection without multiplication or cayley table?

Answer 1

Both groups have $4$ elements, and both groups are commutative. Furthermore all groups $U(p)$ are cyclic, see this duplicate. Hence $U(5)$ must be isomorphic to $\mathbb{Z}/4$. Mapping a generator to a generator yields an isomorphism.

Answer 2

To explicitly give an isomorphism:

Let $\phi: \mathbb{Z}_4\rightarrow\mathbb{Z}_5^\times$ such that $\phi(n)=2^n$. Then $\phi(a+b)=2^{a+b}=2^a2^b=\phi(a)\phi(b)$.

• Assume $\phi(n)=1$, then $2^n\equiv_51$ so $n\equiv_40$, since $2$ is a generator for $\mathbb{Z}_5^\times$. $\phi$ is injective.
• If $a\in\mathbb{Z}_5^\times$ then $a=2^m$, so $\phi(m)=a$. $\phi$ is surjective.