Showing that $(\mathbb{Z}_6,+)$ and $(U(\mathbb{Z}_7), \cdot)$ are isomorphic groups

Question

We have to prove that $(\mathbb{Z}_6,+)$ and $(U(\mathbb{Z}_7), \cdot)$ are isomorphic.

I know that they are and the isomorphism is as follows: $f(0)=1, f(1)=3, f(2)=2, f(3)=6, f(4)=4, f(5)=5$.

It is obvious that it is bijective, but how do I prove that this is a homomorphism?

Answer 1

The group of units of $\mathbb{Z}/7\mathbb{Z}$ has six elements and is abelian,but the only abelian group having six elements is $\mathbb{Z}/6\mathbb{Z}$,another way (following your approach) is to prove that $3$ has order 6,since homomorphisms of cyclic groups are entirely determined by the image of generator.

Answer 2

By construction, your bijection $f$ sends $n1\in\Bbb Z_6$ to $3^n\in U(\Bbb Z_7)$ so it is an homomorphism.