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Show that $(U_{7}, \times) \cong (Z_6, +)$. ($U_{7}$ is the set of units in $Z_7$, and $Z_6$ is the set of integers modulo 6)

I understand this in theory, that we want to define a function $\phi : U_{7} \rightarrow Z_{6}$ that's a homomorphism, and is one-to-one and onto. But there's a lot of things about this I feel like I don't understand or that haven't been explained to me fully.

How do you decide how to define the function?

Is it necessary to prove separately that the function is a homomorphism, and that it's one-to-one, and that it's onto?

How do you prove that a function is a homomorphism? How do you prove a function is onto?

Edit: Following HallaSurvivor's advice I figured out a function that I think should work: $\phi([x]_6)=[3^x]_7$

So now I'm left with my other questions: How do I prove this is a homomorphism and that this is onto, and is it even necessary to do so?

d.v.
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  • For the first question, it usually helps to look at the identity element in each group as well as the set of generators of each group. – RyanK Nov 04 '20 at 01:51
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    Hint: Seven is prime and zero is not a unit. – Shaun Nov 04 '20 at 01:55
  • Hint: Try going the other direction. $\mathbb{Z}/6$ is a really easy group to work with, and once you pick $f(1)$, you're forced into $f(2) = f(1+1) = f(1) \times f(1)$ (since $f$ has to be a homomorphism). Similarly, $f(3) = f(1)^3$, etc. – HallaSurvivor Nov 04 '20 at 02:04
  • By the way, the group $U(p)$ is cyclic, see this duplicate and has $\phi(p)=p-1$ elements. So we obtain an isomorphism $U(p)\cong C_{p-1}\cong \Bbb Z/(p-1)$. Now take $p=7$. – Dietrich Burde Nov 04 '20 at 11:40

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$\textbf{Hint}$ : Think of a map $f$ : $\mathbb{Z}_6$ $\rightarrow$ $U_7$ sending a $generator$ to a $generator$ of respective group. Rest is algebraic manipulation.

Swarup Das
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