[Math] Show that Z(7)\0 under modular multiplication is isomorphic to C(6)

cyclic-groupsgroup-theory

The problem:

I have (G, *) = (Z(7)\0, multiplication under modular arithmetic)

Specifically: G = {1,2,3,4,5,6}

I'm supposed to show that G is isomorphic to C(6) (the cyclic group with 6 elements).

This means that I need to show:

  1. There is a generator x with the order 6

  2. There is a bijection f: G -> C(6) such that f(x*x') = f(x) + f(x) where * is the operation for G and + is the operation for C(6).

What I'm wondering about:

Is G = <3> (that is: is 3 a generator for G?)

I think it is, since 3 repeatedly multiplied with itself gives the series:

3,2,6,4,5,1

How exactly do you prove a bijection like that? Especially, how do you manipulate the RHS? Are there specific rules for + that hold that doesn't hold for groups?

Do you do something like:

f(x*x') = f(3^x * 3^x') = f(3^(x+x')) = … something here …

Best Answer

You're really done. Proving that $\Bbb Z_7\setminus\{0\}$ is generated by $3$, you've shown that it is cyclic and has $6$ elements, and therefore must be isomorphic to $C_6$. You don't need to say more.

If you want to say more, let $f:\Bbb Z_7\setminus\{0\} \to C_6$ be the homomorphism with $f(3) = g$, where $g\in C_6$ is a generator of the cyclic group. Then we must have $f(2) = g^2$, $f(6) = g^3$, $f(4) = g^4$, $f(5) = g^5$ and $f(1) = g^6 = e$. It's rather easy to verify that this is indeed a homomorphism, although to make the workload lighter, it's probably best to rewrite $2 = 3^2, 6 = 3^3$ and so on. Then you can just point out that $f(3^i\cdot 3^j) = f(3^{i+j}) = g^{i+j} = g^i\cdot g^j = f(3^i)\cdot f(3^j)$, and you don't have to verify one entry at the time in the multiplication table..

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