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Prove that the additive group $Z_6$ is isomorphic to the multiplicative group of nonzero elemnets in $z_6$ is isomorphic to multiplicative group of nonzero elements of $Z_7$

Just make a table and show that it is homomorphic, 1-2-1 and onto? With some mapping or is there some theorems that could be used?

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    Constructing a map would be easy with such a small group. Remember, if $\varphi:(\mathbb{Z}/6\mathbb{Z})^{+}\rightarrow (\mathbb{Z}/7\mathbb{Z})^{\times}$ is a homomorphism, we must have $$\varphi(0+6\mathbb{Z})=1+7\mathbb{Z}.$$ We must also have, $$\varphi((a+6\mathbb{Z})+(b+6\mathbb{Z}))=\varphi(a+6\mathbb{Z})+\varphi(b+6\mathbb{Z}).$$ – 2017-02-15

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You are trying to build some $f: (\mathbb Z_6, +) \to (\mathbb Z_7^\times, \cdot)$ which is an isomorphism.

If $f(1)=a$ then it is easy to show that $f(n)=a^n$. Thus, to get an isomorphism, all you need is to find some element $a$, such that $a,a^2, a^3,a^4, a^5, a^6$ are the six elements of $\mathbb Z_7^\times$, i.e. an element of order $6$.

Since there are $6$ elements in $\mathbb Z_7^x$, and you already know the order of $\pm 1$, there are only $4$ choices left to try.

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    $|5|=6$ in $Z_7$ – 2017-02-15
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Let us prove a more general statement:

Proposition. Let $G$ be a finite subgroup of the multiplicative group of a field $K$, then $G$ is cyclic.

Proof. Let $N$ be the exponent of $G$, then each element of $G$ is a root of $X^N-1$ in $K$, since $K$ is a field, the polynomial $X^N-1$ has at mots $N$ roots in $K$, then: $$\#G\leqslant N.$$ However, using Lagrange's theorem, one has: $$N\leqslant\#G.$$ Finally, $N=\#G$ and $G$ is cyclic. $\Box$

Using Bézout's theorem, $\mathbb{Z}/7\mathbb{Z}$ is a field, then using the proposition: $(\mathbb{Z}/7\mathbb{Z})^\times$ is cyclic of cardinality $6$.

The problem then boils down to prove that there is only one cyclic group of order $n$ up to isomorphism, which is done sending a generator of the first group onto the other one.