$(\mathbb Z_7^{*},\cdot)$ is isomorphic to $(\mathbb Z_6,+)$ Just a short question: Is $(\mathbb Z_7^{*},\cdot)$ isomorphic to $(\mathbb Z_6,+)$?
I would say "yes", since both group have order 6, hence the left hand side only can be isomorphic to $S_3$ or $\mathbb Z_6$. However, the left group is abelian, hence must be isomorphic to $\mathbb Z_6$. Correct?
 A: Your answer and reasoning is correct. If you want an explicit isomorphism, you just need to show that $\mathbb{Z}_7^\times$ is cyclic. Then if you find a generator, you get an isomorphism by mapping that generator to $1 \in \mathbb{Z}_6$. But be careful, because there are some elements that aren't generators.
A: The argument you gave is correct. However, it might be good to be aware of the fact that for $p$ a prime number $\mathbb{Z}^{\times}_p$ is always isomorphic to $(\mathbb{Z}_{p-1},+)$, even though there can be commutative groups other than that group of order $p-1$.
A: In a general way, it's good to know any finite subgroup of the multiplicative group of a field is cyclic, whether this field is finite or not.
Indeed, let $G$ be such a finite subgroup and $r$ be the exponent of $G$. All elements of $G$ are roots of the polynomial $x^r-1$. In the field, this polynomial has at most $r$ roots, hence $\lvert G\mkern1mu\rvert\le r$. However, r is a divisor of $\lvert G\mkern1mu\rvert$, and thus $\lvert G\mkern1mu\rvert=r$.
Now, in an abelian group with exponent $r$, there exists an element of order $r$. This element is a generator of $G$.
