GCD's and how they generate groups

Question

I was reading something today an it was talking about $U_{15}$, all the integers relatively prime to 15, and how it was generated by the set {7,11}. I understood it all, but I thought that if the $(a,m)=1$ for $U_{m}$, then $a$, by itself generates the group. I am wrong in this case, but am I thinking of cyclic groups or something else entirely here?

Answer 1

The subgroups of $\mathbb{Z}_{15}^\times = \{1,2,4,7,8,11,13,14\}$ generated by $7$ and $11$ are $$ \langle 7 \rangle = \{1,7,4,13\}, \quad \langle 11 \rangle = \{1,11\}. $$ You need both of them to generate $\mathbb{Z}_{15}^\times$. Indeed, since $\mathbb{Z}_{15}^\times \approx \mathbb{Z}_2 \times \mathbb{Z}_4$ is not cyclic, you need two generators.

Answer 2

For the group $\mathbb Z_n$, every $1\le k < n$ with $\operatorname{gcd}(k,n)=1$ is a generator of the group, and moreover, these are precisely the generators. For the group $U_m$ however, the situation is more complicated. In particular, not every $U_m$ is cyclic.

Answer 3

$\mathbb{Z}/n$ is a cyclic group under addition. However, $U_n$ is the multiplicative group of elements in $\{1, 2, \dotsc, n-1\}$ that are coprime to $n$.

Indeed, generators of $\mathbb{Z}/n$ are integers smaller than $n$ and prime to it. But the operation in $U_n$ is different, so you shouldn't expect the same answer. In fact $U_n$ is very often not a cyclic group.