Similar to how one can find a primitive root element of $\mathbb{Z}/p\mathbb{Z}$, I was wondering if, for any group of size $N$ that can be written as $(k\backslash\{0\},\cdot)$ where $k$ is a field, one can find an element of the group such that its order is $N$.
The sketch of my attempted reasoning is as follows:
Lemma 1: Legendre's Theorem says the order of every element of a group divides the size of the group, $N$.
Lemma 2: The order of every element divides the largest order of an element (Order of any element divides the largest order.).
Lemma 3: The maximum number of solutions to $x^n=1$ in a group given by $(k\backslash\{0\},\cdot)$, where $k$ is a field, is $n$ (https://wstein.org/edu/2007/spring/ent/ent-html/node28.html#prop:dsols, Does the maximum number of roots in a field directly imply the maximum number of solutions in a group).
From Lemma 2 and 3, if the largest order is $n$, then the maximum number of solutions is $n$, but $N$ elements must solve it, so $n\geq N$. From Lemma 1, $n|N$, so $n=N$.
Does this mean that every group that can be defined as $(k\backslash\{0\},\cdot)$, where $k$ is a field, is cyclic??? Or am I missing something? Is there an existing proof of this that has a name?
