Suppose equation $x^{12} = 1$ has $14$ solution in some group. Show that this group is not cyclic. 
Suppose equation $x^{12} = 1$ has $14$ solution in some cyclic group.
  Show that this group is not cyclic.

Any help would be appreciated. I was trying to show this by contradiction, but I didn't go too far.
Attempt:
Suppose that equation has $14$ solutions in some cyclic group $C_n$. then if $a^k \in C_n$ is one of the solution we have that $(a^k)^{12} = 1 \implies a^k$ is a generator of $C_n$. From the condition of the problem there are $14$ different $k$'s such that $a^k = 1 \implies$ there are $14$ generators $\implies \varphi (n) = 14$.
At first glance, it seems to me that I have to find the smallest $n$ such that $\varphi (n) = 14$, but in class our teacher haven't mentioned anything about the inverse of the totient, so maybe I'm wrong with what I've done.

Here is another take.
There is only one finite cyclic group of each order, up to isomorphism.
$\mathbb C^\times$ contains a copy of the cyclic group of order $n$: it is the subgroup of $n$-th roots of unit.
The equation $x^{12} = 1$ has at most $12$ solutions in $\mathbb C$ and so cannot have $14$ solutions.
