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.

Suppose the group is cyclic of order $n$ and generated by $t$ so that $x=t^k$ for some positive integer $k$.
Then we have $t^n=1$ and $t^{12k}=1$. Now $n$ is the least integer for which $t^n=1$ so that $12k=rn$ for some integer $r$.
Suppose $k\gt n$ so that $k=n+l$, then $x=t^l$ and we don't have a distinct solution. So we can restrict solutions to $0\le k \lt n$ in which case $0\le rn\lt 12n$ and $0\le r\lt 12$. There are therefore at most twelve possible values for $r$ and hence at most twelve possible values for $x$.
