# 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.

Let $G$ be a cyclic group of order $n$.

Then $x^{12}=1$ iff $x^{d}=1$, where $d=\gcd(n,12)$.

Moreover, the solutions of $x^{d}=1$ form the cyclic subgroup of order $d$.

This implies that $d$ cannot be $14$, because $14$ is not a divisor of $12$.

I suggest proving instead that for any $r$, the number of solutions of equation $x^{r}=1$ in a cyclic group is at most the number of divisors of $r$.

• This can't be right because for $r$ prime you'd get at most $2$ solutions but there'll be $1$ or $p$ solutions. I think you mean "at most $r$".
– lhf
Commented Apr 27, 2017 at 13:06

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.

• this is a nice way of going about things. Commented Jun 12, 2020 at 8:38

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 $$x$$.