TITLE: Group with exactly 2 elements of order 10.
QUESTION [9 upvotes]: Does this exist?  I don't think it does.  For any cyclic group the totient function of 10 is 4, so there is 4 of them.  But also if one element is of order 10, say $a$, then $a^3$, $a^7$ is also of order 10 but thats three of them.  Now I understand what I just did is mainly for cyclic groups, but can I apply that idea to non cyclic groups such as dihedrals or symmetric groups?

REPLY [3 votes]: If $x$ is an element of order $10$, then also $x^3, x^7, x^9$ are all different elements of order 10. So it is impossible to have exactly two of them.
In general, if $n$ is a natural number and $G$ is a group with exactly $2$ elements of order $n$, then $n=3, 4$ or $6$. This follows from solving the equation $\varphi(n)=2$, where $\varphi$ is Euler's totient function. Each of the cases can be realized: take $G \cong D_n$, with $n=3,4$ or $6$.
Note that $D_3\cong S_3$ and $D_6\cong S_3 \times C_2$.