# Group with exactly 2 elements of order 10.

Does this exist? I dont 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 that 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?

• You answered completely your own question. Check it carefully. +1 – Timbuc Oct 7 '14 at 12:29
• i wanted to make sure in because I havent really worked with dihedrals or symmetric groups that much. – Jack Armstrong Oct 7 '14 at 12:33
• what you know and mention about cyclic groups is more than enough since any group having an element of order ten has a cyclic subgroup of order ten... obviously. No dihedral or symmetric stuff required. – Timbuc Oct 7 '14 at 12:36
• It has nothing to do with dihedral or symmetric groups - see the answers. And moreover $\phi(10)\neq 2$. – Dietrich Burde Oct 7 '14 at 12:39

## 2 Answers

can I apply that idea to non cyclic groups such as dihedrals or symmetric groups?

Yes. If you have an element of order 10 in any group, it generates a subgroup of order 10, which is cyclic by definition. Since this subgroup has four generators, each of those is an element of order 10 in the subgroup, and is also of order 10 in the original group.

• What do you mean with "yes"? The OP asks if a group exists with exactly two elements of order 10 ... – Nicky Hekster Oct 7 '14 at 12:33
• There is only one question, which is "but can I apply that idea to non cyclic groups such as dihedrals or symmetric groups?". I answered that. – fkraiem Oct 7 '14 at 12:34
• Not clear to me ... – Nicky Hekster Oct 7 '14 at 12:35
• Well, I added a direct quote. – fkraiem Oct 7 '14 at 12:37

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 $D_6\cong S_3 \times C_2$.