Give an example of a non-Abelian group that has exactly four elements of order 10.

Would $D_{10}$ still work even though its not cyclic? I know that since $10$ is a positive divisor of $20$ the number of elements of order $10$ in a cyclic group of order $20$ is $\phi(10)=4$

Is there a better approach to find non-abelian groups with exactly $k$ elements of order $d$?

  • $\begingroup$ There can be exactly four elements of order 10 because any element $a$ of order 10 has its generated set $\langle a\rangle$ having exactly 4 generators: $a$, $a^3$, $a^7$, $a^9$. $\endgroup$ – Kenny Lau Sep 18 '17 at 5:51
  • $\begingroup$ So in this case there is actually only one cyclic subgroup of order 10. $\endgroup$ – Kenny Lau Sep 18 '17 at 5:51
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    $\begingroup$ "even though its not cyclic" any cyclic group would be abelian, so it is actually required that the group is not cyclic. $\endgroup$ – Kenny Lau Sep 18 '17 at 5:52
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    $\begingroup$ How about taking the product of $Z_{10}$ with a suitable non-Abelian group? $\endgroup$ – Lord Shark the Unknown Sep 18 '17 at 6:32
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    $\begingroup$ I am finding the comments confusing! The answer to your question is yes, $D_{10}$ (dihedral group of order $20$) works. $\endgroup$ – Derek Holt Sep 18 '17 at 8:17

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