If $G$ is an odd cyclic group of order $n$, then each element $g$ of $G$ has another element $h$ such that $h^2=g$. This is because $2 x = y \mod n$ is solvable for $x$. (Note this is not the same as solving $x^2=y \mod n$.)

What other finite groups have this property?

  • 2
    $\begingroup$ Groups of odd order? $\endgroup$ – Angina Seng Apr 8 '18 at 17:55
  • $\begingroup$ Oh yes, any cyclic group where 2 is invertible. Modifying my question. $\endgroup$ – abnry Apr 8 '18 at 17:56

If $G$ is a finite group of even order, then it has an element of order $2$, $a$ say. Then $a^2=e=e^2$, and the squaring map is not injective. By finiteness, the squaring map is not surjective: there are elements in $G$ which aren't squares.

If $G$ has odd order, then for each $b\in G$, $b^{(|G|+1)/2}$ is a square root of $b$.

| cite | improve this answer | |
  • 1
    $\begingroup$ Ah, very nice! Thanks! $\endgroup$ – abnry Apr 8 '18 at 17:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.