What finite groups always have a square root for each element?

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?

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

1 Answer

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 $$b$$.

• Ah, very nice! Thanks! – abnry Apr 8 '18 at 17:59