# Does there exists a finite abelian group $G$ containing exactly $60$ elements of order $2$?

Suppose there exists such a group. Then Lagrange's theorem assures that the group is of even order.

But I conclude from this and this that such a group has odd number of elements of order $2$. Giving us contradiction.

Hence there does not exist a finite abelian group $G$ containing exactly $60$ elements of order $2$.

More strongly there does not exist a finite group $G$ containing even number of elements of order $2$.

Is my understanding correct?

• Did you get this question from TIFR entrance ?? – Anik Bhowmick Oct 8 '18 at 3:56
• Yes GS 2017 . {}{}{}{}{} – blue boy Dec 7 '18 at 6:00

Consider the relation $\sim$ on $G$ (having even order, otherwise it has no element of order $2$) defined by $$a\sim b \quad\text{if and only if}\quad (b=a\text{ or }b=a^{-1})$$ This is easily seen to be an equivalence relation. The equivalence classes have either one or two elements. If you remove the two-element equivalence classes, you are dropping an even number of elements from $G$, so what remains is an even number. Drop also the class consisting of $1$ and you remain with an odd number of one-element equivalence classes: these elements are precisely those having order $2$.