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?

Clearly, the order of $G$ cannot be an odd, it's obvious. Suppose the order of $G$ is $2n$. Since order of identity is always $1$ i.e. $|e|=1$. So we have left only $2n-1$ element which is odd in number. Elements which are not their own inverses, these elements and their inverses exist in pairs.
Then it should be even in number but we have odd number here. In that case we must have atleast an element which must be self inverse. Elements of order $2$ are exactly those elements which are self inverse. Thus number of element of order $2$ must be in odd number. But we have given $60$ number of elements of $2$ which is absurd.
