I know this is question has been asked several times on here, only hints given ,but just want to check if I have the right idea.
My attempt: Suppose G is non abelian finite group and $|Z(G)| \gt \frac {1}{4} |G|$. Since Z(G) is a subgroup of G then it’s order must divide that of G , i.e it could be of size |G|/3 , |G|/2 or |G| .That means the quotient group G/Z has order 1 ,2 or 3 . If equal to 1 that implies Z(G)=G so abelian , and also since 2 and 3 are prime numbers that means G/Z is cyclic which also implies it’s abelian. Which leads to a contradiction. Is that right? Thanks