A group with 100 elements of order 5 is not abelian

Question

I'm struggeling with the proof, i used sylow's theorem to prove the order of the group G is divisible by 25, i don't know if it will help. I have no idea how to approach this

Answer 1

If a group is abelian, its set of elements of order 5 together with the identity forms a subgroup. In your case, that subgroup has 101 elements. That's impossible, for it has elements whose order does not divide its size.

Answer 2

A totally different approach: from the McKay proof of Cauchy's Theorem (see also here) it follows that in general for any prime $p$ dividing $|G|$ that $\#\{g \in G: \text{order}(g)=p\} \equiv -1$ mod $p$. Clearly $100 \equiv 0$ mod $5$. Hence such a group, whether abelian or not, does not exist at all.