For weeks I tried to solve the following question on Brilliant:
Fill in the blank: "Every group of order ___ is abelian."
And these are the possible answers I get: 15, 16, 20, 21, 27.
Using Sylow's theorems I tried my best to get to a conclusive answer, but unfortunately all I could do is guess that it is probably 15, because the others seem unlikely. So I clicked on 15 and submitted my answer and it turns out that my gut feeling was right. Hooray! But what was the explanation?
Obviously a group G of order 15 has Sylow subgroups of orders 3 and 5. It's also quite obvious they have to be normal in G, because of Sylow's third theorem. This means that their product is isomorphic to G. And since they are both abelian, so is G.
Hang on?! Any group of order 3 obviously must be abelian, but how was I supposed to know that a Sylow 5-subgroup is abelian as well? It is possible for a group of order 5 to be non-abelian, so what piece of information am I missing here?
Are Sylow 5-subgroups always abelian and if so, how was I supposed to know that?
Best Answer
No, it is not possible.
Any group of prime order is cyclic (as shown here) and thus (as shown here) Abelian.