Is a Group of Order 15 Always Abelian?

[Jupiter]

I need to prove that a group of order 15 is abelian. I have (tried to) attached my work but it's too long. Basically, I'm looking at the class equation and considering all possibile orders of the center Z(G). I've successfully eliminated the cases where |Z|=3,5. I'm having trouble with coming up with a contradiction when assuming |Z(G)|=1.
I'm also interested in seeing alternative proofs (perhaps something more elegant), and also a proof that G is in fact cyclic.
Is there any decent generalization of this problem?

 

[Kalimaa23]

Try and look up the fundamental theorem of finite abelian groups. I think it has to do with the fact that 15 is the product of two primes, and that 3*5 is the only decomposion possible. So there is only group of order 15, which is the product of two cyclic groups of order 5 and order 3. Since 3 and 5 are relatively prime, The group of order 15 has an element of order 15 generating it and thus is cyclic.

Was that helpfull?

 

[siddhuiitb]

Hey Dimitri!

15 = 3*5
So, there will exist elements of orders 3 and 5 say 'a' and 'b'. that's correct. But then how can u conclude that there will exist an element in the group of order 15.
If group were abelian then this fact was true, bcoz then o(ab)= lcm (3,5)=15.
But here we have to prove that group is indeed Abelian.