Proving That a Group of Order 5 is Abelian

  • Thread starter Thread starter murshid_islam
  • Start date Start date
  • Tags Tags
    Group
AI Thread Summary
A group of order 5 is proven to be Abelian because it is cyclic, as every group of prime order is cyclic. The discussion emphasizes the importance of Lagrange's Theorem, which states that the order of a subgroup must divide the order of the group. By selecting any non-identity element in the group, one can show that it generates the entire group, confirming its cyclic nature. The conversation also highlights the necessity of understanding cosets, which are introduced in later sections of their study material. Mastering these concepts will simplify the proof that groups of prime order are Abelian.
murshid_islam
Messages
468
Reaction score
21
hi! i am new in abstract algebra.
how can i prove that a group of order 5 is Abelian?

thanks in advance.
 
Physics news on Phys.org
Are you familiar with Lagrange's Theorem?

Start picking some elements and generating cyclic subgroups, see what happens. What happens with coprime group orders? (say, 6). What happens with prime ordered groups?
 
Every group of prime order is cyclic and thus abelian.
 
Would you prove it was cyclic by proving the fact that all subgroups generated by an element have the order n/(n,s)=n or 1, where s is the generator and n is the number of elements in the group?
 
The fact that its cyclic is trivial. Pick any element s (not the 1). And consider the group that it generates. It has to generate the whole group because otherwise it would generate a subgroup. But the order of a subgroup must divide the order of the group.Since only 1 and p divide p (if p is prime) it must generate the whole group. Thus 1 element generates the whole goup and by defenition this means the group is cyclic.
 
"The order of a subgroup must divide the order of the group"--what if one hasn't seen that before, is there an easy proof, besides the psuedo-method i proposed?
 
philosophking said:
"The order of a subgroup must divide the order of the group"--what if one hasn't seen that before, is there an easy proof, besides the psuedo-method i proposed?

The above is known as Lagrange's theorem. It's fundamental to group theory. When I was studying group theory, I learned it before I knew the defintion of "cyclic". You really won't be able to go far without the theorem so I suggest you look up the theorem. Any textbook on goup theory will have it. The proof is not hard, but envolves the concept of cosets. Have you met that yet??
 
No, haven't met cosets. Yet. We're working out of Fraleigh's book, which is an excellent book. I just looked it up, and it's the next section we're doing. We just finished cycles. Can't wait!
 
Yeah its fun stuff ahead mate! Trust me, one you learn Lagrange's theroem, the question you asked will seem really easy to you.
 
Back
Top