Proving That a Group of Order 5 is Abelian

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Homework Help Overview

The discussion revolves around proving that a group of order 5 is Abelian, situated within the context of abstract algebra and group theory.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of Lagrange's Theorem and the properties of cyclic groups. There are attempts to clarify the relationship between subgroup orders and group orders, as well as questions about foundational concepts like cosets.

Discussion Status

The discussion includes various perspectives on proving the cyclic nature of groups of prime order. Some participants suggest looking into Lagrange's Theorem as a critical component, while others express enthusiasm about upcoming topics in their studies.

Contextual Notes

One participant notes that they have not yet encountered cosets, indicating a potential gap in foundational knowledge that may affect their understanding of the discussion.

murshid_islam
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hi! i am new in abstract algebra.
how can i prove that a group of order 5 is Abelian?

thanks in advance.
 
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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 definition 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.
 

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