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Thread: If the order of a group is prime number then the group is cyclic?

  1. #1
    Senior Member x3bnm's Avatar
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    If the order of a group is prime number then the group is cyclic?

    Let $\displaystyle G$ be a group with a prime number $\displaystyle p$ of elements. If $\displaystyle a \in G$ where $\displaystyle a \neq e$,
    then the order of $\displaystyle a$ is some integer $\displaystyle m \neq 1$.

    But then the cyclic group $\displaystyle \langle a \rangle$ has $\displaystyle m$ elements.

    By Lagrange's theorem, $\displaystyle m$ must be a factor of $\displaystyle p$.

    But $\displaystyle p$ is a prime number, and therefore $\displaystyle m = p$.

    It follows that $\displaystyle \langle a \rangle$ has $\displaystyle p$ elements, and is therefore all of $\displaystyle G$!

    Conclusion:

    If $\displaystyle G$ is a group with a prime number $\displaystyle p$ of elements, then $\displaystyle G$ is a cyclic group. Furthermore,
    any element $\displaystyle a \neq e$ in $\displaystyle G$ is a generator of $\displaystyle G$.



    The author said that:

    "But then the cyclic group $\displaystyle \langle a \rangle$ has $\displaystyle m$ elements."

    I understand that the cyclic group has $\displaystyle m$ elements. No problem there. I understand this statement is true for the generator $\displaystyle a$.

    But why the group $\displaystyle G$ has to be definitely cyclic at the end?


    Why the cyclic group $\displaystyle \langle a \rangle = G$?


    Can't there be situations where $\displaystyle aaaa....a \notin G$ because we didn't say that $\displaystyle \langle a \rangle$ is a subgroup of $\displaystyle G$?

    So how did he come to the conclusion that $\displaystyle G$ is cyclic?
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  2. #2
    Senior Member x3bnm's Avatar
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    Re: If the order of a group is prime number then the group is cyclic?

    Don't worry I found the solution. The answer is on Pinter's Abstract Algebra book page-111:

    If $\displaystyle G$ is any group and $\displaystyle a \in G$, it is easy to see that:
    .......

    c) the set of all the powers of $\displaystyle a$ is a subgroup of $\displaystyle G$

    This subgroup is called the cyclic subgroup of $\displaystyle G$ generated by $\displaystyle a$.


    So the answer to my problem in my last post:

    $\displaystyle \text{For all element } a \in G, \langle a \rangle \text{ is a subgroup of } G$

    That's it. That's all I wanted to know.
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  3. #3
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    Re: If the order of a group is prime number then the group is cyclic?

    to answer your earlier question as asked:

    no, for any element a of G, any power of a is ALSO in G, by closure.
    Thanks from x3bnm
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  4. #4
    Senior Member x3bnm's Avatar
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    Re: If the order of a group is prime number then the group is cyclic?

    Quote Originally Posted by Deveno View Post
    to answer your earlier question as asked:

    no, for any element a of G, any power of a is ALSO in G, by closure.
    Thanks Deveno.
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