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  1. #1
    MHF Contributor Drexel28's Avatar
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    Hello everyone. In a recent post a member asked this question

    Let p be a prime and n\ge1. Prove that if G is a group with |G|=p^n that G contains an element g with |g|=p
    .

    TheEmptySet provided a solution that was of course satisfactory, but I came up with an alternative proof. Now the proof seems "too simple" and I am moderately sure I made a fatal assumption. Can someone validate this?

    Proof: Let g\in G. If |g|=p we are done, otherwise by Lagrange's theorem |p|=p^\ell for 2\le \ell\le n. But this would mean that \left|\langle g\rangle\right|=p^\ell and since this subgroup is cyclic and p\text{ }|\text{ } \left|\langle g\rangle\right| wouldn't this mean that there is some g'\in\langle g\rangle such that \left|g'\right|=p and \langle g'\rangle\le\langle g\rangle?
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    Quote Originally Posted by Drexel28 View Post
    Hello everyone. In a recent post a member asked this question

    .

    TheEmptySet provided a solution that was of course satisfactory,


    I don't have the faintest idea what you mean by this.


    but I came up with an alternative proof. Now the proof seems "too simple" and I am moderately sure I made a fatal assumption. Can someone validate this?

    Proof: Let g\in G. If |g|=p we are done, otherwise by Lagrange's theorem |p|=p^\ell for 2\le \ell\le n. But this would mean that \left|\langle g\rangle\right|=p^\ell and since this subgroup is cyclic and p\text{ }|\text{ } \left|\langle g\rangle\right| wouldn't this mean that there is some g'\in\langle g\rangle such that \left|g'\right|=p and \langle g'\rangle\le\langle g\rangle?

    I can't see how you deduce that there exists some g'\in\langle g\rangle such that \left|g'\right|=p unless you already know and can use the fact that a finite cyclic group has one unique sybgroup of any order dividing the group's order...

    I'd rather go like this in this part: let g\in G\Longrightarrow ord(g)=p^k, by Lagrange's theorem. If k =1 we're done, otherwise g^{k-1} is an element of order p and again we're done.

    Tonio
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tonio View Post
    unless you already know and can use the fact that a finite cyclic group has one unique sybgroup of any order dividing the group's order...
    Yes. That is in fact what I was refering to. The fundamental theorem of cylic groups. So basically my proof was based on two things

    1. p is prime, therefore by LT any element g of G must have order p^m\quad 1\le m \le n.

    2. Using that fact we know that there are three possiblities for any g\in G. Namely: |g|=p or |g|=p^\ell \quad 1<\ell<n , |g|=n where 1<\ell<n. If the first case is true then we are done. If the second is true then \langle g\rangle is a cyclic subgroup of G of order p^\ell and using the fundamental theorem of cyclic groups we can conclude there is a subgroup of \langle g\rangle of order p, which of course means that the gnerator of that group is of order p. If the third case is true then G itself is cyclic, in which case we may apply the fundamental theorem again.

    Is there something wrong with that argument?
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    Quote Originally Posted by Drexel28 View Post
    Yes. That is in fact what I was refering to. The fundamental theorem of cylic groups. So basically my proof was based on two things

    1. p is prime, therefore by LT any element g of G must have order p^m\quad 1\le m \le n.

    2. Using that fact we know that there are three possiblities for any g\in G. Namely: |g|=p or |g|=p^\ell \quad 1<\ell<n , |g|=n where 1<\ell<n. If the first case is true then we are done. If the second is true then \langle g\rangle is a cyclic subgroup of G of order p^\ell and using the fundamental theorem of cyclic groups we can conclude there is a subgroup of \langle g\rangle of order p, which of course means that the gnerator of that group is of order p. If the third case is true then G itself is cyclic, in which case we may apply the fundamental theorem again.

    Is there something wrong with that argument?

    I don't think so, though the |g|=n surely was meant to be |g|=p^n

    Tonio
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tonio View Post
    I don't think so, though the |g|=n surely was meant to be |g|=p^n\color{red}\text{yes. Thank you}

    Tonio
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