1. ## Subgroups

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 $|g|=p$
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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 $\langle g'\rangle\le\langle g\rangle$?

2. Originally Posted by Drexel28
Hello everyone. In a recent post a member asked this question

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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

3. Originally Posted by tonio
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 , $|g|=n$ where $1<\ell. 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?

4. Originally Posted by Drexel28
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 , $|g|=n$ where $1<\ell. 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

5. Originally Posted by tonio
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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