0
$\begingroup$

Let $p$ be prime and let $G$ be a group of order $p^n$ for some $n \in \Bbb N$. Show that $G$ has an element of order $p$. (Hint: Choose an element $1 \neq a \in G$ and consider the group $\left<a\right>$. Then find an element of order $p$ in this group.)

Approach: It was already proven that $\left<a\right>$ is a subgroup of $G$, so by Lagranges theorem $\def\ord{\operatorname{ord}}\ord(a)$ divides $|G|$. This implies $\ord(a)\mid p^n$. We can say $p^{n}=\ord(a)k$ for some $k \in \Bbb Z$, so $p^n/k=\ord(a)$.

$\left<a\right>=\{1_G,a,a^2,....,a^{\ord(a)-1}\}$, so we have to find an $1\leq l \leq\ord(a)-1$ such that $\ord(a^l)=\frac{{p^n}/k}{\gcd({p^n/k},l)}=p$. More specific, we have to find an $l$ such that $\gcd({p^n/k},l)=p^{n-1}/k$

This is my approach. What do you think?

$\endgroup$
  • $\begingroup$ "Question about orders of groups in abstract algebra" is an uninformative title. 1. of course it's a "question": this is a question and answer site. 2. you could ask thousands of things about the "orders of groups": which thing are you asking? 3. of course it's "in abstract algebra": that is, after all, the domain in which we talk about groups. $\endgroup$ – symplectomorphic Aug 13 '16 at 6:34
1
$\begingroup$

Note first that the result as stated is not true, as it fails for $n=0$ (a group of order $p^0=1$ clearly has no element of order$~p$). This is not a big deal as it suffices to add $n>0$ to the hypotheses, but it does point to a strange aspect of the problem statement: what really matters for the conclusion is the presence of at least one prime factor $p$ in the factorisation of $|G|$ (assumed finite), not the absence of other prime factors, which is really all that the given hypothesis states. In fact Cauchy's theorem says that whenever $|G|$ is divisible by a prime number$~p$, there is a element of$~G$ with order$~p$.$\def\ord{\operatorname{ord}}$

This being said, the absence of other prime factors does make the proof easier, since one does not have to go searching for an element whose order contains a prime factor$~p$: every element $a\neq e$ will satisfy that property (and such$~a$ exists provided that $n>0$), since its order cannot contain any other prime factors. This is what the hint and application of Lagrange's theorem give; so far your approach is fine.

Where you go somewhat astray is in analysing the situation inside the cyclic group $\left<a\right>$. It is a general fact that in searching for an element of order$~d$, it matters little if one actually finds an element$~a$ whose order is a multiple of$~d$: if $\ord(a)=md$ for some $m\in\Bbb Z$, then $\ord(a^m)=d$ as one easily checks, and one can take $a^m$ instead of$~a$. Stated differently: a cyclic group of a given order$~k$ contains a cyclic subgroup of every order dividing $k$ (indeed it contains a unique such subgroup, though that is not used here). Concretely, if your element $a$ has $\ord(a)=p^l$ (the only possibility, since all divisors of $p^n$ are again powers of$~p$) with $l>0$, then $\ord(a^{p^{l-1}})=p^l/p^{l-1}=p$, so the element $a^{p^{l-1}}$ answers the question.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Pick any non-identity element $g$, by lagrange's theorem its order is $p^k$ for some $k\leq n$. consider the element $h=g^{p^{k-1}}$ it is not $e$ because $p^{k-1}$ is smaller than the order of $g$.

On the other hand $h^p=(g^{p^{k-1})^p=^{p^k}}=e$. So $h$ is the element we wanted.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ In my version of lagrange's, it divides $p^k$ $\endgroup$ – TheMathNoob Aug 13 '16 at 5:43
  • 1
    $\begingroup$ The order of the subgroup divides $p^n$, and for prime $p$, the only numbers which divide $p^n$ are $p^k$ where $k\leq n$. $\endgroup$ – florence Aug 13 '16 at 5:48
  • $\begingroup$ I think that's a theorem in number theory. Can you point it out somewhere? $\endgroup$ – TheMathNoob Aug 13 '16 at 5:50
  • $\begingroup$ If $ord(a)$ had some other prime factorization, then there would be a prime $q \neq p$ such that $q | ord(a)$. However, if $q | ord(a)$ and $ord(a) | p^k$, then $q | p^k$ which is nonsense since the only prime that divides $p^k$ is $p$. Thus, $ord(a)$ must be of the form $p^k$ for $k\neq n$. $\endgroup$ – benguin Aug 13 '16 at 6:03
  • $\begingroup$ the only way that can happen is if $q=1$, but $a \neq 1$ $\endgroup$ – TheMathNoob Aug 13 '16 at 6:11
0
$\begingroup$

The order of $G$ can be more general. See Cauchy's theorem

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.