# showing that a group of order $p^n$ has an element of order $p$

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?

• "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. – symplectomorphic Aug 13 '16 at 6:34

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 $$\left$$. 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.

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.

• In my version of lagrange's, it divides $p^k$ – TheMathNoob Aug 13 '16 at 5:43
• The order of the subgroup divides $p^n$, and for prime $p$, the only numbers which divide $p^n$ are $p^k$ where $k\leq n$. – florence Aug 13 '16 at 5:48
• I think that's a theorem in number theory. Can you point it out somewhere? – TheMathNoob Aug 13 '16 at 5:50
• 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 $k\neq n$. – benguin Aug 13 '16 at 6:03
• the only way that can happen is if $q=1$, but $a \neq 1$ – TheMathNoob Aug 13 '16 at 6:11

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