# Thread: Group whose order is a power of a prime contains an element of order prime

1. ## Group whose order is a power of a prime contains an element of order prime

I don't know how to go about this. So the group has $p^{n}$ elements and I have to somehow use cosets or Langrage's theorem, but I really don't understand much of this. And I can't use Cauchy's Theorem, it uses concepts I haven't covered yet. Any help would be appreciated.

Edit: So I know by Lagrange's theorem that the order of a subgroup of $G$ must divide the order of $G$, so given $g\in G$, the cyclic subgroup $$ has order that divides $p^{n}$. At this part, I'm stuck.

2. So since $p$ is prime, the order of $g$ is either $p$ or some power of $p$, say $p^m$. If $|g|=p$, then we're done, so assume that $|g|\neq p$. So an element $g$ has order $p^m$ if $p^m$ is the smallest positive integer with the property that $g^{p^m}=1$.

Now $g$ generates a cyclic subgroup of $G$ that consists of $p^m$ elements, each of which is a power of $g$.

So let $g^k$ be an element in $\langle g\rangle$ where $k$ is a positive integer and $k\leq p^m$. Now $g^k$ generates a subgroup of $\langle g\rangle$, and the order of $g^k$ must divide $p^m$. Now since $k\leq p^m$, is there some value of $k$ such that $k^p=p^m$?

Check out the section on cyclic subgroups in Artin's Algebra. There is a proposition that talks about the GCD of the order of a group and the order of an element in the group. makes the rest of the proof a snap.

3. I followed you until you brought up $k^{p} = p^{m}$, but I think I got a solution:

Assume $g\ne 1_{G}$ If the order of $$ is $p^{n}$, then $G = $ and so the element $g^{p^{n-1})$ has order $p$ since $(g^{p^{n-1}})^{p} = g^{p^{n}} = 1$, and since $p$ is prime, the claim follows. Now if the order of $$ is not $p^{n}$, then it's order must divide $p^{n-1}$, since the order of $$ must be an element of $P = \{ p, ..., p^{n-1}\}$. So let $p^{m} = ord()$. Then $p^{m}= \frac{p^{n}}{p^{n-m}}$. Then $g^{\frac{p^{n-1}}{p^{n-m}}$ has order $p$. Q.E.D.

4. In fact there are at least p-1 such elements. One of my favorite proofs in algebra is McKay's proof of Cauchy's theorem, which is outlined in Dummit and Foote:

Consider the set S of p-tuples of elements in G whose product is the identity: $S=\{(g_1,\dots,g_p)\mid \prod g_i=1\}$. Note there are $|G|^{p-1}$ of these, since we can choose the first p-1 elements however we like, then choose the last to be the inverse of their product.

Next show that the cyclic group $\mathbb{Z}/p\mathbb{Z}$ acts on the set of p-tuples by cyclic permutation, i.e. if the action of 1 on a p-tuple is to shift every entry one place to the right, and move the last entry into the first place, then the product of the elements remains the identity.

Now by the orbit-stabilizer theorem, the size of each orbit under this action divides p, so every orbit has size 1 or p, and the union of all orbits is the set S. Since p divides |S|, the number of size-1 orbits is divisible by p. A size-1 orbit is one that is unchanged by cyclic permutations, i.e. a p-tuple consisting of just one element: (g, g, ..., g). If there's a tuple like that, it means $g^p=1$, and so if g is not the identity, then |g|=p as desired. But there has to be a nonidentity element like that, since (1, 1, ..., 1) is in S, and so there are at least (p-1) more with orbit size 1.