# p-groups

#### ibnashraf

##### New member
Question:

"Let G be a finite group. Prove that G is a p-group iff every element of G has order a power of p."

this was my attempt at the question:

Suppose G is a p-group
$\Rightarrow |G|=p^n$
, where
$n\in Z$

Let g
$\in G$
$\Rightarrow |G|/|g|$

$\therefore p^n/|g|$

Hence, every element of G has order a power of p.

Conversely, suppose that every element of G has order a power of p
$\Rightarrow |g|=p^n$

But
$|G|/|g|$

$\therefore p^n$
divides
$|G|$

Hence, G is a p-group.

can someone look over this and tell me what mistakes i made if any please ?

#### Sudharaka

##### Well-known member
MHB Math Helper
Question:

"Let G be a finite group. Prove that G is a p-group iff every element of G has order a power of p."

this was my attempt at the question:

Suppose G is a p-group
$\Rightarrow |G|=p^n$
, where
$n\in Z$

Let g
$\in G$
$\Rightarrow |G|/|g|$

$\therefore p^n/|g|$

Hence, every element of G has order a power of p.

Conversely, suppose that every element of G has order a power of p
$\Rightarrow |g|=p^n$

But
$|G|/|g|$

$\therefore p^n$
divides
$|G|$

Hence, G is a p-group.

can someone look over this and tell me what mistakes i made if any please ?
Hi ibnashraf,

The definition of a p-group is,

Given a prime $$p$$, a $$\mbox{p-group}$$ is a group in which every element has order $$p^k$$ for some $$k\in\mathbb{Z}\cup\{0\}$$.

The statement that is to be "proved" is in fact the definition of a p-group.

Kind Regards,
Sudharaka.

#### Swlabr

##### New member
Hi ibnashraf,

The definition of a p-group is,

Given a prime $$p$$, a $$\mbox{p-group}$$ is a group in which every element has order $$p^k$$ for some $$k\in\mathbb{Z}\cup\{0\}$$.

The statement that is to be "proved" is in fact the definition of a p-group.

Kind Regards,
Sudharaka.
I presume the definition of $p$-group the OP has is "$G$ is a $p$-group if the order of $G$ is $p^n$ for some prime $p$". Then the result follows from Lagrange's theorem, which says that if $H$ is a subgroup of $G$ then $|H|$ divides $|G|=p^n$. If $g\in G$ then the set $\{g, g^2, \ldots, g^m=1\}$ where $m$ is the order of $g$ has $m$ elements. It is also a subgroup of $G$. Thus, $m$ divides $G$ and so is a prime power, and we are done.

#### Deveno

##### Well-known member
MHB Math Scholar
but you have only proved half of what needs to be proved:

|G| = pn → if g is in G, then |g| = pk, for some 0 ≤ k ≤ n.

that is the "trivial" part. the other part:

(for all g in G: |g| = pk for some k ≥ 0) → |G| = p​n, for some n in N, is far less trivial (and the OP's "proof" is entirely erroneous).

what we DO know, from a corollary to Lagrange, is that |G| = mpn.

without loss of generality, we may assume p does not divide m.

let q be a prime dividing m. by Cauchy's theorem, G has an element of order q. but this contradictions our assumptions on G. thus there can BE no such prime q.

but the only positive integer m that has no prime divisors is 1, and NOW we're done.