# p-group

• June 17th 2012, 02:57 PM
ibnashraf
p-group
Hi, can someone help me understand the following 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."
• June 17th 2012, 04:00 PM
ibnashraf
Re: p-group
Quote:

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

ok 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 ?
• June 18th 2012, 06:19 AM
Deveno
Re: p-group
one implication is obvious:

if G is a p-group, (so that |G| = pn) then the only possible divisors of |G| are powers of p, and since |g| divides |G|, |g| = pk, for some 0 ≤ k ≤ n.

the other implication is not so obvious: if for any g in G,|g| = pk, for some non-negative integer k, we know that |G| = pnm, for some positive integer m with gcd(p,m) = 1, and some positive integer n.

but it is not clear that we must have m = 1.

however, by Cauchy's theorem, if q ≠ p is a prime that divides m, we must have an element x of G of order q. but there are no such elements, since every element of G has order a power of p. therefore, there are no prime numbers dividing m, so m = 1.