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."

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- Jun 17th 2012, 01:57 PMibnashrafp-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." - Jun 17th 2012, 03:00 PMibnashrafRe: p-group
ok this was my attempt at the question:

Suppose G is a p-group

$\displaystyle \Rightarrow |G|=p^n$, where $\displaystyle n\in Z$

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

$\displaystyle \therefore p^n/|g|$

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

Conversely, suppose that every element of G has order a power of p

$\displaystyle \Rightarrow |g|=p^n$

But $\displaystyle |G|/|g|$

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

Hence, G is a p-group.

can someone look over this and tell me what mistakes i made if any please ? - Jun 18th 2012, 05:19 AMDevenoRe: p-group
one implication is obvious:

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

the other implication is not so obvious: if for any g in G,|g| = p^{k}, for some non-negative integer k, we know that |G| = p^{n}m, 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.