Thread: 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."

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
, where
Let 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

But
divides
Hence, G is a p-group.


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

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| = 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| = 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.