Prove that G is a p-group iff every element of G has order a power of p

[ibnashraf]

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

, 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 ?

 

[Sudharaka]

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

, 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 ?

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.

 

[Swlabr1]

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]

but you have only proved half of what needs to be proved:

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

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

(for all g in G: |g| = p^k 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.

 

[blue_raver22]

Your attempt at the question is mostly correct, but there are a few minor mistakes and areas for improvement.

Firstly, your proof assumes that p is a prime number, which is not explicitly stated in the question. It would be better to start by stating that p is a prime number.

Secondly, when you say "Let G be a p-group", it would be more accurate to say "Let G be a finite p-group". This is because the definition of a p-group requires the group to be finite.

Thirdly, when you say "Let g be an element of G", it would be clearer to state that g is an arbitrary element of G. This is because you will be using this element to prove the statement for all elements of G, not just this specific one.

Next, when you say "Hence, every element of G has order a power of p", it would be better to explain why this is the case. This can be done by stating that since G is a p-group, the order of every element must be a power of p.

In the second part of your proof, you start by saying "Conversely, suppose that every element of G has order a power of p". It would be better to start by saying "Assume G is not a p-group". This is because you are trying to prove that G is a p-group by contradiction.

When you say "But p does not divide the order of G", it would be clearer to say "By assumption, p does not divide the order of G". This makes it clear that you are using the assumption made in the previous sentence.

Finally, when you say "Hence, G is a p-group", it would be better to state the conclusion more explicitly by saying "This is a contradiction, therefore G must be a p-group". This makes it clear that you have reached the conclusion you were trying to prove.

Overall, your proof is mostly correct and the minor mistakes can easily be fixed. Just make sure to be clear and explicit in your statements and to state all necessary assumptions.