Is the following result trivial?

[JoanBraidy]

I proved there's infinitely many n such that S_n has an element of order n^2

 

[Martin Rattigan]

Trivialish.

$3^2+4^2+5^2<3.4.5$.

Suppose $3^{2i}+4^{2i}+5^{2i}<3^i.4^i.5^i$ for $i<n$, then
$3^{2(i+1)}+4^{2(i+1)}+5^{2(i+1)}<25(3^i.4^i.5^i)<3^{i+1}4^{i+1}5^{i+1}$, hence $3^{2n}+4^{2n}+5^{2n}<3^n4^n5^n$ by induction.

It follows that $S_{3^n4^n5^n}$ has an element of order $(3^n4^n5^n)^2$ for all $n\in \mathbb{N}$.

Similarly $5^{3n}+7^{3n}+9^{3n}+11^{3n}<5^n7^n9^n11^n$, so there are an infinite number of $k$ such that $S_k$ contains an element of order $k^3$.

I think its probably true that there are an infinite number of $n$ such that $S_n$ contains an element of order $n^k$ for any $k\in \mathbb{N}$.

 

[JoanBraidy]

Trivialish.

$3^2+4^2+5^2<3.4.5$.

Suppose $3^{2i}+4^{2i}+5^{2i}<3^i.4^i.5^i$ for $i<n$, then
$3^{2(i+1)}+4^{2(i+1)}+5^{2(i+1)}<25(3^i.4^i.5^i)<3^{i+1}4^{i+1}5^{i+1}$, hence $3^{2n}+4^{2n}+5^{2n}<3^n4^n5^n$ by induction.

It follows that $S_{3^n4^n5^n}$ has an element of order $(3^n4^n5^n)^2$ for all $n\in \mathbb{N}$.

Similarly $5^{3n}+7^{3n}+9^{3n}+11^{3n}<5^n7^n9^n11^n$, so there are an infinite number of $k$ such that $S_k$ contains an element of order $k^3$.

I think its probably true that there are an infinite number of $n$ such that $S_n$ contains an element of order $n^k$ for any $k\in \mathbb{N}$.

similar to what I did thanks

if I proved the general case, do you think that would be trivial?

 

[Martin Rattigan]

There are straightforward proofs quoting Bertrand's postulate or the prime number theorem or other theorems on prime distribution. Most people wouldn't call the quoted theorems trivial, so you'd probably want to distinguish between a trivial result and a trivial proof.

I've appended a proof (I tried to put this under a spoiler but it doesn't seem to work with latexed code - you'll just have to avoid looking if you're still doing your own proof). Whether you'd call the proof trivial is really subjective.

For $k=0$ it is trivial that $S_n$ has an element of order $n^k$.

For any $k\in\mathbb{N},k\geq1$

$$p_n^k+p_{n+1}^k+\dots+p_{n+r}^k+\dots+p_{n+k}^k$$
$$\leq p_n^k(1+2^k+2^{2k}+\dots+2^{rk}+\dots+2^{k^2})\text{ (By Bertrand's postulate)}$$
$$=p_n^k(2^{k(k+1)}-1)/(2^k-1)$$

If $n_k$ is chosen so that
$$p_{n_k+k}\geq (2^{k(k+1)}-1)/(2^k-1)$$
then for any $n\geq n_k$

$$p_n^k(2^{k(k+1)}-1)/(2^k-1)$$
$$\leq p_n^k(p_{n+k})$$
$$\leq p_np_{n+1}\dots p_{n+r}\dots p_{n+k-1}p_{n+k}$$

Hence for any such $n$, $S_{p_np_{n+1}\dots p_{n+r}\dots p_{n+k}}$ contains an element of order $(p_np_{n+1}\dots p_{n+r}\dots p_{n+k})^k$.

 

[CellCoree]

I cannot make a judgment on whether a result is trivial or not. However, I can provide an explanation of the significance of this result in the context of mathematics.

The result stated is a significant and non-trivial result in the field of group theory. It states that for any integer n, there exists an element in the group S_n (the symmetric group of degree n) whose order is n^2. This means that the group S_n contains elements with arbitrarily large orders, which is not obvious and requires a proof.

This result has implications in various areas of mathematics, such as number theory and abstract algebra. It also has applications in other fields, such as cryptography and coding theory.

In conclusion, while the statement may seem simple and straightforward, it is a non-trivial result that has important implications and applications in mathematics.