Hello, this question is related to number theory. In case there are any improvements required, kindly let me know.
Let $n\neq p^m,pq$ where $p, q$ are primes, $m\in \mathbb N$ is a positive integer. Let $d_1<d_2<\cdots<d_k$ be the list of all positive proper divisors of $n$.
Consider two positive proper divisors $d_i,d_j$ of $n$. Is it true that there always exists another positive proper divisor $d_k$ of $n$ such that exactly one condition holds?
Condition 1. $d_i\mid d_k$ or $d_k\mid d_i$ but neither $d_j\mid d_k$ or $d_k\mid d_j$.
OR
Condition 2. $d_j\mid d_k$ or $d_k\mid d_j$ but neither $d_i\mid d_k$ or $d_k\mid d_i$ .
My try:
I feel that this is always true.
Suppose I take $n=12$. Then its positive proper divisors are $2<3<4<6$.
If we take $d_i=2,d_j=3$, then we can consider $d_k=4$. Again if we take $d_i=2,d_j=4$, then $d_k=6$ works.
Lets take $d_i=2, d_j=6$, then $d_k=4$ works.
If we take $d_i=3, d_j=4$, then $d_k=2$ works and so on.
I tried with several other $n$. I found it works for every $n$ which is not of the form $n=p^m,pq$.
I am confused on how to prove this in general.
May I please request some help from you all.
Best Answer
I thought this in group-theoretic way.
Think $n$ in terms of cyclic group $C_n$ of order $n$.
Then divisors of $n$ are in bijection with subgroups of $C_n$.
Since $n$ is not prime power, this exactly means that subgroups of $C_n$ do not fit in a common chain.
Lemma If $G$ is a cyclic group of non-prime power order, then for any $H$ with $1<H<G$, there exists $K\le G$ such that $H\nsubseteq K$ and $K\nsubseteq H$.
Proof: If $H$ is of order $p^k$, take $K$ subgroup of order prime $q\neq p$.
If $H$ is not of prime power order, then, due to properness, some Sylow subgroup $P$ of $G$ is not contained in $H$; it is also clear that $H$ can not be in $P$ (since $H$ is not $p$-group. Thus, $K:=P$ works.
Proof of your result: Let $C_n$ be a cyclic group which is not of prime power order.
Let $C_{d_i}$ and $C_{d_j}$ be non-trivial proper subgroups.
If $C_{d_i}$ is not in $C_{d_j}$ or viceversa, then we can take $d_k=d_i$.
Assume with no loss, that $C_{d_i}$ is (properly) contained in $C_{d_j}$.
If $C_n/C_{d_i}$ is not of prime power order, apply lemma to get a subgroup $K/C_{d_i}$ such that $K\nsubseteq C_{d_j}\nsubseteq K$.
If $C_n/C_{d_i}$ is of prime power order, then $C_n/C_{d_j}$ is also of prime power order, say $p^k$; then $C_{d_j}$ can not be $p$-group, but $p$ divides its order (since $|C_{d_j}/C_{d_i}|$ is power of $p$. )
Thus, $|C_{d_j}|=p^a.m$. Then apply Lemma to get subgroup $K$ of $C_{d_j}$ such that $K\nsubseteq C_{d_i}\nsubseteq K$.