Is the following true in number theory? 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.
 A: 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 $K\nsubseteq C_{d_i}\nsubseteq K$.
A: Disclaimer, the proof is based on Condition 1, and the statements' vice versa is true for Condition 2

Let's take two arbitrary proper divisors of $n$. Let's denote them by $a$ and $b$.
If there is a prime $p\neq{a}$, such that $p|a$ but $p\nmid{b}$, $p$ will fulfill the condition.
If there is no such prime, $p$, and $a\neq{b}$, there are two possible cases :

*

*$a=p$, $p|a$ but $p\nmid{b}$

*$p|a,b$ and $p^m|a$ but $p^m\nmid{b}$
Let's look at the second case first. If $a\neq{p^m}$, ${p^m}$ satisfies the condition. If $a = p^m$, because all the primes that divide $a$ divide $b$, and $n\neq{p^m}$, there should be another prime divisor of $n$, let's say $q$, and $pq$ satisfies the condition.
Let's look at the first case. Let $q$ be a prime divisor of $b$, and $b\neq{q}$. Then, $pq$ will satisfy the condition. If $b=q$, because that $n\neq{pq}$ , there must be another prime $r$, or a higher power of $p$ or $q$ that divides $n$. If there is another prime $r|n$, $pr$ satisfies the condition. If there is a $p^m|n$ or $q^m|n$ such that $m>1$, $p^m$ or $q^m$ satisfies the condition.
A: We take prime divisors $p,q$  and then unknown $M,N$  divisors that are coprime to $p,q,$  and are allowed to be $1$ sometimes.  The exponents are $i,j,k,l \geq 1$
We have distinct divisors $A,B$ where we do permit one to divide the other sometimes. Then we display a divisor $C$ that is related to one of $A,B$ but not the other, where two numbers being `related' will mean one number divides the other, in one order or the other.
We will need this LEMMA: if two numbers $x,y$ divide  some number $z,$   then $\operatorname{lcm} (x,y) $ also divides $z.$
$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$
The first pattern, with exponents $i < j,$ divisors $A,B,$ is
$$  A = p^i u \; , \; \; \;  B = p^j v \; .    $$
This includes, whenever $M=N=1,$ the case $  A = p^i  \; , \;   B = p^j  \; .    $
We know that there is a second prime divisor. We take
$$   C = p^i q u $$
so that $A|C. \; \;$  In this one case, we need to confirm that $C$ is a proper divisor: as $j > i$  we see that $C | p^j q u $  Note that $C = \operatorname{lcm} (q,A) $ and thus divides the given number, called $n.$
$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$
The next pattern is divisors
$$  A = p^i q^j u \; , \; \; \;  B = p^k v \; .    $$
We take
$$   C =  q  $$ so that $C| A$
$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$
The third pattern is divisors, with $1 \leq j < k,$
$$  A = p^i q^j u \; , \; \; \;  B = p^i q^k v \; .    $$
We take
$$   C =  q^k  $$ so that $C| B$
$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc     \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$
The last pattern is divisors, with $1 \leq i < k,$
$$  A = p^i q^j u \; , \; \; \;  B = p^k q^l u \; .    $$
We take
$$   C =  p^k  $$ so that $C| B$
$$  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc       \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc \bigcirc  \bigcirc $$
