Divide The Set

We claim that the minimum value of $n$ is $2^5$ . Let $A,B$ be two groups for $S_n$ that does not satisfy the condition $ab=c$ , WLOG assume $2 \in A$ .

If $n\ge2^5$ , then

if $2^2 \in A$ , $2,2,2^2 \in A$ , which is impossible, so $2^2 \in B$

if $2^4 \in B$ , $2^2,2^2,2^4 \in A$ , which is impossible, so $2^4 \in A$

if $2^3 \in A$ , $2,2^3,2^4 \in A$ , which is impossible, so $2^3 \in B$

However we cannot place $2^5$ in any group that does not satisfy the condition, since if $2^5 \in A$ , then $2,2^4,2^5 \in A$ and if $2^5 \in B$ , then $2^2,2^3,2^5 \in B$ . Thus, when $n\ge2^5$ , the set $S_n$ must satisfy the condition, so we know $n\leq2^5$ .

Now, consider the following $A,B$ ,

$A'=\{2,3,16,17,...,31\}$ and $B'=\{4,5,...,15\}$ , it does not satisfy the condition neither, and notice that for $n<2^5$ there are two groups $A\cap A'$ and $B\cap B'$ , where $A,B$ are whatever groups we chose, which does not satisfy the condition.

Thus, the minimum value is $2^5=32$ .

Bonus, can you generalize if $S_n={k,k+1,...,n}$ for some $k>1?$