It is known that in the worst case comparison sort takes $\Omega(nlgn)$ of time for n elements. Let's say I'm given $n$ elements as input to sort. Each input sequence of $n$ elements has $n/k$ subsequences with each subsequence having $k$ elements. Each subsequence is less than its subsequent sequence and greater than its preceding sequence. How do I find a lower bound $\Omega(nlgk)$ on the number of comparisons needed to solve this problem?
I tried to think about it if $n$=9,$k$=3 So I have 3 subsequences of 3 elements. I know that each subsequence has a worst case lower bound of $\Omega(klgk)$. I could just add these three up to get the overall lower bound, right?
If the answer to the above question is yes, then in the general case that there are $n/k$ subsequences I would have $\Omega((n/k)klgk)$?