I want to ask a question, and I found it here: Why is the direct product of a finite number of nilpotent groups nilpotent?
But I am struggling to understand how can we take the product of two normal series and still have a normal series, that is, for the case of two subgroups, if $H\triangleleft K$ and $H'\triangleleft K'$ why can we deduce that $HK\triangleleft H'K'$ ?
Another question is to show the containment in the original post in the link above, many answers show the result differently, it should be possible to show the containment directly no? Using the definition of nilpotence by a central series (not by an upper or descending central series).
Let me answer your first question.
This is a situation where good notation really helps.
Let me use direct product notation: $K \oplus K'$ instead of $KK'$, and similarly $H \oplus H' < K \oplus K'$ instead of $HH' < KK'$.
Also, let me express each element of $K \oplus K'$ as an ordered pair $(k,k')$ with $k \in K$ and $k' \in K'$, and with group operations given coordinate wise, namely $$(k_1,k'_1) (k_2,k'_2) = (k_1 k_2, k'_1 k'_2) \qquad (k,k')^{-1} = (k^{-1},k'{}^{-1}) $$ So now take any $(h,h') \in H \oplus H'$ and any $(k,k') \in K \oplus K'$ and let's calculate: $$(k,k')^{-1} (h,h') (k,k') = (khk^{-1},k'h'k'^{-1}) $$ From the assumption that $H$ is normal in $K$ and $H'$ is normal in $K'$ it follows that $khk^{-1} \in H$ and $k'h'k'{}^{-1} \in H'$ and therefore $(khk^{-1},k'h'k'^{-1}) \in H \oplus H'$. This proves that $H \oplus H'$ is normal in $K \oplus K'$.