Show that $G' = \bigcap_{C\subseteq N \triangleleft G} N$

Question

I am asked to show that $G' = \bigcap_{C\subseteq N \triangleleft G} N$, where $G'$ is the commutator subgroup of $G$, and $C :=\{aba^{-1}b^{-1}\mid a,b\in G\}$. Showing $\bigcap_{C\subseteq N \triangleleft G} N \subset G'$ is not difficult as $C \subset G' \triangleleft G$. So $x \in \bigcap_{C\subseteq N \triangleleft G} N \Rightarrow x \in G'$. However, I am having a difficult time showing the other direction, namely $G' \subset \bigcap_{C\subseteq N \triangleleft G} N$. First I tried showing $x\in G' \Rightarrow x \in C$ but I realized this may not be true. So I am wondering how to take advantage of the additional $C \subset N$ condition.

Answer 1

Here is a way to do it: $G'$ is the subgroup of $G$ generated by the set $C = \{ aba^{-1}b^{-1} | a, b \in G \}.$ Thus $G' = \bigcap\{H | C \subseteq H \leq G\}$ (using the definition of a subgroup generated by a subset of $G$). Now note that, $\{ H | C \subseteq H \leq G\} \supseteq \{ N | C \subseteq N \trianglelefteq G\} \Rightarrow \bigcap \{ H | C \subseteq H \leq G\} \subseteq \bigcap \{ N | C \subseteq N \trianglelefteq G\}.$

Answer 2

Let $P$ be a poset, $X\subseteq Y\subset P$. $\,$If $\,\min Y\,$ exists and is $\,\in X$, then $\,\min X=\min Y$.

Applies here with $Y$ the subgroups containing $C$, and $X$ the normal such. Note that in any collection of sets ordered by inclusion, the $\min$ operator coincides with intersection $\bigcap$.

(This is Krish's answer expressed differently.)