Let $G'=[G,G]$, $G$ be a group. Let $N \triangleleft G$ and $N \cap G'= e$, then $N < Z(G)$.

Question

I want to prove the following statement:

Let $G'=[G,G]$, G group. Let $N \triangleleft G$ and $N \cap G'= e$, then $N < Z(G)$.

My solution is:

Let $gkg^{-1}k^{-1} \in N \cap G'$ with $g \in G$ and $k \in N$, because $gkg^{-1} \in N$ for all $k \in N$(because $N\triangleleft G$) then $$gkg^{-1}k^{-1}=e$$ so $$gk=kg$$ i.e., $N < Z(G)$.

But im not sure if this solution is correct.

Answer 1

It's better to start with an element $n\in N$ then show that $n$ commutes with any $g\in G$, like so.

Since $N$ is normal and $g$ is arbitrary in $G$, we have $gng^{-1}\in N$. Since $N$ is a subgroup of $G$, we have $$h:=\underbrace{gng^{-1}}_{\in N}n^{-1}\in N.$$ But $h$ is the commutator $[g, n]$ of $g$ with $n$, so $h\in N\cap G'=\{e_G\}$, whence $h=gng^{-1}n^{-1}=e_G$, i.e., $$gn=ng.$$ Thus $n\in Z(G)$ (the centre of $G$). Hence $$\boxed{ N\subseteq Z(G)}.$$

This is only half the battle! We need to show that it is a subgroup of $Z(G)$, not just a subset.

Can you carry on from here?

Hover over, tap, or click the box below for a spoiler.

Use either the one- or two-step subgroup lemma. I recommend the former. Also $N\neq \emptyset$ since $e_N=e_G=e_{Z(G)}$.