I found the following problem in an old exam:
If $N$ is normal in a solvable group $G$, then $N'\lhd G$.
I don't understand what the assumption of solvability is for.
We have $N'\operatorname{char} N\lhd G.$
If $\sigma$ is an inner automorphism of $G$, then $\sigma (N)=N$ by $N\lhd G.$ So $\sigma|_N$ is an automorphism of $N$. Therefore $\sigma|_N(N')=N'$ by $N'\operatorname{char} N.$ But obviously $\sigma|_N(N')=\sigma(N'),$ which ends the proof.
Is this proof incorrect or does the assumption of solvability allow one not to use the notion of a characteristic subgroup?
EDIT I'm sorry. Apparently, I missed part of the problem. The whole problem reads:
Let $G$ be solvable and $N\lhd G.$ Show that
(a) $N'\lhd G;$
(b) If, for $H\subset N,\,H\lhd G$ implies $H=\{e\}$, then $N$ is abelian.
But I still don't think solvability is needed. If $H\subset N,\, H\lhd G$ implies $H=\{e\},$ then $N'=\{e\}$ by (a), and $N$ is abelian...
EDIT 2 I've edited the inclusions above to be $\subset$ instead of $\subseteq$, per Arturo's answer, as this is indeed what was written in the problem. I thought the symbol meant $\subseteq$ as it often does and I changed it to make it what I thought to be more precise.