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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.

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This looks right to me. The notion of a characteristic subgroup seems even more basic than that of solvability, so it seems odd that one would want to avoid the former in favor of the latter. – Dylan Moreland Jun 21 '12 at 15:49
this is the second time today I've the $N^{\prime}$ notation but I don't know what it is. Can someone clue me in ? – mike Jun 21 '12 at 16:38
your proof is OK, the result is true whether or not $G$ is solvable. @mike $N^{\prime}$ is the derived subgroup of $N$. This is the smallest normal subgoup $K$ of $N$ such that $N/K$ is Abelian. There are several euqivalent definitions. – Geoff Robinson Jun 21 '12 at 16:46
thanks ${}{}{}{}$ – mike Jun 21 '12 at 16:48
@mike See the Wikipedia article on the commutator subgroup. – Jim Belk Jun 21 '12 at 17:12

1 Answer 1

up vote 1 down vote accepted

The commutator subgroup is a verbal subgroup, so in particular it is fully invariant (more than just characteristic). For any homomorphism $f\colon A\to B$, we have that $f([A,A])\subseteq [B,B]$.

In particular, if $N$ is a subgroup of any group $G$ (solvable or not), and $\phi\in\mathrm{Aut}(G)$ is such that $\phi(N)\subseteq N$, then $\phi([N,N])\subseteq [N,N]$. For inner automorphisms and $N\triangleleft G$, this yields $\phi([N,N])\subseteq [N,N]$, so $[N,N]$ is normal. In fact, it is characteristic in $G$ if $N$ is characteristic in $G$, and fully invariant if $N$ is fully invariant. No assumption of solvability is needed.

Added. For part (b), I suspect the inclusion in the premise is meant to be a proper inclusion. That is:

If $H\subset N$ and $H\triangleleft G$ implies $H=\{e\}$, then $N$ is abelian.

Note that if we don't omit the $H=N$ case, then the implication can only hold if $N=\{e\}$ making this a somewhat silly statement.

Now, given the condition as amended, part (a) implies that either $N'=N$, or that $N'=\{e\}$. But if $G$ is solvable, $N'=N$ cannot occur, since $N$ must be solvable as well, which gives the desired conclusion and shows why you are assuming $G$ is solvable.

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Right! Thank you very much. I have edited my question once again. – user23211 Jun 21 '12 at 20:03

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