Abelian Subgroup of Normal Subgroup of Solvable Group

Question

This is Exercise #11 from Section 3.4 of Dummit and Foote (and is a common question in various texts), but I have not found a satisfactory answer on this site (and in some cases people misstate the definition of a solvable group given in D&F). My hope is that I will not only understand a solution to this problem but in doing so also see how to combine various big ideas, such as products of groups, the isomorphism theorems, and normal subgroups.

The problem as set forth in D&F is "Prove that if $H$ is a nontrivial normal subgroup of the solvable group $G$ then there is a nontrivial subgroup $A$ of $H$ with $A \underline{\triangleleft} G$ and $A$ abelian."

The D&F definition is "A group $G$ is $solvable$ if there is a chain of subgroups

$1 \, = G_0 \, \underline{\triangleleft} \, G_1 \, \underline{\triangleleft} \, G_2 \, \underline{\triangleleft} \ldots \underline{\triangleleft} \, G_s \, = \, G$

such that $G_{i+1}/G_i$ is abelian for $i = 0, \, 1,\ldots, \,s-1$."

My approach has been to attempt to find something isomorphic to $G_{i+1}/G_i$ because then I would have an isomorphism to an abelian. Based on ideas from other postings, I considered letting $G_i$ be the largest index such that $G_i \cap H = {1}$ and then we would have $G_{i+1} \cap H \neq {1}$. Perhaps letting $A = G_{i+1} \cap H$ might lead me somewhere.

This is the place where I am trying to combine the second (i.e., triangle) isomorphism with normal subgroups and am getting stuck. Mightily stuck, I might add.

It seems to me that $H \underline{\triangleleft} \, G_{i+1}$ and by definition $G_i \underline{\triangleleft} \, G_{i+1}$ so that $G_{i+1}H/H \cong G_{i+1}/(G_{i+1} \cap H)$. That did not seem to work, but it got me to consider another product.

I decided to try $G_i (G_{i+1} \cap H)$ and put that at the top of the diamond (for the second isomorphism theorem). Then $G_i$ is on the left and $(G_{i+1} \cap H)$ is on the right with $G_i \cap (G_{i+1} \cap H)$ at the bottom.

Now my isomorphism is $G_i (G_{i+1} \cap H)/G_i \cong (G_{i+1} \cap H)/ (G_i \cap (G_{i+1} \cap H))$.

Here I observed that $G_i \cap (G_{i+1} \cap H) = (G_i \cap H) = {1}$. Recall that I am using $A = (G_{i+1} \cap H)$. That means my isomorphism is actually

$G_i A/G_i \cong A/{1}$.

This seems ever so close. If I knew that $G_i A/G_i$ is abelian then I would be there.

Am I on the right track? Am I completely in the wrong direction? Is there a gap (simple or not) that I am missing? Lastly, please excuse the length of this posting, but I wanted to be thorough with the statement of the problem, the related definitions, and my thinking.

Answer 1

Nice question. You're almost there. $G_iA/G_i$ is abelian, because its a subgroup of the abelian group $G_{i+1}/G_i$.

Answer 2

@DerekHolt wrote "With your current formulation I don't believe that it is possible to prove that A is normal in G, and it might not be true in general." That prompted me to look at my $A$ to see what I could do, especially in view of the commutators.

We are given that $H \underline{\triangleleft} G$, which means that $ghg^{-1} \in H \; \forall \; g \in G, h \in H$. From that we have $ghg^{-1}h^{-1} \in H$, so $[g,h] \in H$.

Now I decided to reformulate my $A$ by using the commutators of $G$ and let the group $C = \langle [x,y] \rangle = \{ xyx^{-1}y^{-1} | x,y \in G \}$ be the commutator group of elements in $G$. Lastly, I set $A = C \cap H$.

We know that $C \underline{\triangleleft} G$ and are given that $H$ is normal in $G$ so their intersection is normal too, which means $A \underline{\triangleleft} G$. Because $A$ contains commutators of $G$, $A$ is abelian too.