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Question: Let $G$ be a solvable group, and let $H$ be a nontrivial normal subgroup of $G$. Prove that there exists a nontrivial subgroup $A$ of $H$ that is Abelian and normal in $G$.

[ref: this is exercise 11 on page 106 of [DF] := Dummit and Foote's Abstract Algebra, 3rd edition]

UPDATE: I cannot use the derived series of a group (the concept is not introduced in [DF] until some 90 pages later), and the use of a commutator subgroup is questionable.

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Nonabelian simple groups are not solvable: the only subnormal series you can build is the one that you gave, and the factor group there is not abelian. The abelian simple groups are the cyclic groups of prime order, and for those you can take $H = A = G$. (I interpret nontrivial to mean "not the identity".) –  Dylan Moreland Jan 19 '12 at 20:36
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@Rick Hence I didn't write the above as an answer. Have you seen Prof Magidin's hint below? –  Dylan Moreland Jan 19 '12 at 20:41
    
So do I: for me, nontrivial means not the identity, which is part of the confusion here. Otherwise, how do I prove the general statement above (forgetting about simple groups)? I need a small nudge of a hint. –  Rick Jan 19 '12 at 20:42
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possible duplicate of Normal abelian subgroup of a solvable group –  Jack Schmidt Jan 19 '12 at 22:46

2 Answers 2

up vote 6 down vote accepted

A simple solvable group must be cyclic of prime order (since it must be abelian, and so cannot have proper [normal] subgroups). But a simple solvable group would not contain nontrivial normal subgroups, so the proposition would be true for such a group by vacuity (the hypotheses are never satisfied). Alternatively, if you allow the whole group to be a "nontrivial normal subgroup", your $H$ can only be $G$ itself, which is already abelian, so you can set $A=H=G$; either way, the proposition is true for such a group.

(If $G$ is solvable, then $[G,G]$ is a proper subgroup of $G$; since it is always normal in $G$, if $G$ is also simple, then we must have $[G,G]=\{1\}$, hence $G$ is abelian).

Hint for the question. If $H$ is abelian, you are done. If not, then $[H,H]$ is nontrivial; use the fact that $H^{(n)}\subseteq G^{(n)}$, where $G^{(k)}$ is the $k$th term of the derived series of $G$, to show that $H$ is solvable, and use the fact that $H\triangleleft G$ to show $[H,H]\triangleleft G$. Then replace $H$ with $[H,H]$, lather, rinse, and repeat.

Different hint. Let $1=G_0\triangleleft G_1\triangleleft\cdots\triangleleft G_s = G$, with $G_i/G_{i+1}$ abelian. By Problem 8, you can pick the $G_i$ normal in $G$. Look at $H_i=H\cap G_i$.

Added. Sigh. I didn't notice that Problem 8 assumes $G$ is finite; the result is true, as witnessed by the derived series, but again it's probably not what you want.

Added 2. Okay, this should work; it takes some of the ideas of the hint in Problem 8 of the same page, so it should be "reasonable". Let $i$ be the largest index such that $G_i\cap H$ is a proper subgroup of $H$. Then $H\subseteq G_{i+1}$, hence $G_i\cap H\triangleleft H$. Moreover, $H/(H\cap G_i) \cong (HG_i)/G_i \leq G_{i+1}/G_i$, so $H/(H\cap G_i)$ is abelian. As in problem 8, this means that $x^{-1}y^{-1}xy\in H\cap G_i$ for all $x,y\in H$. Show that this is also true for all $G$-conjugates of $H\cap G_i$, hence their intersection, which is normal in $G$, contains all $x^{-1}y^{-1}xy$ with $x,y\in H$. If this is trivial, then $H$ is abelian and we are done. If it is not trivial, then this intersection is nontrivial, and normal in $G$. Replace $H$ with this intersection, and note that the largest index $j$ such that the intersection with $G_j$ is proper is striclty smaller than $i$; so you can set up a descent. Lather, rinse, and repeat.

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I think the parenthetical second paragraph is also a good hint: What does it mean to be abelian, in terms of the commutator subgroup? –  Dylan Moreland Jan 19 '12 at 20:46
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Thank you Arturo. I've thought about this, too; but, I feel there must be some other way, because [DF] do not define derived series until much later. They do mention the commutator subgroup in an exercise of an earlier section; however, if this exercise were to use anything of the kind, they are generally kind enough to point the reader to the very exercise, which here they have not done. –  Rick Jan 19 '12 at 20:48
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Hm. The nice thing about these $H^{(k)}$ is that they are characteristic in $H$ and hence normal in $G$. If you use some other series then it isn't so clear what to do. –  Dylan Moreland Jan 19 '12 at 21:22
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@Dylan: They are better than characteristic, they are verbal. But a previous problem asks to show that you can find a series that is witness to the solvability and in which every term is normal in $G$, and that's enough. –  Arturo Magidin Jan 19 '12 at 21:28
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@Dylan: Sigh; the equivalence is true for arbitrary groups (not the entire problem, since it also asks you to show that you can find a series to witness solvability with every factor cyclic, which is false for $G=\mathbb{Q}$, for instance). But I keep running into the commutator subgroup in order to prove it... –  Arturo Magidin Jan 19 '12 at 22:47

Hints (for you to prove):

1) It is true that

$$H\geq H'\geq\ldots\geq H^{(n)}=1\,\, ,\,\, \text{for some}\,\,\,n\in\Bbb N$$

2) Show that $\,H^{(n-1)}\triangleleft G\,\;\;$ (Yes, not only in $\,H\,$ ...!)

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