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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".)2012-01-19
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    @Rick Hence I didn't write the above as an answer. Have you seen Prof Magidin's hint below?2012-01-19
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    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.2012-01-19
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    possible duplicate of [Normal abelian subgroup of a solvable group](http://math.stackexchange.com/questions/59254/normal-abelian-subgroup-of-a-solvable-group)2012-01-19

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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?2012-01-19
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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.2012-01-19
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    Hm. The nice thing about these $H^{(k)}$ is that they are [characteristic](http://en.wikipedia.org/wiki/Characteristic_subgroup) in $H$ and hence normal in $G$. If you use some other series then it isn't so clear what to do.2012-01-19
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    @Rick: Added; I think you need to use Problem 8 in the same page.2012-01-19
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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.2012-01-19
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    @ArturoMagidin Oh, that's interesting. I didn't even know that this was possible. Looking at a preview, it seems like Problem 8 assumes that $G$ is finite. Is that unnecessary for the needed implication?2012-01-19
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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...2012-01-19
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    @Rick: Sigh; I'm on my way out. Let me see if I can figure out a way that doesn't use, explicitly or implicitly, the derived series.2012-01-19
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    @ArturoMagidin, thank you for your inimitable dedication!2012-01-19
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    @Rick: Okay, I think I have something that works, using some of the ideas mentioned in the hint to problem 8.2012-01-20
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    @ArturoMagidin: Splendid! I think it's worked! Thank you!2012-01-20
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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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    In point 1) , is it *any* sequence of subgroups? That is, will $H\ge 1$ will also work?2018-11-04