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Ok, so abelian groups are solvable.

And Thm II.8.5 of Hungerford says A group is solvable iff it has a solvable series. (The group may be finite or infinite.)

However, I can't seem to find a solvable series for $\mathbb{Z}$, for example $\mathbb{Z},2\mathbb{Z},6\mathbb{Z},\ldots$ will not terminate in the identity group.

Someone said that $\mathbb{Z},\left\{0\right\}$ is a solvable series for $\mathbb{Z}$. Is this a definition set by Hungerford since $\mathbb{Z},\left\{0\right\}$ is not even a composition series : $\mathbb{Z}/\left\{0\right\}$ is not a simple group.

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Crosspost from MO. –  Brandon Carter Oct 8 '11 at 5:01
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It was OffT at MO so I posted it here. It's already deleted there. –  Ken Gonzales Oct 8 '11 at 5:04
    
What is Hungerford's definition of a solvable series? The Wikipedia definition for solvable group is just that the derived series $G, G', G'', \ldots$ terminates at the trivial group - which clearly happens for any abelian group since $G'$ is trivial if $G$ is abelian. –  Ted Oct 8 '11 at 6:36
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Are you sure they don't? I haven't checked, but it seems that the tower given by $i_k: 2^k \mathbb{Z} \to 2^{k-1} \mathbb{Z}$ has $0$ as its limit. Indeed, for any $z \in \mathbb{Z}$ we have either $z = 0$ or there exists a $k$ large enough so that $z \not\in 2^k \mathbb{Z}$ –  Alexei Averchenko Oct 8 '11 at 6:37
    
@AlexeiAverchenko: The tower has the zero group as the limit, but solvable series are required to be finite. Groups for which you $\cap G^{(n)} = \{e\}$ but $G^{(k)}\neq\{e\}$ for any $k$ are sometimes called "hypoabelian groups", but they are not solvable. –  Arturo Magidin Oct 8 '11 at 20:41
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2 Answers

While I'm not familiar with Hungerford's book, I imagine that he defines a solvable series to be a (finite) sequence of subgroups of $G$ such that $$1=G_0\unlhd G_1\unlhd\cdots\unlhd G_n=G$$ where each $G_i/G_{i-1}$ is abelian: there's no requirement that these quotients should be simple. In other words a solvable series is not necessarily a composition series.

So the series you allude to in your last paragraph ($G_0={0}$ and $G_n=G_1=\mathbb{Z}$) will do the trick.

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This is indeed the definition given in Hungerford's Algebra. –  Chris Eagle Oct 8 '11 at 20:54
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We don't need the quotients to be simple. If the quotients are simple then you have obtained a composition series of the group. A solvable group has a composition series iff it is finite. Therefore in your case it won't happen.

Don't confuse solvable series with composition series.

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