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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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    @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.2011-10-08

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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.2011-10-08
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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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Z is solvable as it has a normal series where the factors are abelian as Z being abelian ( which is the definition of solvable series) but as Z is not finite so it can't have a composition series ( which follows from the result - an abelian group has a composition series if and only if it is finite ) .