Ok, so abelian groups are solvable.
And Thm II.8.5 of Hungerford saysA 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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$\begingroup$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.
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ $\begingroup$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 ) .
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