6
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Crosspost from MO. $\endgroup$ – Brandon Carter Oct 8 '11 at 5:01
  • 3
    $\begingroup$ It was OffT at MO so I posted it here. It's already deleted there. $\endgroup$ – Ken Gonzales Oct 8 '11 at 5:04
  • $\begingroup$ 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. $\endgroup$ – Ted Oct 8 '11 at 6:36
  • 2
    $\begingroup$ 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 $z \not\in 2^k \mathbb{Z}$ $\endgroup$ – Alexei Averchenko Oct 8 '11 at 6:37
  • $\begingroup$ @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. $\endgroup$ – Arturo Magidin Oct 8 '11 at 20:41
9
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ This is indeed the definition given in Hungerford's Algebra. $\endgroup$ – Chris Eagle Oct 8 '11 at 20:54
2
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 ) .

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.