<p>Here's an argument handling a special case. Consider a fixed poset $P = (S,\leq)$. Say that a (well-founded) forest $F$ is a <em>branching-representation</em> of $P$ if:</p>

<ol>
<li><p>the nodes of $F$ are bijectively labelled by the elements of $P$;</p></li>
<li><p>for each node $u$ of $F$, its children are ???</p></li>
</ol>

<p>(SORRY, CAN'T FIGURE THE RIGHT DIFINITION NOW, I'M PUTTING THIS ON STANDBY)</p>

<p>We may then assign an ordinal $o(u)$ to each node of $T$ inductively: (i) if $u$ is a leaf then $o(u) = 1$, (ii) if $u$ has children $(u_i)$, then $o(u) = \sum_i o(u_i)$. The order type of $T$ is then defined as $o(T) = o(r)$ where $r$ is the root. </p>

<p>Is the following true?</p>

<blockquote>
  <p>Conjecture: the order type of $P$ equals the minimum of $o(T)$ over all possible branching-representations $T$ of $P$.</p>
</blockquote>

<p>Suppose this is true. Consider a poset $P = (S,\leq)$, and let $P^{\natural} = (S',\leq')$ where $S'$ consists of the finite antichains of $S$ and where the order satisfies $X \leq Y$ iff there is an injection $\phi : X \rightarrow Y$ s.t. $u \leq \phi(u)$ (f.a. $u \in X$). Suppose that $ot(P) = \lambda$ with $\lambda$ a limit ordinal, it should then follow that $ot(P^{\natural}) = \omega^{\lambda}$: starting with a b.p. $T$ of $P$, we would construct a b.p. $T'$ for $P^{\natural}$ by 'iterated substitution'.</p>

<p>This might help to analyze the case of <em>ordered free groups</em>. In this case, we have an ordered alphabet $P = (S,\leq)$, and we consider the free group $P^{\sharp}$ over $\{S \cup S^{-1}\}^*$, equipped by the relation $u \preceq v$ iff $u_i \leq v_{\phi(i)}$ for some injection $\phi$. Now, given an element $B = \{b_1,\ldots,b_k\} \in (P^{\sharp})^{\natural}$, we may define the subgroup $\langle B \rangle$ of $P^{\sharp}$ generated by $\cup_{i = 1}^{k} (b_i \downarrow)$.</p>

<p>It follows that the map $\chi : (P^{\sharp})^{\natural} \rightarrow Sub(P^{\sharp})$ defined by $\chi(B) = \langle B \rangle$ is surjective. Consider the equivalence over $(P^{\sharp})^{\natural}$ s.t. $B \sim B'$ iff $\chi(B) = \chi(B')$. Assuming that the equivalence classes of $\sim$ are finite (which seems intuitively true), I think that it would imply $ot(Sub(P^{\sharp})) = ot((P^{\sharp})^{\natural}) = \omega^{\omega^{\omega^{\lambda}}}$ (possibly up to some lower degree terms).</p>

<p>So if the above is correct, you've got at least the ordinals of the form $\omega^{\omega^{\omega^{\lambda}}}$ in question (1).</p>

CC BY-SA 3.0
2014-04-15T02:55:13.720Z
Out of curiosity: are you and HeloLobo different people?
Sasho Nikolov
CC BY-SA 3.0
2014-04-15T16:38:16.450Z
Let's assume that we are the same person. If I answer positively I run the risk of being called a 'liar' by certain people based on the profile colors, and if I answer negatively I run the risk of being called a 'liar' by some other people that may have inside info on the users. Therefore, I'd rather not answer :) By the way, it's still a mystery to me how these colors are attributed, is it explained somewhere on the site?
Super8
CC BY-SA 3.0
2014-04-28T19:43:22.910Z
The profile colors, as far as I know, are assigned randomly.  It's against the spirit of this site to use multiple accounts, as the preference is to even use real names.
Lev Reyzin
CC BY-SA 3.0
2014-04-11T13:39:37.397Z
2014-04-14T13:06:55.323Z
2014-04-14T13:06:55.323Z
Super8
<p>A (possibly infinite) group $G$ is <em>noetherian</em> if it satisfies the following equivalent conditions:</p>

<p>(1) every subgroup of $G$ is finitely generated,</p>

<p>(2) there is no infinite strict ascending chain of subgroups,</p>

<p>(3) any non-empty collection of subgroups has a maximal element under inclusion.</p>

<p>Note that this property is hereditary: if $G$ is noetherian then so are all its subgroups. Also, being noetherian is equivalent to require that the poset of subgroups $(Sub(G),\subseteq)$ is a well-order.</p>

<p>The <em>order type</em> of a well-order $(S,\leq)$ is defined as the supremum of the order type of its linear extensions. By extension, we may define the <em>order type</em> of a noetherian group $G$, denoted by $ot(G)$, as the order type of $(Sub(G),\subseteq)$. We may then ask the following questions:</p>

<p>(1) What ordinals $\alpha$ can be obtained as the order type of a noetherian 
group?</p>

<p>(2) Is it possible to reduce that question to noetherian <em>simple</em> groups? For instance, if $H$ is a normal subgroup of $G$, is it true that $ot(G) = ot(H) + ot(G/H)$?</p>

CC BY-SA 3.0
2014-04-10T12:19:38.103Z
Have you got a communicable intuition what the order-type of a noetherian group may be, or why we can restrict our attention to simple noetherian groups?
Martin Berger
CC BY-SA 3.0
2014-04-10T12:49:18.103Z
Not really, but I would expect the question to be either very easy or very difficult depending on how you interpret it.. As I'm not you I can't tell :)
NisaiVloot
<algebra><gr.group-theory>
Classifying noetherian simple groups by order type?
NisaiVloot
# Classifying noetherian simple groups by order type?

(Asked by: username_0 on 2014-04-10)

A (possibly infinite) group $G$ is *noetherian* if it satisfies the following equivalent conditions:

(1) every subgroup of $G$ is finitely generated,

(2) there is no infinite strict ascending chain of subgroups,

(3) any non-empty collection of subgroups has a maximal element under inclusion.

Note that this property is hereditary: if $G$ is noetherian then so are all its subgroups. Also, being noetherian is equivalent to require that the poset of subgroups $(Sub(G),\subseteq)$ is a well-order.

The *order type* of a well-order $(S,\leq)$ is defined as the supremum of the order type of its linear extensions. By extension, we may define the *order type* of a noetherian group $G$, denoted by $ot(G)$, as the order type of $(Sub(G),\subseteq)$. We may then ask the following questions:

(1) What ordinals $\alpha$ can be obtained as the order type of a noetherian
group?

(2) Is it possible to reduce that question to noetherian *simple* groups? For instance, if $H$ is a normal subgroup of $G$, is it true that $ot(G) = ot(H) + ot(G/H)$?

**Tags:** algebra, gr.group-theory

> **Comments:**
> **username_1**: Have you got a communicable intuition what the order-type of a noetherian group may be, or why we can restrict our attention to simple noetherian groups?  
> **username_0**: Not really, but I would expect the question to be either very easy or very difficult depending on how you interpret it.. As I'm not you I can't tell :)  

---

## 1 answers

### Answer by username_2 on 2014-04-11. Score: 1

Here's an argument handling a special case. Consider a fixed poset $P = (S,\leq)$. Say that a (well-founded) forest $F$ is a *branching-representation* of $P$ if:

1. the nodes of $F$ are bijectively labelled by the elements of $P$;
2. for each node $u$ of $F$, its children are ???

(SORRY, CAN'T FIGURE THE RIGHT DIFINITION NOW, I'M PUTTING THIS ON STANDBY)

We may then assign an ordinal $o(u)$ to each node of $T$ inductively: (i) if $u$ is a leaf then $o(u) = 1$, (ii) if $u$ has children $(u\_i)$, then $o(u) = \sum\_i o(u\_i)$. The order type of $T$ is then defined as $o(T) = o(r)$ where $r$ is the root.

Is the following true?

> Conjecture: the order type of $P$ equals the minimum of $o(T)$ over all possible branching-representations $T$ of $P$.

Suppose this is true. Consider a poset $P = (S,\leq)$, and let $P^{\natural} = (S',\leq')$ where $S'$ consists of the finite antichains of $S$ and where the order satisfies $X \leq Y$ iff there is an injection $\phi : X \rightarrow Y$ s.t. $u \leq \phi(u)$ (f.a. $u \in X$). Suppose that $ot(P) = \lambda$ with $\lambda$ a limit ordinal, it should then follow that $ot(P^{\natural}) = \omega^{\lambda}$: starting with a b.p. $T$ of $P$, we would construct a b.p. $T'$ for $P^{\natural}$ by 'iterated substitution'.

This might help to analyze the case of *ordered free groups*. In this case, we have an ordered alphabet $P = (S,\leq)$, and we consider the free group $P^{\sharp}$ over $\{S \cup S^{-1}\}^\*$, equipped by the relation $u \preceq v$ iff $u\_i \leq v\_{\phi(i)}$ for some injection $\phi$. Now, given an element $B = \{b\_1,\ldots,b\_k\} \in (P^{\sharp})^{\natural}$, we may define the subgroup $\langle B \rangle$ of $P^{\sharp}$ generated by $\cup\_{i = 1}^{k} (b\_i \downarrow)$.

It follows that the map $\chi : (P^{\sharp})^{\natural} \rightarrow Sub(P^{\sharp})$ defined by $\chi(B) = \langle B \rangle$ is surjective. Consider the equivalence over $(P^{\sharp})^{\natural}$ s.t. $B \sim B'$ iff $\chi(B) = \chi(B')$. Assuming that the equivalence classes of $\sim$ are finite (which seems intuitively true), I think that it would imply $ot(Sub(P^{\sharp})) = ot((P^{\sharp})^{\natural}) = \omega^{\omega^{\omega^{\lambda}}}$ (possibly up to some lower degree terms).

So if the above is correct, you've got at least the ordinals of the form $\omega^{\omega^{\omega^{\lambda}}}$ in question (1).

> **Comments:**
> **username_3**: Out of curiosity: are you and HeloLobo different people?  
> **username_2**: Let's assume that we are the same person. If I answer positively I run the risk of being called a 'liar' by certain people based on the profile colors, and if I answer negatively I run the risk of being called a 'liar' by some other people that may have inside info on the users. Therefore, I'd rather not answer :) By the way, it's still a mystery to me how these colors are attributed, is it explained somewhere on the site?  
> **username_4**: The profile colors, as far as I know, are assigned randomly.  It's against the spirit of this site to use multiple accounts, as the preference is to even use real names.  

---

