1
vote
0answers
51 views

Order Preserving Isomorphism

If abelian group G has an archimedean order then there is an order preserving isomorphism $\phi$ of G onto a subgroup of $\mathbb{R}$. Here we can say that G is archimedean totally ordered abelian ...
0
votes
2answers
31 views

Order of the subgroup <g>H of G

This is an old exam question I'm trying to solve: Having a group $G$ and $H$ a normal subgroup of $G$ with order $n$ and taking $g$ in $G$ to be such that $gH$ has order $m$ in $G/H$, I wish to prove ...
0
votes
1answer
19 views

Look for the group members $\left ( (\mathbb Z/24\mathbb Z)^*,\underset{24}{\odot} \right )$ and calculate their orders.

Look for the group members $\left ( (\mathbb Z/24\mathbb Z)^*,\underset{24}{\odot} \right )$ and calculate their orders. The elements are $\overline1,\overline2,\ldots,\overline{23}$. Have the ...
0
votes
1answer
20 views

Evidence about the group $\left ( (\mathbb{Z}/p\mathbb{Z})^*,\underset{p}{ \odot} \right )$

Be $p$ an odd prime number. Show that the group $\left ( (\mathbb{Z}/p\mathbb{Z})^*,\underset{p}{ \odot} \right )$ has a unique element of order $2$, namely $\overline{p-1}$, and show that ...
1
vote
1answer
21 views

Be $G=\{ e,g_1,g_2,\ldots, g_n \}$, $|G|=n+1$. Suppose $G$ has a unique element of order $2$, say $g_1$. Show that $eg_1g_2\ldots g_n=g_1$.

Be $G=\{ e,g_1,g_2,\ldots, g_n \}$ an abelian group of order $n+1$. Suppose $G$ has a unique element of order $2$, say $g_1$. Show that $eg_1g_2\ldots g_n=g_1$. I have serious difficulties with ...
0
votes
1answer
21 views

Prime order of a number

Prove that if $a^p=e$ where $p$ is a prime number, then $a$ has order $p$ ($a\ne e$). Perhaps I should prove this using the contrapositive? Assume that $p$ is not the order of $a$, meaning there ...
3
votes
1answer
63 views

isomorphism between divisible, totally ordered, abelian groups

Let $G$, $H$ be divisible, abelian, linearly ordered groups, whose cardinalities are equal and satisfy $\mu := |G|=|H|>\aleph_{0}$. These are supposed to be (order!) isomorphic. And just about ...
3
votes
2answers
53 views

Embedding $\omega_1$ into the direct sum/product of $\Bbb R$'s and $\Bbb N$'s

I'm trying to find a way to visualize the first uncountable ordinal $\omega_1$. This is rather difficult, as the visualization tactic that I often use for countable ordinals - namely, the "matchstick" ...
0
votes
2answers
51 views

What is the order of $f(x) = \frac{2}{2-x}$ as a permutation?

$f(x) = \frac{2}{2-x}$ defines a permutation of the set $A={\bf R}\setminus \{0,1,2\}$. What is its order in the permutation group $S_A$? I don't know how to find orders of functions. Please help!
1
vote
1answer
36 views

Proof that the sum of the order of the orbits of a set is = the order of the set?

I understand that the order of a set is the number of elements it has, however I don't understand the relationship between this number and the orbits of the set. As I understand it, the orbit of an ...
2
votes
1answer
47 views

In the transition from set theory to order theory, what is the appropriate generalization of “group”?

Given a set $X$, the collection of all bijective endofunctions on $X$ forms a group, called the symmetric group on $X.$ Furthermore, Cayley's theorems says that every group embeds into a subgroup of ...
1
vote
1answer
44 views

What motivates the definition of “Periodic” group action

Consider a group $G$ acting on a set $\Omega$. For example, let $G=\{g\in A(\mathbb R):(\alpha +1)g=\alpha g+1\}$ for all $\alpha\in\mathbb R$, where $A(\mathbb R)$ are the order-preserving ...
1
vote
1answer
58 views

Example for a partially ordered Abelian group

Definition 1: Let $(G,\leq)$ be a nonzero partially ordered Abelian group with order unit $u$. (Recall that $u\in G$ is a order unit if, for every $g\in G$, there exists $N\in\mathbb N$ such that ...
0
votes
1answer
106 views

union of cyclic groups

Let $p$ be a prime integer. Let $C(p^n)$ be the cyclic group of order $p^n$ ($C(p^n)=\{z \in C\mid z^{p^n}=1\}$ and let $C(p^\infty)$ be the union of the groups $C(p^n)$, i.e., $C(p^\infty)=\{z\in ...
1
vote
1answer
56 views

Question about primitive group actions

In Glass' Partially Ordered Groups Corollary 7.4.4 says: If $G$ is an ordered group and $(G,G)$ is the right regular representation, then $(G,G)$ is primitive if and only if $G$ is ...
3
votes
0answers
50 views

How can we define “trivially orthogonal” groups?

In any lattice-ordered group, we say that two elements are orthogonal if their meet is 1. I've been thinking of groups who have only "trivial" orthogonal relations, i.e. $x\perp y\implies x=1$ or ...
1
vote
1answer
75 views

Show that the stabilizer is a prime subgroup

We define a subgroup $H$ as being convex if $g\in H\implies h\in H$ for all $1\leq h\leq g$. A convex subgroup $P$ is prime if for any two convex subgroups $X,Y$: $X\cap Y \subseteq P$ implies that ...
1
vote
0answers
63 views

Intuition behind prime subgroups

In any lattice ordered group, we say that a convex subgroup $P$ is prime if for any two convex subgroups $X,Y$: $X\cap Y \subseteq P$ implies that $X\subseteq P$ or $Y\subseteq P$. This is analogous ...
0
votes
1answer
25 views

Show that $\langle G^+\rangle=G$ in a directed group

Lemma 2.1.8 of Glass' Partially Ordered Groups states: $G$ is a directed group if and only if $\langle G^+\rangle=G$ (where $G^+=\{x:x\geq1\}$) This doesn't make any sense to me. For example, ...
7
votes
0answers
195 views

What do linearly ordered abelian groups look like?

Recently a post on MathOverflow connected a theorem about ordered abelian groups to the axiom of choice, and it made me want to try and play with these objects a bit. But it appears to me that I ...
5
votes
0answers
54 views

Why are the p-adic integers a linearly ordered group? [duplicate]

In a previous question, someone suggested the p-adic integers as an example of a non-archimedean linearly ordered group. I'm not sure why these are linearly ordered - specifically, it doesn't seem to ...
3
votes
0answers
56 views

Can all non-archimedean groups be written as a product of archimedean groups?

All the non-archimedean groups I know of can be written as the product of archimedean groups. I'm wondering if this is generally true. I think I've found a proof, but I haven't heard this theorem ...
2
votes
2answers
53 views

Conditions for linearly ordered groups

One standard definition of a linearly ordered group is that the order $\leq$ obeys the law (1) $a\leq b\implies ac\leq bc$. Suppose (1) only holds for positive c. Must it also hold for negative ...
4
votes
2answers
162 views

Order-preserving isomorphism between $\mathbb{R}^n$ and $\mathbb{R}$

Suppose we have a linearly ordered group over $\mathbb Z^n$ where the ordering goes left-to-right, i.e. when deciding if $(x_1,x_2,\dots)<(y_1,y_2,\dots)$ we first check if $x_1< y_1$, if it is ...
3
votes
1answer
339 views

Isomorphism from $\mathbb{R}$ to $(-1,1)$

There are many bijective functions that map $\mathbb{R}$ to $(-1,1)$, in particular: $$f\left(x\right)=\frac{e^{2x}-1}{e^{2x}+1}$$ (Of course there are others, such as ...
4
votes
2answers
347 views

partially ordered group, positive cone, quotient (exercise)

Definitions: A partially ordered group or po-group is a po-set $(G,\leq)$, such that $G$ is a group and $\forall x,y,a,b\!\in\!G\!:x\!\leq\!y\Rightarrow axb\!\leq\!ayb$, i.e. a po-set that is a group ...
4
votes
5answers
199 views

Are there methods to well order a finite group in a meaningful way?

Can some finite groups be well ordered in a "meaningful" way? I mean, it is clear that we can trivially find a bijection between $\{1,...,n\}$ and a finite group $G$ with $n$ elements, but I am ...
5
votes
3answers
939 views

Isomorphisms: preserve structure, operation, or order?

Everyone always says that isomorphisms preserve structure... but given the (multiple) definitions of isomorphism, I fail to see how the definitions equate with the intuitive meaning, which is that two ...
1
vote
2answers
67 views

Showing that Ab(G) need not be complete

What's the easiest example to show that $Ab(G)$, the set of Abelian subgroups of a group $G$, need not be complete? I heard that $D_4$ was a good example, but $Ab(D_4)=Sub(D_4)$, which is complete. ...
7
votes
1answer
326 views

Why do torsion-free abelian groups admit linear orders?

I have read a theorem that says that every torsion-free abelian group admits a linear order. The proof used tensor products and so was above my head. I tried to find another proof on the web and I ...
4
votes
2answers
254 views

Two order related questions for subgroups of infinite groups

For finite groups what I am asking is really trivial. In finite groups if we pick a non trivial subgroup $H$ of $G$ we can always find (there could be multiple choices) a subgroup that is an immediate ...