It is well known that any non-trivial totally ordered group is infinite.
Is it true that any totally ordered magma with more than one element is infinite too?
My attempt to prove the statement:
Let's say a magma $M(\cdot)$ is totally ordered if it has a total order $<$ compatible with the operation:
$a < b \implies ac < bc$ and $ca < cb$ for any elements $a$, $b$, $c$ from $M$.
Applying the compatibility for any two distinct elements $a < b$ of $M$: $aa < ab < bb$;
Therefore:
- For any two distinct elements $a$ and $b$ of $M$ the elements $aa$ and $bb$ are distinct;
- For any two distinct elements $aa < bb$ of $M$ there is an element $ab$ such that $aa < ab < bb$;
Assuming $M$ has $n > 1$ elements $\{ m_1 … m_n \}$;
- $M' = \{ m_1m_1, m_2m_2, …, m_nm_n \}$ is a subset of $M$;
- All the elements $\{ m_1m_1, m_2m_2, …, m_nm_n \}$ are distinct (from 1);
- $M = M'$ since $M'$ is a subset of $M$ with $n$ elements;
For any two consecutive elements $m_im_i < m_jm_j$ from $M$ there is an element $m_im_j$ between them (from 2);
Contradiction.
Is this correct?
Where can I read about a linear and a cyclic order on a magma?
Best Answer
Your argument is correct but maybe a bit too long.
Suppose your totally ordered finite "magma" is $M=\{a_1,a_2,\dots,a_n\}$ where $a_1\lt a_2\lt\cdots\lt a_n$.
For any $c\in M$ we have $a_1c\lt a_2c\lt\dots\lt a_nc$, whence $a_ic=a_i$. That is, $M$ satisfies the identity $xy=x$.
A similar argument shows that $M$ satisfies $xy=y$ and therefore $x=y$, that is, the "magma" is trivial.