Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Define a permutation or autojection A as a function which bijects from a set X to itself.

Define a binary algebra or magma as a set with a binary operation on it.

Let un-subscripted Hindu-Arabic numerals to denote binary operations.

Conjecture: For every binary algera $N_{1}$=(N, 1) on a set N with with n elements, where n belongs to {2, 3, ...} (equivalently, n equals any natural number greater than 1), there exists an algebra $N_{2}$=(N, 2) (not necessarily distinct from $N_{1}$) such that if A indicates any autojection on N, then A qualifies as an automorphism between $N_{1}$ and some $N_{2}$ (not between every $N_{1}$ and every $N_{2}$).

Question 1: Is this conjecture correct?

Question 2: If correct, how does one prove this conjecture?

This seems true intuitively to me, and it comes as true by definition that A indicates an autojection. So, it would seem that only the homomorphic equation xy1A=xAyA2 would need verified. But, how does one do verify it here, or ensure that the homomorphic equation holds with A as an autojection?

share|improve this question
6  
@Doug Spoonwood: Personal choices as to the names of things are only useful when one is talking to oneself. A similar comment could be made about personal notational choices. –  André Nicolas Jun 19 '11 at 1:04
9  
"Autojection"? You seem to be very, very determined to be not understood both with your vocabulary and your notation: you will probably be happy to know that you are doing great in that front :) –  Mariano Suárez-Alvarez Jun 19 '11 at 2:38
3  
Doug Spoonwood: If we are actually serious about the solution of a problem, we will ask a question in the language of the community. I assume that you are indeed serious about the things that you do, and am willing to give some attention to real problems. But not to trivialities hiding under non-standard notation. –  André Nicolas Jun 19 '11 at 5:07
6  
I am not joking! The Number One Rule Of Communication is know who you are talking to, and you surely don't think Jan Lukasiewicz is reading this site... Take the time to browse a bit in this site and please show me one example of anyone using reverse polish notation and integers to denote operations! I have been involved in mathematics for almost two decades now, and I have never seen anyone alive do that. –  Mariano Suárez-Alvarez Jun 20 '11 at 2:46
8  
@Doug: Plenty of people did you the extreme favor of spending more than they ought of their time trying to puzzle you out. That you would take this as evidence that you did the right thing is just one more piece of evidence that the workings of your brain are essentially disjoint from mine. You might want to ask them if they found it easy or difficult to puzzle you out or not before bringing them forth as witnesses on your behalf. –  Arturo Magidin Jun 21 '11 at 3:27

3 Answers 3

up vote 5 down vote accepted

What is true is that given any autojection there is a binary algebra so that the autojection is an automorphism of binary algebras. You simply define the operation so that it works with the autojection. For instance, suppose you have the binary algebra $(N, \star)$, and an autojection $A: N \rightarrow N$. Then define a new operation $\bullet$ by $a \bullet b = A^{-1}(A(a)\star A(b))$ for all $a, b \in N$.

share|improve this answer
    
I think I see how that works. But since it works, and since A is an autojection it has the inverse function which you gave above, so it follows that we have A(a.b)=A(a)*A(b). Now say we have . defined and * left unknown. Then the autojection takes the whole table for . to members of N. I think the question I have can thus get restated as does the set of pairs (A(a), A(b)) exhaust the set of pairs of NxN? If they do, then * can get defined here by an autojection. –  Doug Spoonwood Jun 19 '11 at 2:29
    
@Doug: Yes, given any two $a,b \in N$, that formula tells you what their product should be, and it is well-defined since A is a bijection. What your conjecture (now a theorem) would be is that every autojection induces a binary operation that makes the autojection into an automorphism. –  Joe Jun 19 '11 at 2:35

No, the conjecture is not correct. Let $N=\{a,b\}$, and define the operation $\star\,\,:N\times N\rightarrow N$ to be the constant function to $a$. Suppose $\bullet\,\,:N\times N\rightarrow N$ is any other operation on $N$.

If the image of $\bullet$ is all of $N$, then there cannot be any isomorphisms from $(N,\star)$ to $(N,\bullet)$.

If the image of $\bullet$ is $\{a\}$ (i.e. $\bullet=\star$), the autojection $p:N\rightarrow N$ with $p(a)=b$ and $p(b)=a$ is not an isomorphism from $(N,\star)$ to $(N,\bullet)$, because $$p(a\star a)=p(a)=b\neq a= p(a)\star p(a)=p(a)\bullet p(a).$$

If the image of $\bullet$ is $\{b\}$, then the identity autojection is not an isomorphism from $(N,\star)$ to $(N,\bullet)$ for a similar reason.

share|improve this answer
    
@Zev I'll use "1" to denote the star operation above, and "2" to denote the dot operation above. If the autojection p is such that p(a)=b, p(b)=a, then p(1(x, y))=p(a)=b. So, "2" has image of {b}. So, p induces a magma from 1. If the autojection i is such that i(a)=a, i(b)=b, then "2" has image of {a}. Parts two and three just tell us that in such a case the autojection can't be p or i respectively. However, such an autojection still exists. –  Doug Spoonwood Jun 19 '11 at 2:02
1  
However the way you qualified your statement, it has to work for any autojections. Specifically it must work for p. –  Joe Jun 19 '11 at 2:10
1  
@Doug: Why don't you use symbols for operations? It's much easier to read. At any rate, your question was asking whether, for any magma $(N,\star)$, there existed another magma structure on $N$, $(N,\bullet)$, for which any autojection of $N$ was an isomorphism from $(N,\star)$ to $(N,\bullet)$. It doesn't matter if there exists another magma structure $(N,\bullet)$ for which there is some autojection that is an isomorphism from $(N,\star)$ to $(N,\bullet)$; what matters is whether all of them are. –  Zev Chonoles Jun 19 '11 at 2:12
1  
The problem is that integers are often used in additive structures to denote repeated addition, so it could get confusing. –  Joe Jun 19 '11 at 13:16
1  
@Doug Spoonwood: The first substantial investigation was done by the linguist Yngve. I did a quick search using embedding depth, linguistics, Yngve. Lots of hits. When long ago I bumped into the Yngve hypothesis, I realized, like presumably hundreds of others, that it explains the total failure of Polish notation in logic. Nowadays it is used only by a few philosophers who never need to deal with a real mathematical problem. But computers love Polish notation. Mathematical notation, and typography, are designed to make the structure of formulas visually graspable, by people. –  André Nicolas Jun 19 '11 at 21:20

Given any two sets $A$ and $B$, and a bijection $f\colon A\to B$, any $\alpha$-ary operation $\tau$ (for $\alpha\in\mathrm{Ord}$) yields an $\alpha$-ary operation $\tau'$ on $B$ defined by $$\tau'\left(\{b_i\}_{i\in\alpha}\right) = f\left(\tau\left(\{g(b_i)\}_{i\in\alpha}\right)\right),$$ where $g\colon B\to A$ is the set-theoretic inverse of $f$. Under this definition, $f\colon (A,\tau)\to (B,\tau')$ respects the operation; i.e., $$f\left(\tau(\{a_i\}_{i\in\alpha})\right) = \tau'\left(\{f(a_i)\}_{i\in \alpha}\right).$$

(This includes binary, ternary, $n$-ary, $\omega$-ary, unary, nullary, etc. operations; and of course, $B$ can be $A$).

This easily extends to any family (or even proper class) of operations defined on $A$, so that given a bijection $f\colon A\to B$ and a class $\Omega$ of operations on $A$, we can define on $B$ a family of operations of the same types as $\Omega$ that make $f$ into an $\Omega$-algebra isomorphism.

And stripped of all the formalism and all the neologisms, what this says is basically that if you change the names of all the elements, but keep the operations the same, you get an isomorphic structure. E.g., addition in the natural numbers in English is isomorphic to addition in the natural numbers in Spanish (just be careful in the translation; remember that $\text{one billion}\neq\text{un billón}$, just like $\text{library}\neq\text{librería}$.)

share|improve this answer
1  
In particular, when $\alpha = 2$, this yields my construction. –  Joe Jun 19 '11 at 2:58
1  
@Matt: Indeed; it's a standard thing, though it can be hard to tease out the meaning when one is slogging through neologisms and idiosyncratic notation. –  Arturo Magidin Jun 19 '11 at 3:08
1  
@Arturo: True enough. I am new, and I was unsure how you (the math SE community) handle cases where users coin their own phrases. I assumed it was OK as he did at least define all the words he invented. –  Joe Jun 19 '11 at 3:11
3  
@Doug You can look at it whatever way you want to look at it (which no doubt will not be the way I look at it). Doesn't matter whether you think of the names of the elements as changed and the operations as the same, or as the names of the operations changed and the names kept the same. Either way, it's a triviality once you strip it of the confusing neologisms and the idiosyncratic notation. –  Arturo Magidin Jun 19 '11 at 18:32
2  
@Doug This is a simple instance of transport of structure. A less trivial example is transporting the class group structure of quadratic fields from ideals to primitive binary quadratic forms - so greatly simplifying Gauss's presentation of composition of forms. –  Bill Dubuque Jun 21 '11 at 5:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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