4

I know that we have cyclic groups.

Maybe the groups which in the most easy-to-access way can explain the concept of a generator.

But do there also exist groups of two or more loops?

For example $\cases{{g_1}^{n_1} = e\\{g_2}^{n_2} = e}$

With $$\{g_k,{g_k}^{1},\cdots,{g_k}^{n_k}\} \text{ distinct}$$

and $${g_1}^{m}{g_2}^n={g_2}^n, \forall n\neq kn_2$$

In other words the only way to enter a loop is through $e$ and trying to loop in the other direction if we already are in another loop will do nothing.

Kind of like a state machine that starts counting one direction and then cannot count in the other direction until we have looped a whole circle around to $e$.

Is this possible or will this violate some axiom for groups?

Is there some other algebraical concept which it would be possible for?

mathreadler
  • 26,534
  • 1
    Can you proofread your two expressions in the middle? They look like they have typos – Milten Mar 15 '21 at 20:38
  • 2
    There's the free product of groups — in particular, the example given shows a free product of cyclic groups that would seem to cover what you're looking for, though without a clearer sense of what you're saying it's hard to be sure. You can always 'count' in another direction, because the product of two elements has to be an element. – Steven Stadnicki Mar 15 '21 at 20:38
  • 3
    Note $g_1^m g_2^n= g_2^n$ implies $g_1^m=e$. So what is $m$? – anon Mar 15 '21 at 20:41
  • 1
    $\forall n\ne kn_2$ is still unclear. But runway44 is right, $g_1^mg_2^n = g_2^n$ implies $g_1^m=e$ no matter what. So I don't think you'll find exactly what you're looking for. The closest thing is probably the direct product of to cyclic groups, $\langle g_1\rangle\times\langle g_2\rangle$. – Milten Mar 15 '21 at 20:45
  • @runway44 yes, I think you are right. So I wonder what kind of algebraic concept I should be looking for if I want to be able to construct something like it. – mathreadler Mar 15 '21 at 20:50
  • Sounds like you're describing all finitely generated abelian groups. – pancini Mar 15 '21 at 20:51
  • 4
    Careful: "loop" is a term of art in abstract algebra! It refers to a structure which is weaker than a group (a set with a binary operation on it, but we put fewer conditions than we put on a group). – Arturo Magidin Mar 15 '21 at 20:54
  • You want a group that is a union of cyclic groups, $G=\cup C_i$, and where $C_i\cap C_j={e}$ if $i\neq j$? You cannot do it for two, because no group is a union of two proper subgroups, unless one of them contains the other. On the other hand, $C_2\times C_2$ is equal to the union of the subgroups ${(e,e),(x,e)}$, ${(e,e),(e,x)}$, and ${(e,e),(x,x)}$, which pairwise intersect only at the identity. – Arturo Magidin Mar 15 '21 at 20:55
  • 4
    Note that 'state machines' are also a perfectly valid mathematical object; if your intention is to just have the entities of the form $(g_1)^j$ and $(g_2)^k$ but no 'cross objects' such as $g_1g_2$, then that's really what you're after. What are you trying to 'do' with the object you're looking for? – Steven Stadnicki Mar 15 '21 at 20:59
  • @StevenStadnicki I did not know it existed as a mathematical concept. I thought "state machine" was just an informal name for what we got to learn in engineering classes. – mathreadler Mar 15 '21 at 21:18

1 Answers1

4

Let's examine the structure you propose. This is how I interpret it: The set is $M=\{e, g_1, g_1^2, \ldots, g_1^{n_1-1}, g_2, g_2^2, \ldots, g_2^{n_2-1}\}$. Composition is non-commutative and defined as: $$ ex = xe = x $$ $$ g_i^a g_i^b = g_i^{a+b\bmod n_i}, \quad i\in\{1,2\} $$ $$ g_i^a g_j^b = g_j^b, \quad i\ne j\in \{1,2\} $$ The last one makes our object lose a lot of algebraic structure. It is not associative since $g_1(g_2g_1)=g_1^2$ but $(g_1g_2)g_1=g_1$. It is not cancellative since $eg_2 = g_1g_2$ but $e\ne g_1$. It doesn't have right division since $x g_1 = g_2$ has no solutions in $x$.

What it does have is an identity element, and all elements have a unique two-sided inverse. It also has unique left division for whatever that's worth: E.g. $g_1^a x = g_2^b$ is satisfied exactly by $x = g_2^b$.

I don't think there is a widely used term for this sort of structure, other than just discribing it when you need to. This other answer cites a paper that simply calls it a magma with inverses, but note that this would perhaps more often be called a unital magma with inverses. If I needed an ad hoc name for your specific construction, I might call it something like "the disjoint product of $\langle g_1 \rangle$ and $\langle g_2 \rangle$" (at least if that name isn't taken for something else).

[EDIT/erratum: I erroneously called the structure a loop at first. That's wrong, because a loop is a quasigroup with identity, which means it satisfies the "Latin square property"; in particular, it is cancellative].

But indeed, you should consider Steven Stednicki's comment ("What are you trying to 'do' with the object"?). This algebraic structure might not be what you're actually after, even though it matches your formal discription.

Milten
  • 7,722