Talk:Monoid

From Wikipedia, the free encyclopedia

	This article is within the scope of WikiProject Mathematics.
	Mathematics grading:	B Class	Mid Importance	Field: Algebra
Needs history and more writing in prose Tompw 13:48, 5 October 2006 (UTC)

Contents 1 Why? 2 Computer Examples 3 Uniqueness of identity element 4 Fixed Identity under Mappings 5 Uniqueness of inverses? 6 Why magma in definition? 7 Recursively adding an identity? 8 Every monoid contains a group? 9 unify or clarify the following statements in the article

[edit] Why?

   * Associativity: for all a, b, c in M, (a∗b)∗c = a∗(b∗c)
   * Identity element: there exists an element e in M, such that for all a in M, a∗e = e∗a = a.
One often sees the additional axiom
   * Closure: for all a, b in M, a∗b is in M
though, strictly speaking, this is not necessary as it is implied by the notion of a binary operation.

Why? Doesn't Closure need to be defined for Associativity to be? --Carbonrodney 07:43, 16 November 2006 (UTC)

It's not saying that closure isn't needed, it's saying that we already have it, because we have a binary operation . --Zundark

[edit] Computer Examples

could you give some examples related to computer ?

[edit] Uniqueness of identity element

Under what situation is the identity element guaranteed to be unique? My off-the-cuff guess is that a cancellative monoid has a unique identity. Is there more to it than that? (Thanks!)

The identity is unique in every monoid. If e and f are two identities, then e = ef = f, so they are equal. -- Fropuff 15:13, 2004 Jul 29 (UTC)

[edit] Fixed Identity under Mappings

Can someone show me an example of when a homomorphism would not preserve the identity (assuming that wasn't given as an axiom)?

In my graduate abstract book, it doesn't state that a homomorphism must preserve identity (only an isomorphism) and leaves it as an exercise to construct an example to illustrate. However, in a different class, I was asked to show that a ring isomorphism preserves multiplicative identities (when present, obviously). As far as I can tell, I didn't make use of the fact that my mapping was bijective (1-1 and onto). Since my rings are additive groups and multiplicative monoids, then there should be no difference, unless the distributive law somehow differentiates the two types of monoids. -- Ub3rm4th 13:05, 2005 Jan 25 (UTC-6)

That is very strange. The whole point of a homomorphism is that it preserves the essential properties of the algebraic structure. If a monoid homomorphism doesn't preserve the identity, then it's just a semigroup homomorphism. --MarkSweep 19:24, 25 Jan 2005 (UTC)

Yep. Fairly standard terminology (although I'm sure you'll find somebody who disagrees!) is that a monoid homomorphism is a map between monoids which preserves the multiplication and the identity. Since monoids are also semigroups, there is also a notion of a "semigroup homomorphism of monoids", which as it turns out will preserve multiplication but not necessarily the identity. You can easily show that a homomorphic image of a monoid identity will act as a local identity for the image of the map, which is another way of saying that a surjective semigroup homomorphism always preserves an identity if present; hence the result for rings. Cambyses 05:37, 26 Jan 2005 (UTC)

Thank you very much. From my past experiences with wikipedia, I didn't expect answers so soon, so this is the first I've checked. For some reason, I like the answer that it would have to be a "semigroup homomorphism of monoids". My interpretation of homomorphisms was exactly that they preserve the structure of the system, thus I was deeply disturbed. I don't think my book (Hungeford) differentiates between a monoid homomorphism and a semigroup homomorphism of monoids. Much gratitude! -- Ub3rm4th 10:38, 07 Feb 2005 (UTC-6)

[edit] Uniqueness of inverses?

If an element in a monoid has an inverse is that inverse unique? Also why?

Yes, associtivity guarantees that inverses, if they exist, are unique. Suppose b and c are both inverses of a. Then

c = ce = c(ab) = (ca)b = eb = b

So b = c. -- Fropuff 00:24, 13 January 2006 (UTC)

[edit] Why magma in definition?

The term 'magma' is quite unnecessary in encyclopedic definition of monoid. Very few people use this word outside Bourbaki circles. The idea to insert an obscure term in a definition or an explanation - for students etc - is simply ridiculous.

Of course we need an article for magma, but it does not mean we should stick this word everywhere.

I tend to agree. I'll remove it for now. -- Fropuff 03:23, 28 March 2006 (UTC)

[edit] Recursively adding an identity?

The fourth example stated:

Any semigroup S may be turned into a monoid simply by adjoining an element e not in S and defining ee = e and es = s = se for all s ∈ S. This can be done recursively (i.e. if S is monoid with identity e, one obtains a new monoid S ∪ {e′} with identity e′, and so on).

I removed the last sentence because by adjoining an identity e′ to a monoid with identity e simply gives e = e′ and the monoids are isomorphic. If someone can think of a compelling reason to keep this "recursive" construction method in please say so.

TooMuchMath 21:04, 10 April 2006 (UTC)

[edit] Every monoid contains a group?

Is it true that "every monoid contains a group" as written in the article? For example, if Z+ forms a monoid over multiplication with identity one, what group is made with Z+? I assume that to form a group, that set would need to be augmented with inverses. Maybe the word "contains" is misleading me. I can see how every group contains a monoid, but not vice-versa. - Simian1k 15:24, 13 April 2006 (UTC)

I don't know why the article bothers to mention it, but it's certainly true: every monoid contains at least the trivial group {1}. --Zundark 17:42, 13 April 2006 (UTC)

If you consider all the units in a monoid as a set they form a submonoid (IE the set is closed under *) that is also a group (inverses are present). This is the largest group contained in the monoid.

For some monoids the group of units is trivial, however there are examples where this is not the case. Consider M = < x,y | x⁵ = 1,y⁵ = y²,[x,y] = 1 > . This is the (monoid) direct product of two cyclic monoids. The submonoid < x > is a group isomorphic to Z/5Z and the submonoid < y > is not a group. The set of units of M is given by the submonoid < x > , IE the only elements of M which are invertible are powers of x.

Thank you for clearing that up. --Simian1k 16:01, 14 April 2006 (UTC)

[edit] unify or clarify the following statements in the article

These appear to be two different ways of saying the same thing: 1) A monoid satisfies all the axioms of a group with the exception of having inverses. A monoid with inverses is the same thing as a group; 2)If a monoid has the cancellation property and is finite, then it is in fact a group. drefty.mac 08:15, 28 October 2006 (UTC)

They are not really the same thing. Cancellativity is (in general) a necessary but not a sufficient condition for the existence of inverses. So (1) does not immediately imply (2). And (2) is a statement only about finite monoids (it is false if you remove the hypothesis of finiteness) so it can't imply (1) in the general (possibly infinite) case. (Of course in a strict logical sense, any two statements which are provable imply each other, but you know what I mean.... :-). Best wishes, Cambyses 11:21, 23 November 2006 (UTC)