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I was introduced to semigroups today and had a question. So all the examples of semigroups I was given were either monoids or groups. So I was curious, does there exist a semi-group which is not abelian and does not contain identity?

I tried to construct an example, but every example I tried to construct had an identity element.

Thanks!

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  • $\begingroup$ @bof That's far simpler than the solutions given below (and I was also going to give it.) Why not make it an answer? $\endgroup$ – rschwieb Jan 13 '15 at 23:28
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If $S$ is a set with at least two elements, and multiplication is defined by $x\cdot y=y$, then $(S,\cdot)$ is a noncommutative semigroup with no right identity. (Of course every element is a left identity.)

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Let $A^{\ast}$ be the free monoid on a nonempty set $A$; this is just the set of all words using $A$ as an alphabet. Let $e$ be the identity. Since no element of $A^{\ast}$ has an inverse other than $e$, $A^{\ast}-\{e\}$ is a semigroup without identity, and it is nonabelian provided $A$ has more than one element.

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Take the set of all finite, non-empty sequences of elements in some non-empty set $S$ with operation given by concatenation (commonly called the free semigroup on $S$). That is, take $S^+=\bigcup_{n\geq 1}{S^n}$ where for $(s_1,\ldots, s_n),(t_1,\ldots, t_m)\in S^+$ we have their concatenation given by $$(s_1,\ldots, s_n)(t_1,\ldots, t_m)=(s_1,\ldots, s_n,t_1,\ldots, t_m).$$ This is non-commutative so long as $S$ contains at least two elements; if $x,y$ are such distinct elements, then $(x)(y)=(x,y)\neq (y,x)=(y)(x)$.

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    $\begingroup$ Your last statement isn't true; for instance, if you take $(\mathbb N,\max,0)$ as your monoid, then $\max(0,n)=n$, so $0$ is the identity. However, for every $n\neq 0$ it holds that $\max(1,n)=n$, so the new monoid would have an identity. $\endgroup$ – Milo Brandt Jan 13 '15 at 23:49
  • $\begingroup$ @Meelo Huh, I hadn't even considered a new identity appearing. Thanks for pointing that out! Removed that bit. Do you know of any conditions that stop something like that? $\endgroup$ – Hayden Jan 14 '15 at 2:15
  • $\begingroup$ @Hayden : $\:$ Being cancellative on either side stops things like that. $\;\;\;\;$ $\endgroup$ – user57159 Jan 14 '15 at 6:58
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A simple example would be the set of all non-empty strings (over some alphabet) with the concatenation operator. The empty string would be the identity, but we excluded it from the domain, so there is no identity.

It might be a little unsatisfying to create a semigroup with no identity simply by removing the identity, but if we take any semigroup without an identity and define a new identity element $e$ such that $e\cdot x = x\cdot e = x$ for all $x$ in the semigroup, then we always create a semigroup with an identity - so any example we can possibly give is just a semigroup with its identity removed (and which remains closed under its operation when we removed the identity).

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It depends on exactly what you mean by "does not contain identity". What about one sided identities? What if there is more than one distinct identity? The existence of a single unique identity element is a property of groups. When you are not talking about groups, you have to be more specific about what you are excluding. In this example, every element is a left identity, but there is no right identity or two-sided identity. I copied this example straight from Wikipedia (see the last example in the list), but in my defense I built an identical example myself before I ran across it.

only two elements {e, f}
∗ defined by
e ∗ e = f ∗ e = e and
f ∗ f = e ∗ f = f

both e and f are left identities,
but there is no right identity
and no two-sided identity

The operation is associative, which you can see if you list out all combinations of two operations on three elements, e.g.

(e * f) * e = f * e = e
e * (f * e) = e * e = e

(e * f) * f = f * f = f
e * (f * f) = e * f = f

...

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  • $\begingroup$ This is the same answer as bof's, but I came up with it independently and we posted at the same time. $\endgroup$ – Tony Jan 14 '15 at 0:00
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    $\begingroup$ Instead of listing out all combinations, you could simply note that, if $x*y=y$ for all $x,y$, then $x*(y*z)=x*z=z=(x*y)*z$. $\endgroup$ – bof Jan 14 '15 at 0:01
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Yes, and you can do it with just two elements: Consider the set $\{a, b\}$ with the operation that all products are equal to $a$. This is clearly associative, and neither element is an identity.

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  • $\begingroup$ Ah, I used a similar example but I see where I went wrong with it! Thanks! $\endgroup$ – Jeff Jan 13 '15 at 21:46
  • $\begingroup$ @Jeff That doesn't make $a$ an identity, since $ab \neq b$. Are you looking for a semigroup without identity, or a semigroup such that $xy = x$ never holds for any $x$ and $y$? If you're looking for the latter, something like the strictly positive integers under addition should work. $\endgroup$ – MartianInvader Jan 13 '15 at 21:46
  • $\begingroup$ I realized my mistake which is why I deleted my comment! Thanks for a very simple example! $\endgroup$ – Jeff Jan 13 '15 at 21:48
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    $\begingroup$ The OP wants a "non-abelian" semigroup. I think that means noncommutative. Your example is commutative. $\endgroup$ – bof Jan 13 '15 at 21:48
  • $\begingroup$ I just realized it was abelian, I'm looking for a non-abelian. $\endgroup$ – Jeff Jan 13 '15 at 21:49
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The Cayley table

$$\begin{array}{c|ccccccc} \ast & 0 & 1 & 2 & 3 & 4 & 5 & 6\\\hline 0 & 4 & 2 & 6 & 0 & 4 & 5 & 6\\ 1 & 3 & 5 & 1 & 6 & 4 & 5 & 6\\ 2 & 0 & 5 & 2 & 6 & 4 & 5 & 6\\ 3 & 4 & 1 & 6 & 3 & 4 & 5 & 6\\ 4 & 4 & 6 & 6 & 4 & 4 & 5 & 6\\ 5 & 6 & 5 & 5 & 6 & 4 & 5 & 6\\ 6 & 4 & 5 & 6 & 6 & 4 & 5 & 6\\ \end{array}$$

defines a non commutative semigroup with no (left nor right) identity, over the finite set $\{0,1,2,3,4,5,6\}$ (reference).


Furthermore, as pointed out in the answer by @bof, finite magmas like the one in this answer are associative, non commutative (the Cayley table is not symmetric), have no left identity, hence have no identity.

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You may recall from group theory that every group embeds into a symmetric group. There is an analagous result that every semigroup embeds into the semigroup of functions on some set. So an easy way to construct examples of semigroups is to take a set $S$, and then consider some subset of the functions $S \to S$ that is closed under composition.

For your question in particular, we could take

  • The set of functions $f: \mathbb{N} \to \mathbb{N}$ such that the image of $f$ is finite, under composition. (Or replace $\mathbb{N}$ with any infinite set.)

  • The set of functions $f: \{0,1,2\} \to \{0,1\}$ under composition.

  • The set of bounded, continuous $f: \mathbb{R} \to \mathbb{R}$ under composition.

In all these examples, the idea is to take the set of functions on some set under composition (function composition is always associative), but to further restrict the set in some way that excludes the identity function. Function composition is generally nonabelian, so we should get a nonabelian semigroup without identity.

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