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#### eddybob123

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- Aug 18, 2013

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- Thread starter eddybob123
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- Aug 18, 2013

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- Mar 5, 2012

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Hi eddybob123!

Suppose we pick $G=\{g\}$, $H=\{h\}$, and the following operation tables.

+ | g | h |

g | g | g |

h | h | h |

⋅ | g | h |

g | g | h |

h | g | h |

Then G∪H is closed for + and ⋅.

Furthermore, both operations are associative and also distributive.

However, there is no unit element for ⋅.

Therefore, (G∪H, ⋅) does not have to be a monoid (proof by counterexample).

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- Aug 18, 2013

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Then is it a semigroup?

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- Mar 5, 2012

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Then we first need to examine your assertion:Then is it a semigroup?

I do not know how to interpret this.the operation $\cdot$ is closed, associative, and distributive over $+$ in $G$ and $H$.

Is ⋅ closed over G and also closed over H? But not necessarily over G∪H?

If so, then it

So I guess I should assume you mean that ⋅ is closed over G∪H?

And then also assume that ⋅ is associative over G∪H?

If so, then (G∪H,⋅) is trivially a semi-group, since its required properties were just stated.

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- Aug 18, 2013

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In your first post of this thread, it seems as though you have misunderstood my question. Obviously, there is only one non-isomorphic monoid of order one, and that is the trivial one. If you take the union of two trivial monoids, you obtain another trivial monoid, but with a different operation.

Ok, how about this: if $G$ and $H$ are monoids both under the operation $+$, then determine whether or not $(G\cup H\cup \{e\},\cdot)$ is necessarily a monoid. $e$ is the identity of $\cdot$.

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- Mar 5, 2012

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I seem to be misunderstanding.

In your first post of this thread, it seems as though you have misunderstood my question. Obviously, there is only one non-isomorphic monoid of order one, and that is the trivial one. If you take the union of two trivial monoids, you obtain another trivial monoid, but with a different operation.

Ok, how about this: if $G$ and $H$ are monoids both under the operation $+$, then determine whether or not $(G\cup H\cup \{e\},\cdot)$ is necessarily a monoid. $e$ is the identity of $\cdot$.

If ⋅ is well-defined (i.e. closed) over G∪H∪{e}, is associative, and has an identity e, then (G∪H∪{e},⋅) satisfies all conditions for a monoid.

In other words, then there is no need for an operation + or for distributivity.

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- Mar 22, 2013

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Take $G = H = \mathbb{N}\setminus \{0\}$ with operation usual multiplication and $\cdot$ the usual addition. $\left (\mathbb{N}\setminus \{0\}, \cdot \right)$ is not a monoid.eddybob123 said:Suppose that $(G,+)$ and $(H,+)$ are both monoids and that the operation $\cdot$ is closed, associative, and distributive over $+$ in $G$ and $H$. My question then is whether or not $(G\cup H,\cdot)$ is necessarily a monoid.

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- Mar 5, 2012

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Yes. (Edited)If an operation is distributive and associative, then isn't it automatically closed?

To form rings, you need 1 set that has + and $\cdot$ defined on it.This is what forms rings from abelian groups.

Not 2 different sets.

Huh?Take $G = H = \mathbb{N}\setminus \{0\}$ with operation usual multiplication and $\cdot$ the usual addition. $\left (\mathbb{N}\setminus \{0\}, \cdot \right)$ is not a monoid.

This

Huh?$\cdot$ is closed, associative and distributive over $+$, right?

Since $G$ and $H$ has the same operator $+$, the base set of one is necessarily the subset of the other, hence this trivially follows.

Where did you get that $G$ and $H$ have the same operator $+$?

And even if they do, that does not mean one is a subset of the other.

Consider for instance $G=2\mathbb Z$ and $G=3\mathbb Z$ (the multiples of 2 respectively 3).

Last edited:

- Mar 22, 2013

- 573

$\cdot$ is associative and distributive on $G$ and $H$, but not on $G\cup H$. How can you define associativity on the elements $a, b, c \in G \cup H$ whereas $a \cdot b$ or $b \cdot c$ are not even defined ( assuming $a, c \in G$ and $b \in H$).I like Serena said:Then ⋅ is not closed on G∪H, but it is associative and distributive on G∪H.

There is no identity, so not.I like Serena said:This is a monoid

I am assuming they are the semigroups instead of the base sets of the semigroups of OP. (Which I think contrary of yours).I like Serena said:Where did you get that G and H have the same operator +?

Yes, I agree. Well, in my defense, I was working on several things at once and as well as answering to this post.I like Serena said:And even if they do, that does not mean one is a subset of the other.

- Mar 22, 2013

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- Mar 5, 2012

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You are right. I retract my statement. (I can do that, can't I?)$\cdot$ is associative and distributive on $G$ and $H$, but not on $G\cup H$. How can you define associativity on the elements $a, b, c \in G \cup H$ whereas $a \cdot b$ or $b \cdot c$ are not even defined ( assuming $a, c \in G$ and $b \in H$).

I have removed it from my previous comment.

With regular multiplication, the identity is 1, which is part of the set...There is no identity, so not.

Huh?I am assuming they are the semigroups instead of the base sets of the semigroups of OP. (Which I think contrary of yours).

Obviously I am missing something here, but (G,+) and (H,+) are stated to be monoids.

What else can G and H be other than sets?

No problem.Yes, I agree. Well, in my defense, I was working on several things at once and as well as answering to this post.

- Mar 22, 2013

- 573

How often this happens! You mistook the notation by the property in their my post. If you read carefully, you'll see that I have defined $\cdot$ to be the usualI like Serena said:With regular multiplication, the identity is 1, which is part of the set...

It is a matter of notational inconsistency we are having here. I am simply abbreviating $(G, +)$ and $(H, +)$ as $G$ and $H$.I like Serena said:Obviously I am missing something here, but (G,+) and (H,+) are stated to be monoids. What else can G and H be other than sets?

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- #14

- Aug 18, 2013

- 76

So what you're saying is... addition is distributive over multiplication.Take $G = H = \mathbb{N}\setminus \{0\}$ with operation usual multiplication and $\cdot$ the usual addition. $\left (\mathbb{N}\setminus \{0\}, \cdot \right)$ is not a monoid.

- Feb 15, 2012

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To see what I mean, let $F(A)$ be the free semigroup on the set $\{A\}$ and $F(B)$ be the free semigroup on the set $\{B\}$.

Clearly both semigroups are sub-semigroups of $F(A,B)$. The problem is that $F(A) \cup F(B)$ is not, and adding the empty word to this union doesn't change this, we still have a problem with the product $AB$, for example. This problem exists even though we can embed both semi-groups in the semi-ring $\Bbb N[F(A,B)]$, where your proposed closure and distributive properties both hold for $G = \Bbb N[F(A)]$ and $H = \Bbb N[F(B)]$.

In general, algebraic operations perform poorly with respect to unions (because of closure issues). The union generally isn't "big enough" to be the join in the lattice of substructures.

If you are already stipulating that $G \cup H$ is already closed and associative under $\cdot$, there is nothing to prove, we can always create a monoid by "adjoining a 0", and defining:

$x \cdot 0 = 0 \cdot x = x$ for all $x \in G \cup H$.

IF the semi-groups $G,H$ are part of a larger structure that already HAS a multiplicative identity, $e$, then we can run into problems:

Consider $\Bbb N \times \Bbb N$ with the operations:

$(a,b) + (a',b') = (a+a',b+b')$

$(a,b) \cdot (a',b') = (aa',bb')$

Now we can form a monoid from the union of $G = \{(k,k):k \in \Bbb N\}$ and $H = \{(k,0): k \in \Bbb N\}$, taking the unit to be that of $G$.

The problem is, $H$ is not a sub-monoid of $G \cup H$, that is, the inclusion mapping is no longer a monoid homomorphism, because the identity of $H$ is $(1,0)$ which does not map to the identity $(1,1)$ of $G \cup H$.

If, in the example above, we instead took $G = \{(0,k): k \in \Bbb N\}$, then we again get that $G \cup H$ is a semigroup, and we also have that $G \cup H \cup (1,1)$ is a monoid, but now neither of $G,H$ is a submonoid of $G \cup H \cup (1,1)$.

That is, even though the semi-groups structures of $G,H$ are compatible with the semi-group structure of $G \cup H$, the monoid structures are not. Depending on your point of view, this is either a minor inconvenience, or a fatal flaw.

I suspect less trivial examples could be formed with $\text{End}_F(F^n)$ under matrix multiplication, but alas I don't have the time at the moment to look for some.