What is the relationship between monoids and groups in homomorphisms?

[calvino]

I have the following problems to do:

Problem 1: Let S be a monoid. Find a subgroup G of S with the property that any monoid homomorphism f: H-->S (H any group) has its image in G.

Problem 2: Find a monoid S such that there is no group G that contains S as a submonoid

Problem 3: Let X be a set, and let ~ be the least congruence relation on F_Mon(X) with xy ~ yx for all x, y in X. Prove that F_Mon(X)/~ is a free Abelian monoid over X.

any help?

For problem 1, I was thinking simply the group consisting of the identity of S. since homomorphisms will take e_H (id in H) to e_S, the id in G and S... but i have a feeling also that that is wrong since we are looking for a subgroup that will hold all of H's image (at least that's what i believe the question is asking)

For 2, I was trying to think of some sort of monoid with elements that do not have inverses. that way no GROUP could contain it. I was looking at functions with composition... not sure where to look.

NOTE: i added Problem 3, which I am currently working on. So i'll keep you updated. for now, any insight would be great. thanks

 

[AKG]

For 2, functions with composition is definitely a good place to look. What is it you're not sure about?

For 1, isn't it just entirely trivial? Just let G be the biggest group in S, i.e. the group generated by all elements of S that have inverses.

 

[calvino]

you're right...i thought too much of 1. thanks

as for 2...it's just that i don't know when to give up. it seems as if it could be anything to do with functions.

***added another problem. =/

 

[calvino]

for 2, if we look at any monoid S with multiplication as an operation and O is an element of S, then this works...doesn't it? are there any others? I couldn't find what I was supposed to with composition. =S

 

[AKG]

I don't know what you're talking about. You were on the right track before. You want to find something without inverses, and most functions don't have inverses under composition.

 

[calvino]

my problem becomes..how do i construct such a monoid? (can u give me an example of a non-invertible function?)

 

[AKG]

Can you not think of an example of a non-invertible function?

 

[calvino]

no...is it not true that any non-invertible function can be made invertible? in a sense sine isn't invertible, if you consider the domain of ALL angles. what non-invertible functions were you thinking of? Even if I did know of one, how do i start to assume the set of functions is closed?

as for the last problem. i answered it like so...someone pls let me know if it is the right way to go.

in rough:
every element b of F_mon(X) for some X is of the form @(x_i)^(k_i) for all i, and k_i>=1, x_i <> x_(i+1). Here @ is the product concatenation. now, without loss of generality, say b= x_1^(k_1) x_2^(k_2) x_3^(k_3), and x_1=x_3. then b= x_1^(k_1) x_2^(k_2) x_3 ... x_3, k_3 times. since xy~yx, for all x,y in F_mon(X)/~, when we apply ~, we see that

b= x_1^(k_1) x_2^ x_3 (k_2) x_3 ... x_3, k_3-1 times

this construction continues until we get
b= x_1^(k_1) x_2^(k_2) x_3^(k_3)
= x_1^(k_1) x_3^(k_3) x_2^(k_2)
= x_1^(k_1+k_3) x_2^(k_2)
which means (f_mon(X)/~) =~ (f_abmon(X))//

 

[calvino]

for 2, if we look at any monoid S with multiplication as an operation and O is an element of S, then this works...doesn't it? are there any others? I couldn't find what I was supposed to with composition. =S

is this not correct for 2?

if we take Z equipped with normal multiplication, then no group can contain this monoid since 0 isn't invertible. it's much easier than the function example, to me anyway.

another example is the following: say we have three elements e (identity), a, b<>e in the monoid such that
e a b
---------
e| e a b
a| a e a
b| b a e


it cannot be contained in a group since ab=a -> a^(-1)ab= a^(-1)a -> b=e. are any of the above true satisfactory for the question?

 

[HallsofIvy]

For 1, you need the largest group that is a subgroup of the monoid: the group of all invertible elements of S. Since S is a monoid, it has an identity so that at least is invertible.

 

[matt grime]

no...is it not true that any non-invertible function can be made invertible?

No. It cannot be made invertible.

For 2, in a group G xy=xz implies y=z. And remember ytou're trying to extend S to be a group...

Abelianness in 3 is trivial from the fact that xy~yx, so the freeness must come from the 'least' part...