# More algebra MONOIDS, etc

1. Oct 4, 2006

### 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 im currently working on. So i'll keep you updated. for now, any insight would be great. thanks

Last edited: Oct 4, 2006
2. Oct 4, 2006

### 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.

3. Oct 4, 2006

### 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.

Last edited: Oct 4, 2006
4. Oct 4, 2006

### 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

5. Oct 4, 2006

### 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.

6. Oct 5, 2006

### calvino

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

Last edited: Oct 5, 2006
7. Oct 5, 2006

### AKG

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

8. Oct 5, 2006

### 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))//

9. Oct 5, 2006

### calvino

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?

Last edited: Oct 5, 2006
10. Oct 6, 2006

### HallsofIvy

Staff Emeritus
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.

11. Oct 6, 2006

### matt grime

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.....

Last edited: Oct 6, 2006