Thread: Monoid homomorphisms

Here is the question:

If is a homomorphism of groups then and for all . Show by example that the first conclusion may be false if G, H are monoids that are not groups.

Just in case:

A monoid is a set G with a binary operation and contains a two-sided identity element such that .

I have constructed a couple of monoids with homomorphisms, but I'm not getting anywhere. In addition I have a proof of that does not use the inverses relationship , so I don't see where this would enter the problem. Additionally the proof does not seem to give me any clues as to how might not be equal to .

The proof is short so I will include it. Perhaps someone may be able to see a loophole that is escaping me.

Proof:
Let G be a set with a binary relation * and let there be an element of G, such that for all g in G. Let H be another set with a binary relation and let there be an element such that for all h in H. Further let f be a homeomorphism .

Then
. Since by definition for all h, must be .

-Dan

I had a thought while making my lunch. I'm reconsidering the line in the proof above which states:
and interpreting as because we can compare it with the equation . My thought was that we cannot simply compare the equations and say the inference is correct. So if we can't do the comparison then we can't prove that . (In other words we cannot solve the equation for as the inverse of h need not exist in a monoid.)

But this raises a different problem for me. Take a look at the problem of finding the inverse of an element ab where a and b are in a group G. We can do this: consider the element . I say this is equal to the inverse of the element of ab in G. For . Since we have
and


then by comparison.

So is the comparison a valid method or not?

-Dan

the answer depends on how you define a homomorphism between monoids. some definitions REQUIRE that a monoid homomorphism preserve the identity, and in that case, you can't provide a counter-example.

if, on the other hand, you are only requiring the definition of a homomorphism as given for a group, that is:

f(ab) = f(a)f(b), then it is easy to construct such a map which does not preserve the identity. consider the following two-element monoid N = {e,a}, where a^2 = a, and e is the identity.

for any other monoid M define f:M-->N by f(x) = a, for all x in M. then f preserves products but does not send the identity of M to e.

Thanks, Deveno, that was a nice counterexample.

I came up with a solution to my other quetion in my last post.

If we have
and


Then we know that


Now just multiply both sides by on the right...


And we are left with

-Dan

the humorous way of saying this is: if you want to take off your socks and shoes, take off the shoes first, and then the socks.

Originally Posted by topsquark

Here is the question:
Just in case:
I have constructed a couple of monoids with homomorphisms, but I'm not getting anywhere. In addition I have a proof of that does not use the inverses relationship , so I don't see where this would enter the problem. Additionally the proof does not seem to give me any clues as to how might not be equal to .

The proof is short so I will include it. Perhaps someone may be able to see a loophole that is escaping me.

Proof:
Let G be a set with a binary relation * and let there be an element of G, such that for all g in G. Let H be another set with a binary relation and let there be an element such that for all h in H. Further let f be a homeomorphism .

Then
. Since by definition for all h, must be .

-Dan

Perhaps a more interesting example consider as a monoid with coordinate-wise multiplication, the identity obviously being . Consider then with coordinate-wise multiplication. This is a monoid with identity . This serves to show that a subset of a monoid which itself is a monoid with the same operation need not be a submonoid. Moreover, consider then the embedding this is clearly a homomorphism of semigroups but not a monoid homomorphism since