1. ## Monoid homomorphisms

Here is the question:
If $f: G \to H$ is a homomorphism of groups then $f(e_G) = e_H$ and $f \left (a^{-1} \right ) = f(a)^{-1}$ for all $a \in G$. 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 $e \in G$ such that $ea = ae = a ~\forall a \in G$.
I have constructed a couple of monoids with homomorphisms, but I'm not getting anywhere. In addition I have a proof of $f(e_G) = e_H$ that does not use the inverses relationship $f \left ( a^{-1} \right ) = f(a)^{-1}$, 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 $f(e_G)$ might not be equal to $e_H$.

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, $e_G$ such that $e_G*g = g*e_G = g$ for all g in G. Let H be another set with a binary relation $\times$ and let there be an element $e_H$ such that $e_H \times h = h \times e_H = h$ for all h in H. Further let f be a homeomorphism $f: G \to H$.

Then
$f(a) \times f(e_G) = f(a*e_G) = f(a)$. Since by definition $h \times e_H = h$ for all h, $f(e_G)$ must be $e_H$.

-Dan

2. I had a thought while making my lunch. I'm reconsidering the line in the proof above which states:
$f(a) \times f(e_G) = f(a)$ and interpreting $f(e_G)$ as $e_H$ because we can compare it with the equation $h \times e_H = h$. 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 $f(e_G) = e_H$. (In other words we cannot solve the equation $h \times f(e_G) = h$ for $f(e_G)$ 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 $b^{-1}a^{-1} \in G$. I say this is equal to the inverse of the element of ab in G. For $(ab)(b^{-1}a^{-1}) = a (b b^{-1}) a^{-1} = aea^{-1} = aa^{-1} = e$. Since we have
$(ab)(ab)^{-1} = e$ and
$(ab)(b^{-1}a^{-1}) = e$

then $(ab)^{-1} = b^{-1}a^{-1}$ by comparison.

So is the comparison a valid method or not?

-Dan

3. 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.

4. Thanks, Deveno, that was a nice counterexample.

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

If we have
$(ab)(ab)^{-1} = e$ and
$(ab)(b^{-1}a^{-1}) = e$

Then we know that
$(ab)(ab)^{-1} = e = (ab)(b^{-1}a^{-1})$

Now just multiply both sides by $b^{-1}a^{-1}$ on the right...
$b^{-1}a^{-1}(ab)(ab)^{-1} = b^{-1}a^{-1}(ab)(b^{-1}a^{-1})$

And we are left with $(ab)^{-1} = b^{-1}a^{-1}$

-Dan

5. 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.

6. 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 $f(e_G) = e_H$ that does not use the inverses relationship $f \left ( a^{-1} \right ) = f(a)^{-1}$, 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 $f(e_G)$ might not be equal to $e_H$.

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, $e_G$ such that $e_G*g = g*e_G = g$ for all g in G. Let H be another set with a binary relation $\times$ and let there be an element $e_H$ such that $e_H \times h = h \times e_H = h$ for all h in H. Further let f be a homeomorphism $f: G \to H$.

Then
$f(a) \times f(e_G) = f(a*e_G) = f(a)$. Since by definition $h \times e_H = h$ for all h, $f(e_G)$ must be $e_H$.

-Dan
Perhaps a more interesting example consider $\mathbb{Z}\times\mathbb{Z}$ as a monoid with coordinate-wise multiplication, the identity obviously being $(1,1)$. Consider then $\mathbb{Z}\times\{0\}$ with coordinate-wise multiplication. This is a monoid with identity $(1,0)$. 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 $\iota:\mathbb{Z}\times\{0\}\hookrightarrow\mathbb{ Z}\times\mathbb{Z}$ this is clearly a homomorphism of semigroups but not a monoid homomorphism since $(1,0)\mapsto (1,0)$

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### Example of monoid homorphisM in descrete

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