# $f (e_G )=e_H$ is necessary for Monoid homomorphisms

In abstract algebra $-$ hungerford, a question is given as

If $f: G\to H$ is a homomorphism of groups, then $f (e_G)=e_H$ and $f (a^{-1}) ={ f (a)}^{-1}$ for all $a\in G$ . Show by example that the first conclusion may be false if G and H are monoids that are not a group.

Please look it. Monoid homomorphism

Here $f (e_G )=e_H$ is taken in definition of Monoid homomorphism.

That confuses.

Again , for semigroup homomorphism on two groups, when preserving identify be a group homomorphism, Why this fails for Monoid homomorphism ?

Is there have any geometrical significance?

• For group homomorphisms, $f(e)=e$ is something you can prove, just from $f(ab)=f(a)f(b)$. For monoids, that conclusion may fail, and thus the need to explicitly add it as part of the definition of monoid homomorphism. I think that that's what they mean. At any rate, that's one way to interpret it which makes sense. That being said, not everyone define everything the same way, so you shouldn't trust Wikipedia definitions too much when doing book exercises. Apr 23, 2018 at 19:54
• The key point is that $x^2 = x$ doesn't have to have a unique solution in a monoid. The identity has to go to an element that satisfies that equation. Apr 23, 2018 at 20:09
• The confusion is caused by looking at the definition in a different place than where you found the exercise. I just checked and Hungerford does not include that requirement (since he defined homomorphisms in general for semigroups). Apr 24, 2018 at 7:05
• This is a duplicate of math.stackexchange.com/questions/872916 Apr 24, 2018 at 7:32
• @DerekHolt sorry, I don't found it on my search. But I have an extra question about differences between semigroup homomorphism and Monoid homomorphism. Apr 24, 2018 at 8:16

As you noticed groups have this property: every semigroup-homomorphism between them is also a group-homomorphism.

This result can be generalized, indeed the same proof allows to show that every semigroup-homomorphism from a monoid to a group preserves the unit.

If you look at the proof you can notice that what makes this magic possible is the existence of inverse and the fact that the unit is idempotent.

Since semigroup-homomorphisms preserve the property of being idempotent and since in a group there can be only one idempotent, namely the unit, it follows that the image of every idempotent must be sent in the unit of the group. From this it follows in particular the unit-preservation property.

Since in a monoid an idempotent is not necessarily a unit the above argument does not work and indeed it is possible to provide semigroup-homomorphisms between monoid which are not monoid-homomorphism.

If you want I can provide some counter-example, anyway I hope I made clear what is the reason why this magic works for groups.

• Can we generalise group homomorphism from Monoid homomorphism between two groups? Can we say Monoid homomorphisms also preserve the property of being idempotent? Apr 24, 2018 at 1:00
• @SandipAgarwal I don't understand your first question. About the second, as said above every semigroup-homomorphism preserves the property of being idempotent, hence monoid-homomorphisms preserve the property too. Apr 24, 2018 at 6:55
• In first 3 lines of your answer "As you noticed groups have this property: every semigroup-homomorphism between them is also a group-homomorphism." Is it possible for Monoid homomorphism? That is my first question. Apr 24, 2018 at 8:25
• @SandipAgarwal meaning if for monoids every semigroup-homomorphism is also a monoid-morphism? Then the answer is no, as I pointed out in the answer there are counterexamples. Apr 24, 2018 at 10:32
• If instead you were asking if a semigroup-homomorphism between groups is a monoid homomorphism then the answer is clearly yes: because every group-homomorphism is a monoid-homomorphism as well. Apr 24, 2018 at 10:34

The most important application of this phenomenon, in my opinion, is to ring theory. A ring with $1$ is a monoid under multiplication, and ring homomorphisms do not have to send $1$ to $1$ (unless explicitly defined to do so). One example is the following: $$M=(\Bbb Z\times\Bbb Z,\cdot)\\(a,b)\cdot(c,d)=(ac,bd)$$together with the function $$h:M\to M\\(a,b)\mapsto (a,0)$$ This function clearly satisfies the homomorphism property, but it doesn't send $e_M=(1,1)$ to $(1,1)$.

• Can we a have counter example where monoid is cyclic? Dec 23, 2021 at 19:10