I have from Exercise 1 pg 33 - "Algebra - T. W. Hungerford" following: enter image description here

Question to proof: Show by example that the first conclusione may be false if $G,\, H$ are monoids that are not group.

I thinked:

Proof: let be for example two monoids $(G:=\{\emptyset\}, \bot)$ and $(H:=\{\emptyset, \{\emptyset\}\}, \top)$ with: $$\begin{array}{c|c} \bot & \emptyset \\ \hline \emptyset & \emptyset \end{array} \, \, \, \,\, \,\, \,\, \,\, \,\, \,\, \,\, \,\, \, \begin{array}{c|cc} \top & \emptyset & \{\emptyset\}\\ \hline \emptyset & \emptyset & \emptyset \\ \{\emptyset\} & \emptyset & \{\emptyset\} \end{array}$$ $G$ with $e_G:=\emptyset$ like neutral element and $H$ with $e_H:=\{\emptyset\}$ like neutral element. Let be $f$ a function from $G$ to $H$ with $\emptyset \mapsto \emptyset$ (namely $x \mapsto x$), the function is Homomorphismus in fact $$ \begin{align} {\color{Red}f}{\color{Red}(}{\color{Red}x}{ \color{Red}\bot }{\color{Red}y}{\color{Red})}=&f( \emptyset \bot \emptyset)= {\color{Green}f}{\color{Green}(}{\color{Green}\emptyset}{\color{Green})}= \emptyset \\ {\color{Red}f}{\color{Red}(}{\color{Red}x}{\color{Red})} {\color{Red}\top} {\color{Red}f}{\color{Red}(}{\color{Red}y}{\color{Red})}=&f(\emptyset) \top f(\emptyset)= \emptyset \top \emptyset= \emptyset \\ &\text{with }x,y \in G \end{align}$$ but it means that $$ \begin{align} f(e_G)&\neq e_H \\f(e_G)={\color{Green}f}{\color{Green}(}{\color{Green}\emptyset}{\color{Green})}=\emptyset &\neq \{\emptyset\}= e_H \end{align}$$ Therefore $f$ ist Semigroup-Hom but not Monoid-Hom...

Is it correct?


Yes, it's fine. But you risk losing yourself with the symbols.

Consider the group $G=\{1\}$ and the monoid $H=\{0,1\}$; in the former the operation is obvious; in the second it is the usual multiplication. The map $f\colon G\to H$ defined by $f(1)=0$ is a semigroup homomorphism which is not a monoid homomorphism.

The example is exactly the same as yours, but perhaps clearer.

Another classical example is the map $f\colon\mathbb{Z}\to\mathbb{Z}\times\mathbb{Z}$ defined by $f(x)=(0,x)$, which is a group homomorphism when addition is considered, but only a semigroup and not monoid homomorphism with respect to multiplication. In the codomain, operations are performed componentwise.

  • $\begingroup$ Good... thanks soo much! ;) $\endgroup$ – mle Feb 8 '17 at 0:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.