Question Number 190241 by astridmei last updated on 30/Mar/23
$${show}\:{that}\:{a}\circledast{b}={a}+{ab}+{b}\:{is}\:{a}\:{monoid}\:{when}\:{G}={Z} \\ $$
Answered by mehdee42 last updated on 30/Mar/23
$$\ast:\mathbb{Z}×\mathbb{Z}\rightarrow\mathbb{Z}\:\:;\:{a}\ast{b}={a}+{b}+{ab} \\ $$$$\left(\mathbb{Z},\ast\right)\:{is}\:{monoid}\:{whenever} \\ $$$$\left.\mathrm{1}\right)\forall{a},{b}\in\mathbb{Z}\:\rightarrow{a}\ast{b}\in\mathbb{Z} \\ $$$$\left.\mathrm{2}\right)\forall{a},{b},{c}\in\mathbb{Z}\rightarrow\left({a}\ast{b}\right)\ast{c}={a}\ast\left({b}\ast{c}\right) \\ $$$$\left.\mathrm{3}\right)\exists{e}\in\mathbb{Z}\:;\forall{a}\in\mathbb{Z}\:\rightarrow{a}\ast\:{e}={e}\ast{a}={a} \\ $$$$\therefore \\ $$$$\left.\mathrm{1}\right){a}\ast{b}={a}+{b}+{ab}\in\mathbb{Z}\:\:\checkmark \\ $$$$\left.\mathrm{2}\right)\left({a}\ast{b}\right)\ast{c}=\left({a}+{b}+{ab}\right)\ast{c}={a}+{b}+{ab}+{c}+\left({a}+{b}+{ab}\right){c}={a}+{b}+{c}+{ab}+{ac}+{bc}+{abc} \\ $$$${a}\ast\left({b}\ast{c}\right)={a}\ast\left({b}+{c}+{bc}\right)={a}+{b}+{c}+{bc}+{a}\left({b}+{c}+{bc}\right)={a}+{b}+{c}+{ab}+{ac}+{bc}+{abc}\:\:\:\:\:\:\:\:\Rightarrow\left(\mathrm{2}\right)\checkmark \\ $$$$\left.\mathrm{3}\right)\:\forall\:{a}\in\mathbb{Z}\:\rightarrow{a}\ast\mathrm{0}={o}\ast{a}={a}\:\checkmark \\ $$$$ \\ $$