Definition:
If $S$ is a subset of monoid $[M;∗]$, the submonoid generated by $S$, $⟨S⟩$, is defined by:
- The identity of M belongs to $⟨S⟩$; and $a ∈ S ⇒ a ∈ ⟨S⟩$
- $a, b ∈ ⟨S⟩ ⇒ a ∗ b ∈ ⟨S⟩$
Exercise:
Let $S_1$ = {$1$}, $S_2$ = {$\frac{1}{2}$} and $S_3$ = {$-1$} be subsets of $\Bbb{Q}$. Describe $⟨S_i⟩$ ($i = 1, 2, 3$) when $\Bbb{Q}$ is a monoid with multiplication.
I'm not quite sure how to solve this. For example with the first subset I get that
{$1$, $id_Q$} ∈ $⟨S_1⟩$ ($id_Q$ being the identity of $\Bbb{Q}$)
but that's about as far as I can get. And as applying the definition to the different subsets is definitely not the solution can someone explain me how to proceed or tell me what I'm doing wrong here? Thank you!
Under multiplication, identity element of $ \mathbb{Q} $ is $1$. So, $ 1 \in \langle S_i \rangle $ for any $i$. Furthermore, we have to make sure that the group is closed under the binary operation, multiplication in this case. Consider $ \langle S_1 \rangle $. $1$ is already there and certainly, $ \langle S_1 \rangle = \{ 1 \} $. On the other hand, $ \langle S_3 \rangle$ requires we add $ 1$ to the set. Then, we are done as product of $ 1 $ and $ - 1$ (and no matter how many times or how you operate these two) is always $ \pm 1 $. So, $ \langle S_3 \rangle = \{ \pm 1 \} $. Finally, for $ \langle S_3 \rangle $, you can see that it will have $ 1 $ and $ \frac{1}{2} $ at least. However, $\frac{1}{2}\cdot \frac{1}{2} = \frac{1}{4}$ so we have to make sure it's in the set. By similar argument $ \frac{1}{2^n}$ such that $n \geq 0 $ should be in the set. So, we have $ \langle S_3 \rangle = \{ \frac{1}{2^n} \mid n \geq 0 \} $.