A monoid is defined as a set with associativity and an identity element. So if I am understanding this correctly the set $S=\{3^k\mid k \in \mathbb N \cup \{0\}\}$ would be a monoid under the operation of multiplication? Also it would not be a group because it there are no inverse elements ?
Is $\{3^k\mid k\in\mathbb N\cup\{0\}\}\subseteq \mathbb N$ a monoid under multiplication? If so, is it a group
abstract-algebragroup-theorymonoid
Best Answer
Yes, it is a monoid: the identity is $1=3^0$, and for all $r,s,t\in\Bbb N\cup\{0\}$, we have
$$\begin{align} 3^r(3^s3^t)&=3^r3^{s+t}\\ &=3^{r+(s+t)}\\ &=3^{(r+s)+t}\\ &=3^{r+s}3^t\\ &=(3^r3^s)3^t. \end{align}$$
Since $\frac13$ is not a non-negative power of three, $3$ has no inverse in $S$, so the monoid in question is not a group.