Is $\{3^k\mid k\in\mathbb N\cup\{0\}\}\subseteq \mathbb N$ a monoid under multiplication? If so, is it a group?

Question

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 ?

Answer 1

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.

Answer 2

You are correct. The set $\{ 3^k ~|~ k \in \mathbb{N} \}$ is a monoid under multiplication. In fact, it is isomorphic to the monoid $\mathbb{N}$ with addition (for me $0 \in \mathbb{N}$). Do you see why?

You are also correct that it is not a group. Is there any $k$ such that $3^k \times 3 = 1$? This shows that $3$ has no inverse.

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I hope this helps ^_^

Answer 3

Yes, it would be assuming that your multiplication is the standard multiplication, that is $$3^j\ast 3^i=3^{i+j},$$

for all $i,j\in \Bbb{N}\cup\{0\},$ where $3^0=1$ would be the identity element.