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 ?
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1$\begingroup$ Hi: titles of the form "Question on <thing>" are never good. You should be more descriptive. In your case, you can just use the question. $\endgroup$– rschwiebOct 15, 2020 at 20:20
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$\begingroup$ Yes, this is a free monoid with one variable. $\endgroup$– JakobianOct 15, 2020 at 20:20
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3 Answers
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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.
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$\begingroup$ What do you mean by $3$ has no inverse? I did not understand that part. $\endgroup$– PragsOct 15, 2020 at 22:24
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$\begingroup$ I mean, @Prags, that there is no element in $S$ (as I have edited in just not) that is an inverse of $3$ with respect to multiplication; $\frac13$ is a negative power of $3$. $\endgroup$– ShaunOct 15, 2020 at 22:27
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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.
I hope this helps ^_^
answered Oct 15, 2020 at 20:14
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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.
