# Let $M$ be a monoid and let $M^*$ be the group of invertible elements of $M$. Prove the following…

This is a multipart question. I will post both parts here. I've only attempted the first part. I think if I can get help on the first part, I will be able to figure out the second. I will restate the question below and show my attempt at the first part below that.

Let $M$ be a monoid and let $M^*$ be the group of invertible elements of $M$. Prove the following:

For each $m \in M$ there is exactly one monoid homomorphism $f:\mathbb N \rightarrow M$ such that $f(1)=m$.

For each $m\in M^*$ there is exactly one monoid homomorphism $f:\mathbb Z \rightarrow M$ such that $f(1)=m$.

I copy and pasted the question. In the second part, my professor wrote "there is exactly one monoid homomorphism $f:\mathbb Z \rightarrow M$". Does he mean "there is exactly one monoid homomorphism $f:\mathbb Z \rightarrow M^*$"?

Anyway here is my attempt at the first part:

Let $M$ be a monoid. Next suppose that for each $m \in M$, there are more than one monoid homomorphisms $f: \mathbb N \rightarrow M$ and $g: \mathbb N \rightarrow M$ such that $f(1)=m$ and $g(1)=m$. Then $f(1)=m=g(1)$ and $f(1)=g(1)$. Therefore there is only one monoid homomorphism such that $f(1)=m$.

As always, thank you all for any help.

[EDIT: In the next section (not posted here), there are similar questions related to rings. Again, I think if I can complete these proofs, I should be able to figure out the questions related to rings.]

• "In the second part, my professor wrote 'there is exactly one monoid homomorphism $f:\mathbb{Z} \to M$'. Does he mean 'there is exactly one monoid homomorphism $f:\mathbb{Z}\to M^*$'?" Both are true. Since $\mathbb{Z}$ is a group, its image under any homomorphism must be a group and hence must be contained in $M^*$. – Trevor Gunn Feb 5 '18 at 0:29
• Before I forget. You have shown that the homomorphism is unique but not that it exists. You've proved most of the existence: you know that such a function must be defined by $f(n) = nm$, so you just need to explain why this is a homomorphism. – Trevor Gunn Feb 5 '18 at 1:42
• Thanks @TrevorGunn. I think you mean that I just to to spell out how this fits the definition of a homomorphism, right? Also, can you explain to me how the second part (mapping integers to the inverse elements of $M$) is any different from the first? Do I need to add the fact that the negative part of the integers can account for the inverse elements that are $M^*$. So maybe I need to add something like: from $0_N = 1 - 1$ and $f(0_N)=0_M$, we get $0_M=f(0_N)=f(1-1)=f(1)+f(-1)=f(1)-f(1)$. [EDIT: formatting] – dnem41 Feb 5 '18 at 2:33
• Yeah. You can use the fact that $f(-n) = f(n)^{-1}$ (or $-f(n)$ if you're using additive notation) to extend to the negative integers. – Trevor Gunn Feb 5 '18 at 4:34

## 1 Answer

This is incorrect. You've shown that $f(1) = g(1)$ but you haven't shown that $f = g$. That is, you haven't shown that $f(n) = g(n)$ for all $n \in \mathbb{N}$. Notice that your proof works just the same if you replace $\mathbb{N}$ with any monoid $N$ and $1$ with any element $n \in N$:

Let $M$ be a monoid. Next suppose that for each $m \in M$, there are more than one monoid homomorphisms $f: N \rightarrow M$ and $g: N \rightarrow M$ such that $f(n)=m$ and $g(n)=m$. Then $f(n)=m=g(n)$ and $f(n)=g(n)$. Therefore there is only one monoid homomorphism such that $f(n)=m$.

But it would be too good to be true if there was a unique monoid homomorphism $f : N \to M$ with $f(n) = m$.

Instead, you need to use some special property of $\mathbb{N}$ and $1$ that isn't shared between all monoids. That property is the so called "principal of induction" which says that every element of $\mathbb{N}$ can be written as a sum of $1$'s:

$$1 + 1 + \dots + 1.$$

I'll leave it to you to figure out what this has to do with the principal of induction as it is usually stated and to formally prove (by induction) that if $f(1) = g(1) = m$ then $f(n) = g(n)$ for all $n \in \mathbb{N}$.

• Thanks for pointing out my mistake. Can you tell me if this is sufficient? – dnem41 Feb 5 '18 at 1:07
• Let $M$ be a monoid. Next suppose that for each $m \in M$, there are more than one monoid homomorphisms $f: \mathbb N \rightarrow M$ and $g: \mathbb N \rightarrow M$ such that $f(1)=m$ and $g(1)=m$. Then $f(1)=m=g(1)$ and $f(1)=g(1)$. Because $f$ and $g$ are monoid homomorphisms, we can write $f(1+1)=f(1)+f(1) = m+m=2m = m + m = g(1) + g(1) = g(1+1)$. Suppose $f(k)=km$. Then $f(k+1)=f(k)+f(1)=km+m=(k+1)m$. Next suppose $g(k)=km$. Then $g(k+1)=g(k)+g(1)=km+m=(k+1)m$. By induction, $f(n)=g(n)$. Therefore there is only one monoid homomorphism such that $f(1)=m$. – dnem41 Feb 5 '18 at 1:10
• @dan Yes, that takes care of everything except $0$. – Trevor Gunn Feb 5 '18 at 1:14
• Oi. Ok, so I would just add the fact that in a homomorphism, the identity in $\mathbb N$ maps to the identity in $M$. So $f(0)=e_M$. Maybe instead of $f(1+1)=f(1)+f(1)=2m$, I would say $f(1+0)=f(1)+f(0)=1+e_m=1$ and repeat for $g$. That seems good? – dnem41 Feb 5 '18 at 1:17
• @dan $f(0) = e_M = g(0)$ since homomorphisms preserve identites. $f(1) = g(1) = m$ by assumption (that's the base case). You don't need to show that $f(2) = g(2)$, just the inductive step. – Trevor Gunn Feb 5 '18 at 1:21