1

I'm trying to show that there is only one group, $\mathbb{Z}/3$ with addition, up to isomorphism. I could "brute-force" the multiplication table, which I think is in effect what I'm doing, though I'm hoping my approach below is a bit more "elegant."

Let $G = \{e,x,y\}$ be a group of order $3$ where $e$ is the identity element. Then $e^{-1} = e$. Then either $x$ and $y$ are inverses for each other or their own inverses. Suppose, for the sake of contradiction, that $x = x^{-1}$ and $y = y^{-1}$, and consider $xy$. As inverses in a group are unique, we know $xy \neq e$, so $xy = x$ or $xy = y$. If $xy = x$, cancelling $x$ gives $y = e$, which is a contradiction. If $xy = y$, then cancelling $y$ gives $x = e$, another contradiction. Therefore, we must have $x = y^{-1}$. Having determined the inverse of each element of $G$, the entire multiplication table is determined.

I'm not sure, in particular, whether I'm allowed to say that because I've determined the inverses of each element of $G $, I've determined the entire multiplication table. In this particular case, there are so few elements that I could surely fill in the rest of the table, but I don't know if such a strategy works in general. I'm also not completely sure how to write down an explicit isomorphism with $\mathbb{Z}/3$. Surely I would map $e$ to $0$, but I'm not sure if it matters where I send $x$ and $y$, since the remaining two elements of $\mathbb{Z}/3$ have order $3$ and seem to me to be "interchangeable" in this sense.

Shaun
  • 47,747
Brad G.
  • 2,258
  • 1
    No it doesn't matter; any bijection $\varphi$ such that $\varphi(e) = 0$ is an isomorphism. – balddraz May 28 '23 at 16:32
  • 3
    The same argument goes through with $3$ replaced by an arbitrary prime number $p$: there is only one group of order $p$ up to isomorphism. – KCd May 28 '23 at 16:38

1 Answers1

2

They are interchangeable indeed. Map $e$ to zero, map $x$ to $1$ and map $y$ to $2$. Or, map $y$ to $1$ and $x$ to $2$. It doesn't matter and the isomorphism is not unique. Drawing out multiplication tables definitely doesn't work in general! How would you consider isomorphism problems for infinite groups with that strategy?

Pick a non-identity element of $G$ and call it $x$. As the order of an element divides the order of the group, we see $\langle x\rangle=G$. There is a unique homomorphism $\Bbb Z\to G$ induced by $1\mapsto x$ that, by the above, must have image $G$ and kernel $3\Bbb Z$, so $\Bbb Z/3\Bbb Z\cong G$.

FShrike
  • 46,840
  • 3
  • 35
  • 94