5

Say $G = \langle x \rangle$ and $H = \langle y \rangle$ are cyclic groups of order $n$. These are isomorphic, and the isomorphism has the form $f(x^i) = y^i$.

My question is as follows. I don't want to define this map only for $0 \leq i < n$ because then if I want to prove that $f(x^i y^j) = f(x^i) f(y^j)$, I'd have to construct $x^{i+j}$, and it may be the case that $i + j > n$.

It's clear to me every element of, say $G$, can be written uniquely as $x^i$ if $0 \leq i < n$, but if I allow $i$ to range over all of $\mathbb{Z}$, it is not unique. Would this require me, for a fully rigorous proof, to prove that the map $f$ is well-defined? That is, would I need to show that if $x^i = x^j \in G$, then $y^i = y^j$?

Shaun
  • 44,997
Cardinality
  • 1,097
  • 3
  • 7
  • 2
    Logically I believe you are correct. I don't think it's that hard to show though, so you should be able to carry your strategy out! – Greg Martin Sep 12 '22 at 05:49
  • 1
    Just wondering, have you also considered the case when $n = \infty$? – stoic-santiago Sep 12 '22 at 05:54
  • 3
    There seems to be an error in your question. Do you mean $$f(x^i x^j) = f(x^i) f(x^j)$$ in place of $$f(x^i y^j) = f(x^i) f(y^j)$$ – stoic-santiago Sep 12 '22 at 06:00
  • 1
    It may be easier and clearer to prove that $G$ and $H$ are both isomorphic to $\mathbb Z_n $ by constructing a surjective homomorphism $\mathbb Z \to G$. – lhf Sep 12 '22 at 11:14

2 Answers2

4

You are correct to say that for a fully rigorous proof we must show that $f$ is well-defined, that is, if $i,i'\in \Bbb Z$ and $x^i=x^{i'}$ then $y^i=y^{i'}.$

For any $a,b\in G$ there are unique $i,j\in\{0,...,n-1\}$ such that $a=x^i$ and $y=x^j.$ So $f(a)f(b)=y^{i+j}.$ If $0\le i+j\le n-1$ then $f(a)f(b)=f(ab).$ If $i+j\ge n$ then $i+j-n \in \{0,...,n-1\}$ so by def'n of $f$ we have $f(a)f(b)=y^{i+j}=y^{i+j-n}=f(x^{i+j-n})=f(x^{i+j})=f(ab).$

BTW. You used $y$ to denote a generator of $G'$ but later used $y$ to denote a member of $G$. This is bad style, as it can lead to confusion.

DanielWainfleet
  • 57,985
1

Let $\theta:G\rightarrow H$ be a homomorphism. i am using left notation here, i.e. $x\theta=\theta(x)$ for convenience. $x^i\theta=y^i$. first we prove it is a homomorphism. $(x^i x^j)\theta=x^{i+j}\theta=y^{i+j}=x^i\theta x^j\theta=y^iy^j$.

2 cases arise:

  1. $i+j<n$, then it is true.
  2. $i+j\geq n$. then : $i+j=qn+r,\; 0\leq r<n$ $x^{i+j}=x^{qn}x^r=e x^r=x^r$ since any integer multiplying $n$ gives the identity $e$.

So $(x^ix^j)\theta=x^{i+j}\theta=x^r\theta=y^r$.

$x^i\theta x^j \theta=y^i y^j=y^{i+j}=y^{qn+r}=y^r$. thus $\theta$ is a well defined homomorphism.

it is easy to see that it is 1-1 and onto.

ryaron
  • 1,091