Finite Quotient Groups $\mathbb{Z}/\!\!\sim$

Question

Given an equivalence relation $\sim$ on $\mathbb{Z}$, we can talk about the set $\mathbb{Z}/\!\!\sim\;= \{[x]\ \vert\ x\in\mathbb{Z}\}$ where $[\cdot]$ are the equivalency classes of $\sim$, that is, $[x] = \{y\in\mathbb{Z}\ \vert\ x\sim y\}$. Letting the binary operation $+\colon\ \mathbb{Z}/\!\!\sim\times\ \mathbb{Z}/\!\!\sim\ \to\ \mathbb{Z}/\!\!\sim$ be defined by $[x] + [y] = [x+y]$, we can treat $\mathbb{Z}/\!\!\sim$ as a group. My question is whether or not there exists some $\sim$ such that $\mathbb{Z}/\!\!\sim\;\not\cong \mathbb{Z}/n\mathbb{Z}$ but that $\mathbb{Z}/\!\!\sim$ is finite, cyclic and abelian. Also, can anyone provide examples of $\sim$ such that $\mathbb{Z}/\!\!\sim\;\neq \mathbb{Z}/n\mathbb{Z}$ but that $\mathbb{Z}/\!\!\sim$ is finite, cyclic and abelian if we allow $\mathbb{Z}/\!\!\sim\;\cong \mathbb{Z}/n\mathbb{Z}$?

Answer 1

You're missing a condition on $\sim$ for $\mathbb{Z}/{\sim}$ to support the group structure you'd like: $\sim$ needs to be compatible with $+$ and $-$, in the sense that for all $n,m,p,q$, if $n\sim m$ and $p\sim q$, then $n+p \sim m+q$ and $-n \sim -m$. In the language of universal algebra, we require that $\sim$ be a congruence.

Let $H$ be the $\sim$-equivalence class containing $0$. We make two claims: $n \sim m$ if and only if $n-m \in H$, and $H$ is a subgroup of $\mathbb{Z}$. For the first claim, if $n\sim m$, then $-m \sim -m$ and compatibility of $\sim$ with $+$ implies $n-m \sim m-m = 0$. Conversely, if $n-m \sim 0$, then $m \sim m$ and compatibility of $\sim$ with $+$ implies $n = (n-m)+m \sim 0+m = m$. To prove the second claim, given $n, m \in H$, we have $n \sim 0$ and $m \sim 0$. Compatibility with $+$ implies $n+m \sim 0+0 = 0$, i.e., $n+m \in H$. Compatibility with $-$ implies $-n \sim -0 = 0$, i.e., $-n \in H$.

What this shows is that $\mathbb{Z}/{\sim}$ as a group is the same as the quotient group $\mathbb{Z}/H$. The only nontrivial subgroups of $\mathbb{Z}$ are of the form $n\mathbb{Z}$ for $n>0$, so it follows that $\mathbb{Z}/{\sim}$ is always the group of integers mod $n$ for some $n$.

Answer 2

Any cyclic group is abelian: $g^ng^m = g^{n+m} = g^{m+n} = g^mg^n$. Also, any finite cyclic group $G$ is isomorphic to $\mathbb Z/n\mathbb Z$, where $n=|G|$. To see this, note that $G$ can be written as $$G = \{1,g,\dots,g^{n-1}\}$$ and then the map $G \to \mathbb Z/n\mathbb Z$ given by $g^k \mapsto k+n\mathbb Z$ is a group isomorphism (check this!).

Answer 3

If we require that $+ \colon \mathbb{Z}/{\sim} \times \mathbb{Z}/{\sim} \rightarrow \mathbb{Z}/{\sim} $ is well-defined (that is, it takes the same values no matter which class representatives you pick) you can show that $[0]$ has to be a (normal) subgroup of $\mathbb{Z}$:

• $0 \in [0]$ by definition.
• If $x, y \in [0]$ then $[x+y] = [x] + [y] = [0] + [0] = [0]$ so $x + y \in [0]$.
• If $x \in [0]$ then $[-x] = [0] + [-x] = [x] + [-x] = [0]$ so $-x \in [0]$.

Similarly, you can show that the equivalence classes $[n]$ must be cosets of $[0]$. (Give it a try!)

Therefore, for $\mathbb{Z}/{\sim} $ to be a group with that operation it must be the quotient of $\mathbb{Z}$ by a subgroup.