Let $S$ be the collection of all groups. Define a relation on $S$ by $G \sim H$ iff $G ≈ H$. Prove that this is an equivalence relation. So $S$ is partitioned into isomorphism classes.

Proof: Let $S$ be a relation on S defined by $G \sim H$ iff $G ≈ H$. Show reflexive, symmetric, transitive. It is obvious that $G≈G$ so $G\sim G$. Assume $G\sim H$, thus $G≈H$, which implies that $H≈G$, so $H\sim G$, thus symmetric. Assume $G\sim H$ and $H\sim K$, for $K$ a group in $S$. Thus $G≈H$ and $H≈K$. (I have already shown that if $G≈H$ and $H≈K$, then $G≈K$). From earlier this implies that $G≈K$, so $\sim$ is an equivalence relation. Thus $S$ is partitioned into isomorphism classes. QED.

  • $\begingroup$ Looks good to me, though can't you just say $\approx$ qualifies as an equivalence relation? Using $\sim$ also seems unnecessary, but your preference I guess. $\endgroup$ Commented Oct 12, 2014 at 17:20
  • $\begingroup$ sorry, ≈ means isomorphic and ~ means equivalence class, probably should have made that clear. $\endgroup$ Commented Oct 12, 2014 at 17:37
  • $\begingroup$ I know. An equivalence relation doesn't have to have the symbol $\sim$. You can just directly show that $\approx$ satisfies the conditions of an equivalence relation, thus you can form isomorphism classes with groups. $\endgroup$ Commented Oct 12, 2014 at 17:46
  • $\begingroup$ You've just written out the definitions for $\sim$ to be reflexive, symmetric, and transitive. If those properties are trivial for $\sim$, you don't to write them out. If they aren't trivial, then you need to prove them properly. $\endgroup$
    – anomaly
    Commented Oct 12, 2014 at 18:08

1 Answer 1


The proof doesn't do anything except reformulating. You really have to tell and convince the reader of your proof that your arguments are correct. For example, as for reflexivity, write down an isomorphism $G \to G$. As for symmetry, given an isomorphism $G \to H$, construct an isomorphism $H \to G$. If you have already proven transitivity, that's OK. By the way, there is no need to choose an extra symbol ~ for what is just ≈, just work with ≈ or the classical symbol $\cong$ for isomorphism.


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