Given an equivalence relation $\sim$ on a group $G$, such that $$ a \sim a' \ \text{ and } \ b \sim b' \ \Longrightarrow \ ab \sim a'b' \ , $$ the equivalence class $[e_G]$ of the identity is a normal subgroup of $G$. Moreover, $a \sim b$ if and only if $ab^{-1} \in [e_G]$. Furthermore, this way we can define an equivalence relation $\sim_H$, which "plays nicely" with the group operation, for any normal subgroup $H$.

I am curious whether the converse statement is true: if an equivalence relation on a group $G$ is such that $[e_G]$ is a normal of subgroup of $G$, then $$ a \sim a' \ \text{ and } \ b \sim b' \ \Longrightarrow \ ab \sim a'b' \ ? $$ If not, provide a counterexample. I have spent considerable amount of time thinking about the statement and I think it is false.


2 Answers 2


No, the converse need not hold.

Let $G$ be the Klein $4$-group, $G=\langle x,y\mid x^2=y^2=1, xy=yx\rangle$. Let $\sim$ be the following equivalence relation: $$\sim = \{ (1,1), (x,x), (x,y), (y,x), (y,y), (xy,xy)\}.$$ Then the equivalence class of $1$ is just $\{1\}$, which is of course a normal subgroup of $G$. However, even though $x\sim x$ and $x\sim y$, we do not have $1=xx\sim xy$.

(In fact, any group other than the trivial one and the cyclic group of order $2$ will yield a counterexample: if $G$ contains an element different from its inverse, say $x$, then take the equivalence relation that makes $x\sim x^{-1}$, but all other elements just equivalent to themselves. Then $x\sim x$ and $x\sim x^{-1}$, but $xx\not\sim xx^{-1}$, since $x^2\neq 1$; yet the equivalence class of $1$ is just $\{1\}$. If all nontrivial elements are of order $2$, and there are at least two of them, $x\neq y$, then as above make every element equivalent to itself and $x\sim y$; then $xx\not\sim xy$, even though $x\sim x$ and $x\sim y$.)

What you want is a congruence. An equivalence relation on $G$ will satisfy that $a\sim a'$ and $b\sim b'$ imply $ab\sim a'b'$ if and only if it is a subsemigroup of $G\times G$, viewed as a subset. There is an expansive discussion of this in this answer. See in particular the material between the first and second horizontal lines.

  • $\begingroup$ Thank you so much! Great answer! Indeed, consider cyclic group of order 3, $\{0, 1, 2\}$ and an equivalence relation on it corresponding to the partition $\{ \{0\}, \{1, 2\} \}$. Then $1 \sim 2$, but $1+1 = 2$ is not equivalent to $1+2 = 0$. $\endgroup$
    – Yerbolat
    Dec 9, 2022 at 20:39
  • $\begingroup$ @Yerbolat: Yes, that is the situation in the parenthetical comment, since $2$ is the inverse of $1$. $\endgroup$ Dec 9, 2022 at 20:40

An equivalence relation $∼$ on $G$ satisfying the condition $$ \text{$a ∼ a'$ and $b ∼ b'$} \implies ab ∼ a' b' \qquad \text{for all $a, a', b, b' ∈ G$} $$ is called a congruence relation on $G$. Given a congruence relation $∼$ on $G$, the equivalence class of $e_G$ with respect to $∼$ is a normal subgroup $N$ of $G$, and the equivalence classes with respect to $∼$ are precisely the cosets with respect to $N$.

We can construct counterexamples to your question as follows: Let $G$ be any non-simple group. This means that $G$ admits a normal subgroup $N$ that is neither trivial nor all of $G$. Let $∼$ be the equivalence relation on $G$ corresponding to the partition $$ G = N ∪ \bigcup_{g \notin N} \{ g \} \,. $$ In other words, all elements of $N$ are equivalent with respect to $∼$, and every element outside of $N$ is equivalent to only itself.

The equivalence class of $e_G$ with respect to $∼$ is $N$, which is a normal subgroup of $G$. But no other equivalence class of $∼$ is a coset with respect to $N$, because $N$ is non-trivial. Such a non-coset equivalence class exists because $N$ is a proper subgroup of $G$. The equivalence relation $∼$ is thus not a congruence relation.


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