# Does an equivalence relation on a group play well with the group operation, provided that the equivalence class of the identity is a normal subgroup?

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.

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.

• 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 $1+2 = 0$. Dec 9, 2022 at 20:39
• @Yerbolat: Yes, that is the situation in the parenthetical comment, since $2$ is the inverse of $1$. 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.