How equivalence classes relate to subgroups and cosets in a group

Expert intro: This proposition sits right at the meeting point of equivalence relations and group structure. Once the compatibility condition is in place, the identity class starts behaving like a subgroup, and the coset picture appears naturally.

Detailed walkthrough

Let $\sim$ be an equivalence relation on a group $G$ satisfying condition $(t)$. We prove two claims:

1. The equivalence class of $e_G$ is a subgroup $H$ of $G$.
2. For all $a,b\in G$,

$a\sim b \iff a^{-1}b\in H \iff aH=bH.$

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Step 1: Define the identity class

Let

$H=\{x\in G: x\sim e_G\}.$

Since $e_G\sim e_G$, we have $e_G\in H$, so $H\neq\varnothing$.

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Step 2: Show that $H$ is a subgroup

Take $a,b\in H$. Then

$a\sim e_G \quad\text{and}\quad b\sim e_G.$

Because $\sim$ is an equivalence relation, it is symmetric, so from $b\sim e_G$ we get

$e_G\sim b.$

Using condition $(t)$, multiplying on the left by $b^{-1}$ gives

$b^{-1}\sim e_G.$

Now multiply on the left by $a$ again using $(t)$:

$ab^{-1}\sim a.$

Since $a\sim e_G$, transitivity yields

$ab^{-1}\sim e_G,$

so $ab^{-1}\in H$.

Thus $H$ is closed under the subgroup test, and therefore $H$ is a subgroup.

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Step 3: Show that $a\sim b \Rightarrow a^{-1}b\in H$

Assume $a\sim b$. Multiply on the left by $a^{-1}$ and use $(t)$:

$e_G\sim a^{-1}b.$

That means

$a^{-1}b\in H.$

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Step 4: Show that $a^{-1}b\in H \Rightarrow aH=bH$

If $a^{-1}b\in H$, then $b\in aH$. Hence every element of $bH$ lies in $aH$, and similarly the reverse inclusion holds by symmetry. So

$aH=bH.$

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Step 5: Show that $aH=bH \Rightarrow a\sim b$

If $aH=bH$, then $a\in bH$, so for some $h\in H$,

$a=bh.$

Then

$b^{-1}a=h\in H,$

which means

$e_G\sim b^{-1}a.$

Multiply on the left by $b$ and use $(t)$:

$b\sim a.$

By symmetry, $a\sim b$.

Therefore,

$a\sim b \iff a^{-1}b\in H \iff aH=bH.$

This completes the proof.

💡 Pitfall guide

A subtle place to slip is the subgroup check: many people try to prove closure under multiplication first, but the given argument naturally produces $ab^{-1}\in H$, which is enough for the subgroup test.

Another common mistake is forgetting that the relation condition $(t)$ must be applied after multiplying on the left. That step is doing the real work in the proof.

🔄 Real-world variant

If the relation were only a congruence-like condition but not an equivalence relation, then the identity class might fail to be a subgroup. If $H$ were normal, the cosets $aH$ would also behave well under multiplication of classes, which is the next step in building a quotient group.

🔍 Related terms

equivalence relation, subgroup, coset