Correspondence between normal subgroups and congruence relations

A Normal subgroup $𝑁 ◃ 𝐺$ uniquely defines a congruence relation $\equiv$ on $𝐺$ and vice versa, #m/thm/group such that $[𝑒] \equiv = 𝑁$ , and for any $𝑔 \in 𝐺$ the congruence classes correspond to cosets of $𝑁$ :

[𝑔] \equiv = 𝑔 𝑁 = 𝑁 𝑔

Proof

First, we will prove that $[𝑒] \equiv$ is a subgroup. Clearly $𝑒 \in [𝑒] \equiv$ . Let $𝑎, 𝑏 \in [𝑒] \equiv$ , i.e. $𝑎 \equiv 𝑏 \equiv 𝑒$ . Then $𝑎 𝑏 - 1 \equiv 𝑏 𝑏 - 1 = 𝑒$ and therefore $𝑎 𝑏 - 1 \in [𝑒] \equiv$ . Therefore $[𝑒] \equiv$ is a subgroup by One step subgroup test.

Next, we will show that $𝑁 = [𝑒] \equiv$ is a Normal subgroup. Let $𝑔 \in 𝐺$ and $𝑛 \in 𝑁$ . Then $𝑔 𝑛 𝑔 - 1 \equiv 𝑔 𝑒 𝑔 - 1 = 𝑒$ and thus $𝑔 𝑛 𝑔 - 1 \in 𝑁$ . Hence $𝑔 𝑁 𝑔 - 1 = 𝑁$ for any $𝑔 \in 𝐺$ , Therefore $𝑁$ is normal.

Finally we show the equivalence between (left) cosets of $𝑁$ and congruence classes. For any $𝑔, ℎ \in 𝐺$

ℎ \in [𝑔] \equiv ⟺ ℎ \equiv 𝑔 ⟺ 𝑔 - 1 ℎ \equiv 𝑒 ⟺ 𝑔 - 1 ℎ \in 𝑁 ⟺ ℎ \in 𝑔 𝑁

as required.

As a result of this theorem, normal subgroups may be used to form a Quotient group (following the usual notion of Algebraic quotient) where each coset is taken as a group element.

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