Let (H,∗) be a subgroup of (G,∗) and if R be the relation defined on G by aRb iff ab−1∈H.
Show that R is an equivalence relation on G.
Since a∗a−1=e∈H for any a∈G, R is reflexive relation.
If aRb, then a∗b−1∈H. Since H is a subgroup, b∗a−1=(a∗b−1)−1∈H. Therefore, bRa and R is symmetric relation.
If aRb and bRc , then a∗b−1∈H and b∗c−1∈H. Since H is a subgroup, a∗c−1=(a∗b−1)∗(b∗c−1)∈H . Therefore, aRc and R is transitive relation.
Consequently, R is an equivalence relation.
Taking into account that b∗a−1∈H if and only if there exist h∈H such that b∗a−1=h, i.e. b=h∗a∈Ha, we conclude that bRa iff b is an element of H∗a={h∗a∣h∈H}. Therefore, [a]R={b∈G∣bRa}=H∗a={h∗a∣h∈H}.
Consider subgroup H=⟨(13),(14)⟩={(1),(13),(14),(34),(134),(143)} of the symmetric group S4. Since ∣S4∣=24 , there exist 624=4 different equivalence classes S4 under R. Let us find them:
[(1)]R=H(1)=H={(1),(13),(14),(34),(134),(143)},
[(12)]R=H(12)={(1),(13),(14),(34),(134),(143)}(12)={(12),(123),(124),(12)(34),(1234),(1243)},
[(23)]R=H(23)={(1),(13),(14),(34),(134),(143)}(23)={(23),(132),(14)(23),(243),(1324),(1432)},
[(24)]R=H(24)={(1),(13),(14),(34),(134),(143)}(24)={(24),(13)(24),(142),(234),(1342),(1423)}.
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