Let πΊ be an arbitrary group. Consider the following relations on πΊ: π 1:β¦
2019
Let πΊ be an arbitrary group. Consider the following relations on πΊ:
π 1: βπ, π β πΊ, π π 1π if and only if βπ β πΊ such that π = πβ1ππ
π 2: βπ, π β πΊ, π π 2π if and only if π = πβ1
Which of the above is/are equivalence relation/relations?
- A.
π 1 and π 2
- B.
π 1 only
- C.
π 2 only
- D.
Neither π 1 nor π 2
Attempted by 70 students.
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Correct answer: B
Answer: The conjugacy relation (R1) is an equivalence relation; the relation a = b^{-1} (R2) is not.
R1 (a = g^{-1} b g):
Reflexive: take g = e, then a = e^{-1} a e = a.
Symmetric: if a = g^{-1} b g then b = g a g^{-1}, so swapping a and b is also a conjugation (use g^{-1} as the conjugating element).
Transitive: if a = g^{-1} b g and b = h^{-1} c h then a = (hg)^{-1} c (hg), so a is conjugate to c.
R2 (a = b^{-1}):
Not reflexive in general: reflexivity would require a = a^{-1} for every a, which is false in most groups.
Symmetric: if a = b^{-1} then b = a^{-1}, so the relation is symmetric.
Not transitive in general: if a = b^{-1} and b = c^{-1} then a = c, but the relation would require a = c^{-1}, which need not hold. For example in a cyclic group of order 3, 1 = -2 and 2 = -1, so 1 is related to 2 and 2 to 1, but 1 is not related to itself.
Conclusion: Only the conjugacy relation R1 satisfies reflexivity, symmetry and transitivity, so it alone is an equivalence relation.