GATE Papers >> CSE >> 2019 >> Question No 20

Question No. 20 CSE | GATE 2019

Let $G$ be an arbitrary group. Consider the following relations on $G$:

$ R_1:\forall a,\;b\in G,\;a\;R_1b $ if and only if $ \exists g\in G $ such that $ a=g^{-1}bg $

$ R_2:\forall a,\;b\in G,\;a\;R_2b $ if and only if $ a=b^{-1} $ 

Which of the above is/are equivalence relation/relations?

Answer : (B) $ R_1\;\mathrm{only} $

Solution of Question No 20 of GATE 2019 CSE Paper

$ \mathrm R1:\forall\mathrm a,\;\mathrm b\in\mathrm G,\mathrm a\;\mathrm R^1\;\mathrm b $ if and if $ \exists\mathrm g\in\mathrm G\;\mathrm{such}\;\mathrm{that}\;\mathrm a=\mathrm g^{-1}\mathrm{bg} $

Reflexive: a = g-1 ag can be satisfied by putting g = e, identify "e" always exists in a group. So reflexive

Symmetric: aRb ⇒ a = g-1 bg for some g

⇒ b = gag-1 = (g-1)-1 ag-1

g-1 Always exists for every $ \mathrm g\in\mathrm G. $

So symmetric

Transitive: aRb and bRc ⇒ a = g1-1 bg1 and b = g2-1 cg2 for same g1g2 $\in$ G.

Now a $ =\mathrm g_1^{-1}\mathrm g_2^{-1}{\mathrm{cg}}_2{\mathrm g}_1=\left({\mathrm g}_2{\mathrm g}_1\right)^{-1}{\mathrm{cg}}_2{\mathrm g}_1 $

$ {\mathrm g}_1\in\mathrm G\;\mathrm{and}\;{\mathrm g}_2\in\mathrm G\Rightarrow{\mathrm g}_2{\mathrm g}_1\in\mathrm G $  since group is closed so aRb and aRb ⇒ aRc

Hence transitive

Clearly R1 is equivalence relation.

R2 is not equivalence it need not even be reflexive, aR2a ⇒ a = a-1 $ \forall\mathrm a $  which not be true in a group.

R1 is equivalence relation is the correct answer.

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