pls its prove that questions?
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since G is a group a, b belongs to G implies that ab, (ab) ^-1,a^-1,b^-1 all belongs to G
Now (a.b)(b^-1.a^-1)
=a(b. b^-1)a^-1
=a. e. a^-1,where e is the identity element in G
=a. a^-1
=e
Again (b^-1.a^-1).(a.b)
=b^-1.(a^-1.a).b
=b^-1.e.b
=b^-1.b
=e
Hence by the law of inverse and it's uniqueness it follows that
(ab)^-1=b^-1 a^-1
Now (a.b)(b^-1.a^-1)
=a(b. b^-1)a^-1
=a. e. a^-1,where e is the identity element in G
=a. a^-1
=e
Again (b^-1.a^-1).(a.b)
=b^-1.(a^-1.a).b
=b^-1.e.b
=b^-1.b
=e
Hence by the law of inverse and it's uniqueness it follows that
(ab)^-1=b^-1 a^-1
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