Question

Prove that the inverse of the product of two elements of
a group is the product of the inverses taken in the
reverse order

i.e (ab)'-1 = b'-1 a'-1 ¥a,b€G

​

Answer 1

SOLUTION

TO PROVE

The inverse of the product of two elements of a group is the product of the inverses taken in the reverse order i.e.

$\sf{{(a\circ b)}^{ - 1} = {b}^{ - 1} \circ {a}^{ - 1} }$

PROOF

Here G is the given group

Let e be the identity element

And

$\sf{a, b \in G}$

Therefore

$\sf{(a\circ b) \in \: G \: and \: ( {b}^{ - 1}\circ {a}^{ - 1} ) \in} \: G$

Now

$\sf{(a\circ b)\circ( {b}^{ - 1}\circ {a}^{ - 1} ) \: }$

$= \sf{a\circ(b\circ {b}^{ - 1})\circ {a}^{ - 1} \: } (since \: \circ \: is \: associative)$

$= \sf{a \circ e \circ\: {a}^{ - 1} }(by \: the \: property \: of \: inverse)$

$= \sf{(a\circ e)\circ {a}^{ - 1} \: }$

$= \sf{a \circ{a}^{ - 1} \: }(by \: the \: property \: of \: identity \: element)$

$= \sf{e} \: (by \: the \: property \: of \: inverse)$

$\therefore \: \sf{(a\circ b)\circ( {b}^{ - 1}\circ {a}^{ - 1} ) = e \: } \: \: .....(1)$

Similarly

$\sf{( {b}^{ - 1}\circ {a}^{ - 1} ) \circ \: (a\circ b) = e} \: \: \: .....(2)$

Hence from Equation (1) & Equation (2) we get

$\sf{{(a\circ b)}^{ - 1} = {b}^{ - 1} \circ {a}^{ - 1} }$

Hence proved

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