inverse of a product

Theorem.

If $a$ and $b$ are arbitrary elements of the group  $(G,*)$, then the inverse of $a*b$ is

${(a*b)}^{-1}=b^{-1}*a^{-1}.$(1)

Proof.  Let the neutral element of the group, which may be proved unique, be  $e$.  Using only the group postulates we obtain

$$(a*b)*(b^{-1}*a^{-1})=a*(b*(b^{-1}*a^{-1}))=a*((b*b^{-1})*a^{-1})=a*(e*a^{-1})=a*a^{-1}=e,$$

$$(b^{-1}*a^{-1})*(a*b)=b^{-1}*(a^{-1}*(a*b))=b^{-1}*((a^{-1}*a)*b)=b^{-1}*(e*b)=b^{-1}*b=e,$$

Q.E.D.

Note.  The (1) may be by induction extended to the form

$${(a_1*\mathrm{\cdots }*a_n)}^{-1}=a_n^{-1}*\mathrm{\cdots }*a_1^{-1}.$$

Title	inverse of a product
Canonical name	InverseOfAProduct
Date of creation	2015-01-30 21:19:19
Last modified on	2015-01-30 21:19:19
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	17
Author	pahio (2872)
Entry type	Theorem
Classification	msc 20A05
Classification	msc 20-00
Synonym	inverse of a product in group
Synonym	inverse of product
Related topic	InverseOfCompositionOfFunctions
Related topic	GeneralAssociativity
Related topic	Division
Related topic	InverseNumber
Related topic	OrderOfProducts