# inverse of a product

###### Theorem.

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

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

 $(a_{1}*\cdots*a_{n})^{-1}=a_{n}^{-1}*\cdots*a_{1}^{-1}.$
 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