## Inverse of a composition

Show that in each group holds $$(a\circ b)^{-1}=b^{-1}\circ a^{-1}$$.
Oserve first that $$(b^{-1}\circ a^{-1})\circ(a\circ b)= b^{-1}\circ (a^{-1} \circ a) \circ b= b^{-1}\circ e \circ b= b^{-1}\circ b= e$$.
Hence both $$(a\circ b)^{-1}$$ and $$b^{-1}\circ a^{-1}$$ are inverses of $$a\circ b$$. Therefore they must be equal.