Double inversion
Task number: 2440
Show that in each group holds: \((a^{-1})^{-1}=a\).
Resolution
\((a^{-1})^{-1}=(a^{-1})^{-1}\circ e= (a^{-1})^{-1} \circ (a^{-1}\circ a)= ((a^{-1})^{-1} \circ a^{-1}) \circ a = e \circ a = a\)
Show that in each group holds: \((a^{-1})^{-1}=a\).
\((a^{-1})^{-1}=(a^{-1})^{-1}\circ e= (a^{-1})^{-1} \circ (a^{-1}\circ a)= ((a^{-1})^{-1} \circ a^{-1}) \circ a = e \circ a = a\)