Does the inverse distribute over the group operation o?

[aeonstrife]

G is a group with respect to o.

Another operation # is defined by x#y=(xoy)^-1

Show that G is not a group wrt #

I've gotten that the operation is closed but I can't figure out how to prove associativity because the inverse is a bit confusing.

 

[Office_Shredder]

How else can you write (xoy)^-1?

 

[aeonstrife]

o is an operation. (xoy)^-1 is just the inverse of (xoy) if that makes sense

 

[aPhilosopher]

I think what office shredder was asking is "Does the inverse distribute over the group operation o?" In other words, does (xoy)^-1 = x^-1oy^-1 or does it equal something else?