Does group inverse commute with multiplication for all groups? Is the following a property of $\textbf{all}$ groups?
$a^{-1} \circ b^{-1} = (a \circ b)^{-1}$
As far as I can tell it is true for addition and multiplication, but in the notes that I have come across, I have not seen it stated as a property. 
 A: Look at $S_3$, 
$$(12)^{-1} = (12)$$ and $$(13)^{-1} = (13)$$ but $$((12)(13))^{-1} = (132)^{-1} = (123) \neq (132)$$
A: No it is not true. You have $(a\circ b)^{-1} = b^{-1}a^{-1}$ because
$$(b^{-1}a^{-1})(ab)=b^{-1}(a^{-1}a)b=b^{-1}b=e$$
Your formula holds in abelian groups where $\circ$ is commutative (where $b^{-1}a^{-1}=a^{-1}b^{-1}$)
A: The inverse of $(a\circ b)$ is an element such that when you multiply by it, you get the identity.  In general, the inverse of $(a\circ b)$ is $(b^{-1}\circ a^{-1})$ since:
$$
(a\circ b)\circ(b^{-1}\circ a^{-1})=a\circ(b\circ b^{-1})\circ a^{-1}=a\circ e\circ a^{-1}=a\circ a^{-1}=e.
$$
Since $(b^{-1}\circ a^{-1})$ has exactly the property we're looking for (and inverses are unique), $(a\circ b)^{-1}=(b^{-1}\circ a^{-1})$.
If you tried 
$$
(a\circ b)\circ(a^{-1}\circ b^{-1}),
$$
you would get stuck because $b\circ a^{-1}$ could be almost anything in the group.  The only time that this can be simplified is if the group is abelian so that the order of the elements can be changed.
A: In particular ,take $a=i$  and $b=j$   in quaternion group
than $a^{-1}=-i$ and $b^{-1}=-j$ implies $a^{-1} \circ b^{-1}=k$ 
which is not equal to $(a \circ b)^{-1}=-k$
