# Does group inverse commute with multiplication for all groups? [duplicate]

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.

• No, it is only true for abelian groups. – Michael Burr Jul 21 '15 at 12:14
• Notice the order of operators, you assumed commutativity. – orion Jul 21 '15 at 12:15
• @DietrichBurde - it only answers the question if the question assumes that we are working with abelian groups. – beoliver Jul 21 '15 at 12:45
• The property stated in the link is that a group $G$ is abelian if and only if $(ab)^{-1}=a^{-1}b^{-1}$ holds for all $a,b \in G$ (see the answer of Brian M. Scott, point 1.) and 2.)). – Dietrich Burde Jul 21 '15 at 13:02
• Indeed $a^{-1}b^{-1}=(ab)^{-1}\iff ab=ba$. – Carsten S Jul 21 '15 at 17:39

## 4 Answers

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.

Look at $S_3$, $$(12)^{-1} = (12)$$ and $$(13)^{-1} = (13)$$ but $$((12)(13))^{-1} = (132)^{-1} = (123) \neq (132)$$

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 $b^{-1}a^{-1}=a^{-1}b^{-1}$)

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 $(a \circ b)^{-1}=-k$