# Group Theory - Proving $(a*b)^{-1} = (a^{-1}) * (b^{-1})$

I'm trying to prove the below equation, where $a,b \in G$ and $(G, *)$ is a group.

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

I'm not really sure how to do it though. I tried doing something like

$(a*b)^{-1} * (a*b) = e = a * a^{-1} * b^{-1} * b$

$(a*b)^{-1} * (a*b) = e = a * (a^{-1} * b^{-1}) * b$

However, I can't get the RHS to $(a * b)* (a^{-1} * b^{-1})$ without assuming communitivity.

How would I go about doing this? Any help is appreciated.

Thanks

Edit: Given that correction, would this be correct?

$(a*b)^{-1} * (a*b) = e = a*(b*b^{-1})*a^{-1}$

Then

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

Thus

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

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This isn't true, unless you actually have commutativity. The statement that holds is $(a*b)^{-1}=b^{-1}a^{-1}$. – Kevin Carlson Aug 21 '12 at 4:31
Thanks, is my edited version correct? – user1520427 Aug 21 '12 at 4:50

Most of the answers above tell you how to verify what $(a \ast b)^{-1}$ is once you have found it. Of course this is all very well and good if you "already know the answer", but gives no insight in how "you find the answer". But this is group theory, and groups are in a sense "designed for solving equations". So let's write an equation and solve it.

An inverse for $a \ast b$ is an element $x \in G$ such that:

$(a \ast b) \ast x = e = x \ast (a \ast b)$. Therefore (using the first equation):

$a^{-1} \ast ((a \ast b) \ast x) = a^{-1} \ast e$

$(a^{-1} \ast a) \ast (b \ast x) = a^{-1}$

$e \ast (b \ast x) = a^{-1}$

$b \ast x = a^{-1}$

$b^{-1} \ast (b \ast x) = b^{-1} \ast a^{-1}$

$(b^{-1} \ast b) \ast x = b^{-1} \ast a^{-1}$

$e \ast x = b^{-1} \ast a^{-1}$

$x = b^{-1} \ast a^{-1}$

Having found what any "right-inverse" of $a \ast b$ must be, and since $G$ is a group, it must be that $x$ is also a left-inverse, although it may now be verified directly that $b^{-1} \ast a^{-1}$ is also a left inverse.

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The problem is that what you are trying to prove is false. The correct statement is $(a * b )^{-1} = b^{-1} * a^{-1}.$

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Thanks for the reply, is my edited post correct? – user1520427 Aug 21 '12 at 4:43
@user1520427 It seems to take a very difficult path. It is easy to check if something is indeed the inverse of another element: Y is the inverse of X if XY=YX=e. – Ragib Zaman Aug 21 '12 at 4:58
I was trying to do the derivation of it, not just check that it's true. I guess my wording was a bit ambiguous. – user1520427 Aug 21 '12 at 7:27

To show that $(a\star b)^{-1} = b^{-1}\star a^{-1}$, all you need to do is multiply $a\star b$ and $b^{-1}\star a^{-1}$ and see that they are inverses of one another:

$$\begin{eqnarray} (a\star b)\star( b^{-1}\star a^{-1}) & = & \\ a\star ((b\star b^{-1})\star a^{-1}) & = & \\ a\star (e\star a^{-1}) & = & \\ a\star a^{-1} & = & e \end{eqnarray}$$

so $b^{-1}\star a^{-1}$ is the inverse of $(a\star b)$.

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Assuming your post should be prove $(a*b)^{-1}=b^{-1}*a^{-1}$ for $a,b \in G$ where $(G,*)$ is a group.

Then by definition of inverses $(a*b)*(a*b)^{-1}=e$. Then using left inverse and associative properties we can get the following:

$(a*b)*(a*b)^{-1}=e \implies a^{-1}*(a*b)*(a*b)^{-1}=a^{-1}*e \implies (a^{-1}*a)*b*(a*b)^{-1}=a^{-1} \implies e*b*(a*b)^{-1}=a^{-1} \implies b*(a*b)^{-1}=a^{-1}$

Now use left inverses and associative properties again with $b$ and you'll get the desired result of $(a*b)^{-1}=b^{-1}*a^{-1}$.

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