Show ab is also a unit

Question

Given A, a ring with 1, show that if $a,b \in A$ are units, then $ab$ is also a unit. I proceeded like this: $aa^{-1}=1$ and $bb^{-1}=1$ then $aa^{-1}=b^{-1}b \rightarrow aa^{-1}b^{-1}=b^{-1} \rightarrow baa^{-1}b^{-1}=1$.

My problem with this one is that I'm not allowed to use commutativity to reorder the term $ab$ and, is it possible to assume $a^{-1}b^{-1}=(ab)^{-1}$? Can you provide me with hints?

Answer 1

You need to find an $x$ so that $abx= 1$

So that means $a(bx) = 1$ so what does $bx = ???$.

Can you go on?

$ab = ab$

$ab*b^{-1} = a*1 = a$

$ab*b^{-1}a^{-1} = a*a^{-1} = 1$

So $b^{-1}a^{-1} = (ab)^{-1}$.

Answer 2

Hint: It's the same identity for $(AB)^{-1}$ that works for square matrices.