Check my proof that $(ab)^{-1} = b^{-1} a^{-1}$

The following question is problem Pinter's Abstract Algebra. And to put things in context: $G$ is a group and $a, b$ are elements of $G$.

I want to show $(ab)^{-1}$ = $b^{-1}a^{-1}$.

I originally thought of proving the fact in the following manner: \begin{align*} (ab)^{-1}(ab) &= e \newline (ab)^{-1}(ab)(b^{-1}) &= (e)(b^{-1}) \newline (ab)^{-1}(a)(bb^{-1}) &= (b^{-1}) \newline (ab)^{-1}(a)(e) &= (b^{-1}) \newline (ab)^{-1}(a) &= (b^{-1}) \newline (ab)^{-1}(a)(a^{-1}) &= (b^{-1})(a^{-1}) \newline (ab)^{-1}(e) &= (b^{-1})(a^{-1}) \newline (ab)^{-1} &= (b^{-1})(a^{-1}) \newline \end{align*}

I know this may seem extremely inefficient to most, and I know there is a shorter way. But would this be considered a legitimate proof?

Thanks in advance!

Solution 1:

Your way is absolutely fine. As you note, there is in fact an easier way. It would be enough to show that the element $c$ such that $(ab)c = e$ is in fact $c = b^{-1} a^{-1}$:

$$\begin{align} (ab)b^{-1} a^{-1} &= a (b b^{-1}) a^{-1} \\ &= a e a^{-1} \\ &= a a^{-1} \\ &= e. \end{align}$$

Solution 2:

These questions are standardly done by going straightforward, definition-based. So for the element $ab$ we seek the element $x$ s.t. $abx = xab = e$. Sure. We know such an element is unique (if not - prove this too).

So let's just do it. $ab b^{-1} a^{-1} = a (b b ^{-1}) a^{-1} = a e a^{-1} = a a^{-1} = e$. That's one direction.

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

As for your method above - it looks great. Well done.

Solution 3:

The definition of inverse is a*a-1 = I (ie a operated with inverse should give identity element) a-1*a = I (ie a inverse operated with a should also give identity element)

so here

$(AB) B^{-1} A^{-1} = A(BB^{-1})A^{-1} = AIA^{-1} = AA^{-1} = I$

$B^{-1} A^{-1} (AB) = B^{-1}(A^{-1}A)B = B^{-1} I B = B^{-1}B = I$

Here when $B^{-1}A^{-1}$ Operated to AB on both sides and in both case it given I (Identity Matrix ).

$X Z = I$ means that $Z$ is the inverse of $X$ Similarly If $ABB^{-1}A^{-1}$ is giving identity element then $B^{-1} A^{-1}$ is the inverse of $AB$.

Solution 4:

For a direct proof systematically using the associative property and the fact that $x = xe = xyy^{-1}$ for all $y\in G$ proceed as follows. \begin{align} (ab)^{-1} = & (ab)^{-1}e \\ = & (ab)^{-1}[aa^{-1}] \\ = & (ab)^{-1}\big[(ae)a^{-1}\big] \\ = & (ab)^{-1}\Big[\big(a[bb^{-1}]\big)a^{-1}\Big] \\ = & (ab)^{-1}\Big[([ab]b^{-1})a^{-1}\Big] \\ = & (ab)^{-1}\Big[[ab]\big(b^{-1}a^{-1}\big)\Big] \\ = & \Big[(ab)^{-1}[ab]\Big]\big(b^{-1}a^{-1}\big) \\ = & e\big(b^{-1}a^{-1}\big) \\ = & b^{-1}a^{-1} \end{align}