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!
$\endgroup$ 55 Answers
$\begingroup$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 $c = b^{-1} a^{-1}$:
$$\begin{align} (ab)b^{-1} a^{-1} &= a (b b^{-1}) a^{-1} \\ &= a e a^{-1} \\ &= a a^{-1} \\ &= e. \end{align}$$
$\endgroup$ 2 $\begingroup$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.
$\endgroup$ 2 $\begingroup$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 $AB$.
$\endgroup$ 2 $\begingroup$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}
$\endgroup$ $\begingroup$Consider system of linear equations $Cx =b$.
Let $C = AB$.
$\therefore$ $ABx = b$ .............................................................................................................................(1)
If inverse exists, $x = C^{-1}b$
$\therefore$ $x=(AB)^{-1}b$ ......................................................................................................................(2)
Now, multiply equation (1) by $(A)^{-1}$,
$\therefore$ $A^{-1}ABx$ = $A^{-1}b$
$\therefore$ $Bx$ = $A^{-1}b$ ...........................................................................................................................(3)
Now, multiply equation (3) by $(B)^{-1}$,
$\therefore$ $B^{-1}Bx$ = $B^{-1}A^{-1}b$
$\therefore$ $x = B^{-1}A^{-1}b$ ......................................................................................................................(4)
Comparing (2) & (4),
$(AB)^{-1} = B^{-1}A^{-1}.$
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