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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!

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    Yes, it still comes as a legitimate proof. You might want to know that for any proof which looks long to you and seems inefficient, there exists an infinity of proofs which are longer and even less efficient. There do exist reasons to prefer shorter proofs, but the provability and the truth of the statement is not one of them.2013-05-23

5 Answers 5

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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 $c = b^{-1} a^{-1}$:

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

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    @Mark,JavaMan When doing group theory for the first time-that's how most students begin$a$serious study of algebra,with groups-you don't have to assume the inverse and/or identity is unique. When I first learned it from Nick Metas, his definition of a group required only that it possess AT LEAST ONE left identity and each element had at least one left inverse.We then proceeded to tediously prove the uniqueness of the left and the right identity(inverse) and then ultimately uniqueness. The advantage of JavaMan's approach is it can work within the weaker group axioms.So can Jon's,in fact.2011-12-27
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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.

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    @Mathemagician1234: What you say is a major flaw is not major and does not exist here. mixedmath's answer *shows* that it is a 2 sided inverse rather than assuming it. And the answer also points out that uniqueness is something that ultimately requires proof, although this proof is omitted here.2011-12-28
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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 $AB$.

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    Welcome to MSE! It really helps readability to format answers using MathJax (see FAQ). Regards2013-05-23
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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}

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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),

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