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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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Looks ok to me. Ideally I would like to see the implication arrows between the various equations as well as descriptions of exactly what you did to the previous equation to get the next, but I'm a bit old-fashioned in this respect. If this style of presentations is fine with your teacher, then you have nothing to worry. –  Jyrki Lahtonen Jul 21 '11 at 21:21
    
It seems likely that this is a group theory question - it would be good to have this stated. –  Mark Bennet Jul 21 '11 at 21:28
    
Ah - I implicitly assumed that it was a group as well. –  mixedmath Jul 21 '11 at 21:37
    
@Mark: Thanks for the advice. The post has been edited which makes it clear that this question is a group theory question. –  Student Jul 21 '11 at 22:38
    
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. –  Doug Spoonwood May 23 '13 at 14:30

4 Answers 4

up vote 13 down vote accepted

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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1  
If there is a question about $ab$ being invertible e.g. are we talking about a group here (as seems likely), there is also one about the uniqueness of the inverse. If you have a multiplication table where the product of any pair of elements is the identity, for example, the inverse would not be unique. Also $ab$ may be invertible without either $a$ or $b$ being invertible. If it's a group we don't have these problems. –  Mark Bennet Jul 21 '11 at 21:27
    
@Mark: Good Point. –  JavaMan Jul 21 '11 at 21:34
    
@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. –  Mathemagician1234 Dec 27 '11 at 7:00

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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Note in my response to JavaMan's proof above-your proof has the major flaw in that it assumes the inverse is 2 sided and unique and the definition of a group doesn't require that. But if the definition of a group uses the stronger axioms(2 sidedness and uniqueness are assumed), then it'll work fine.In my experiences,though,the proof method given by Jon above is a bit more straightforward and will probably be easier for an absolute beginner to do on their own. –  Mathemagician1234 Dec 27 '11 at 6:59
    
@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. –  Jonas Meyer Dec 28 '11 at 3:55

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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OK........................! –  JOBBINE JOSEPH May 23 '13 at 12:20
    
Welcome to MSE! It really helps readability to format answers using MathJax (see FAQ). Regards –  Amzoti May 23 '13 at 12:26

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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