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How would I go about proving $(x^2)^{-1} = (x^{-1})^2$ in terms of group theory?

Would I start by multiplying both sides by the inverses of the LHS and RHS and using the property of the identity? Or should I use induction?

Thanks.

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    I think I understand this now, thanks for all the help!2012-11-09

4 Answers 4

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Use the rule that $(ab)^{-1} = b^{-1} a^{-1}$.

To prove this rule, start with $(ab)^{-1} ab = e$ and then successively right-multiply by $b^{-1}$ and then $a^{-1}$.

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By $(x^2)^{-1}$ you mean the inverse of $x^2$. So, you can just show that $(x^2)\cdot (x^{-1})^2$ is the identity.

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    This is all there is to it. Very straightforward answer.2012-11-09
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    But all you've shown is that it's the identity - what would you do about the expression on the RHS?2012-11-09
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    Would it be (x^2) * (x-1)^2 = (x^2) (x^-1 * x^-1) Then e = (x*x)(x^1 * x^1) e = (x*x^-1) * (x*x^-1) = e*e = e2012-11-09
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    If $ab = 1$, then $b$ must be $a^{-1}$. Thus, if $xxx^{-1}x^{-1} = 1$, which it obviously does (I'm not saying that demeaningly, but all you have to do is cancel $xx^{-1}$ twice), then $(x^{-1})^2$ must be the inverse of $x^2$. So, this answer is extremely simple and accomplishes the deed.2012-11-09
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    @amWhy I said it was obvious and then explained why. I didn't just say it was obvious and leave the OP to figure it out. And, the reason I clarified is because you can't hear emotion from typed words so I wanted to state exactly what I meant, which I did. And, part of my comments are not directed toward the OP but toward the general audience to express that this is the best solution to the problem. All other answers amount to using $(ab)^{-1} = b^{-1} a^{-1}$, which is perfectly fine, but it's a lot simpler to just do $xxx^{-1}x^{-1} = 1$. It's a one sentence proof and very clear.2012-11-10
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    @Graphth One of the reasons I used the $(ab)^{-1} = b^{-1} a^{-1}$ was because, while conversing with the OP, s/he mentioned having already proved that. And yes, I upvoted this answer (Joe Johnson's) because it is very neat and clean, and doesn't require anything more than the definitions of group identity and inverse together with exponentiation of a group element.2012-11-10
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Note: $$(x^2)^{-1} = (x\cdot x)^{-1} = (x^{-1}\cdot x^{-1}) = (x^{-1})^2\tag{*}$$

So we have

$$(x^2)^{-1} = (x^{-1})^2.$$

as desired.


Added, for clarification, re comments:

$$(x^2)^{-1} = (x^{-1})^2$$ $$\iff$$ $$(x\cdot x)^{-1} = (x^{-1}\cdot x^{-1})\tag{1}$$ $$\iff$$ $$x^{-1}\cdot x^{-1} = x^{-1}\cdot x^{-1}\tag{2}$$

$(2)$ can certainly be reduced to look "prettier" by, say, left-multiplying by $x$, or by group cancellation laws, to arrive at the obvious equivalence: $e = e$, where $e$ is the identity.

But equation $(*)$ at the top is more straightforward and direct for establishing equality.
Note that $(*)$ uses the fact that $(a\cdot b)^{-1} = (b^{-1}\cdot a^{-1})$, which you've already proven, according to your comment below. In this case $a = b = x$.

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    Multiply the LHS by the inverse i.e. x^2?2012-11-09
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    I'm guessing that the task here is to use properties/definitions you have learned about the inverse of an element, the identity, and the use of exponents when exponentiating a group element.2012-11-09
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    Yep, we've just started group thoery and I think we need to be using the properties of identity, inverse, associativity and closure.2012-11-09
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    Exactly, so what happens if you left-multiply each side by $x$? Can you see the equivalence between $(1)$ and $(2)$?2012-11-09
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    Oh right, if I do that twice then I'd be using the properties of identity and associativity to show that (2) is true - it's essentially like the proof for (ab)^-1=a^-1*b^-12012-11-09
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    Yes, but note that $(ab)^{-1} = b^{-1}a^{-1}$. Here, since we are multiplying x by itself in $(x\cdot x)^{-1}$, no need to worry about order. You need only show equality of the LHS and RHS, so at (2) you are done. If you left multiply each side by x twice, you'll arrive at the equality $e = e$ where $e$ is the identity.2012-11-09
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    Of course, that makes sense. Incidentally, how would you prove it using the group cancellation laws?2012-11-09
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    by group cancellation laws, you can cancel the left-most factor of the LHS and the left-most factor of the RHS.2012-11-09
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    So that would leave (x^1)^-1=(x^-1)^1?2012-11-09
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    Recall, you can left-multiply each side by $x$, and use the definition of an inverse to get $e = e$, right? Recall that each for any group element $a, \; a^1 = a$2012-11-09
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    So $(x^1)^{-1} = x^{-1}$ and $(x^{-1})^1 = x^{-1}$.2012-11-09
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    Yes I think I see what you mean.2012-11-09
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    I can get that result by multiplying each side by x^2 though, in the form of (x*x) and using the property of associativity.2012-11-09
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    Exactly..... :-)2012-11-09
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    @amWhy: I love manipulating the elements in groups structures. :+)2013-08-10
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    No, @Sami, I don't :( But I'd love hear from him!2014-06-27
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Hint: $(a\cdot b)^{-1}=b^{-1}\cdot a^{-1}$, and $x^2=x\cdot x$.

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    So (x^2)^(-1)=(x.x)^(-1)=(x^-1)*(x^-1) Would I then be allowed to assume that the the sum of the indices is 2? Or would I have to do something else before that?2012-11-09
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    @Manasa: Write $y=x^{-1}$, and take it from the RHS of what you wrote above. It will be clearer.2012-11-09
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    Somehow that makes things more confusing...2012-11-09
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    @Manasa: Write $y=x^{-1}$ and you have $x^{-1}\cdot x^{-1}=y\cdot y=y^2=(x^{-1})^2$ as wanted.2012-11-09