# Proving $(x^2)^{-1} = (x^{-1})^2$ in context of group theory.

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! –  Mathlete Nov 9 '12 at 23:26

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.

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

$(2)$ 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? –  Mathlete Nov 9 '12 at 22:48
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. –  amWhy Nov 9 '12 at 22:50
Yep, we've just started group thoery and I think we need to be using the properties of identity, inverse, associativity and closure. –  Mathlete Nov 9 '12 at 22:51
Exactly, so what happens if you left-multiply each side by $x$? Can you see the equivalence between $(1)$ and $(2)$? –  amWhy Nov 9 '12 at 22:54
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^-1 –  Mathlete Nov 9 '12 at 22:59

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. –  Graphth Nov 9 '12 at 23:17
But all you've shown is that it's the identity - what would you do about the expression on the RHS? –  Mathlete Nov 9 '12 at 23:18
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 = e –  Mathlete Nov 9 '12 at 23:20
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. –  Graphth Nov 9 '12 at 23:35
@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. –  Graphth Nov 10 '12 at 17:40

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? –  Mathlete Nov 9 '12 at 22:49
@Manasa: Write $y=x^{-1}$, and take it from the RHS of what you wrote above. It will be clearer. –  Asaf Karagila Nov 9 '12 at 22:52
Somehow that makes things more confusing... –  Mathlete Nov 9 '12 at 23:03
@Manasa: Write $y=x^{-1}$ and you have $x^{-1}\cdot x^{-1}=y\cdot y=y^2=(x^{-1})^2$ as wanted. –  Asaf Karagila Nov 9 '12 at 23:30

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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