2
$\begingroup$

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.

  • 1
    I think I understand this now, thanks for all the help!2012-11-09

4 Answers 4

0

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 $a^{-1}$.

8

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.

  • 0
    @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
3

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 $a = b = x$.

  • 0
    No, @Sami, I don't :( But I'd love hear from him!2014-06-27
1

Hint: $(a\cdot b)^{-1}=b^{-1}\cdot a^{-1}$, and $x^2=x\cdot x$.

  • 0
    @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