# Verification?

• Oct 6th 2011, 03:35 PM
VonNemo19
Verification?
I'm trying to verify my solution to this problem:

Suppose $a$ and $x$ are both elements in group $G$. Solve the following equations simultaneously for $x$:

$x^2=a^2$ and $x^5=e$ where $e$ is the identity element of $G$

OK. I find that $x=(a^4)^{-1}$.

I'm having trouble verifying this result. CAN I verify?

*note: $x*x=x^2$ is an example of the notation employed here.
• Oct 6th 2011, 04:02 PM
Deveno
Re: Verification?
what do you mean by verify?
• Oct 6th 2011, 04:44 PM
Ackbeet
Re: Verification?
Well, you can do this:

$(a^{-4})^{5}=a^{-20}=(a^{2})^{-10}\dots,$ and see if you get $e.$

I would also work by attempting to compute the order of $a.$ That might help you verify what

$(a^{-4})^{2}$ is.
• Oct 6th 2011, 04:47 PM
MonroeYoder
Re: Verification?
$x^2 (a^2)^{-1} = e = x^5$

$(a^2)^{-1} = x^{-1}x^{-1} x^5 = x^3 = x^2 x = a^2 x$

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

Note that $xa^4=(a^{-1}a^{-1}(a^2)^{-1})(a^4)=a^{-1}a^{-1}(a^2)^{-1}a^2aa=e$

So since inverses are unique $x=(a^4)^{-1}$
• Oct 6th 2011, 05:48 PM
Deveno
Re: Verification?
the reason for my question, in my earlier post, is that simultaneous equations like this aren't like what you may be used to, where you "plug in the value for x" and see if it works.

we aren't told what a is (or even what the group operation is), so there's not some kind of computational verification you can do.

if you have derived a value of x in terms of a correctly (using the group axioms properly), that's all the verification you need (as in the post above, and i think there was an earlier thread on this same topic).