1. ## Verification?

I'm trying to verify my solution to this problem:

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

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

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

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

*note: $\displaystyle x*x=x^2$ is an example of the notation employed here.

2. ## Re: Verification?

what do you mean by verify?

3. ## Re: Verification?

Well, you can do this:

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

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

$\displaystyle (a^{-4})^{2}$ is.

4. ## Re: Verification?

$\displaystyle x^2 (a^2)^{-1} = e = x^5$

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

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

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

So since inverses are unique $\displaystyle x=(a^4)^{-1}$

5. ## 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).