I've been reading through a book on introductory group theory/abstract algebra and I've been having trouble understanding how to go about proving a problem that uses this theorem dealing with the order of groups and the law of exponents:
"Theorem 2: Division Algorithm if m and n are integers and n is positive, there exists unique integers q and r such that:
$m = nq + r$
and
$0 \le r \lt n $ "
I have the exercise problem stating that if $\phi$ is a group with identity element $e$, and $x ∈ \phi$; then suppose $|x| = 150$ and that $|x^{15}| = 10$ . Prove this is possible and that it can be proved generally. And i think i have the idea of how it works, but don't think i'm using the correct steps and/or reasoning to my steps
my idea:
$e = x^{150}$ ---------> definition of $|x| = 150$ and that applying the order preserves the footprint
$= (x^{15})^{10}$ -------> rules of exponents
$= e$ --------------> definition of $|x^{15}| = 10$ and that applying the order preserves the footprint
and my thoughts for proving it more generally would be:
$x^{m} = x^{nq}x^{r} = x^{r} ∈ [1,x,…,x_{n−1}]$ with roughly the same steps
$|a| = m$ means i) $a^m = e$ and ii) if $0 < k < m$ then $a^k \ne e$
So if $|x| =150$ then
i) $(x^{15})^{10} = x^{150} =e$ (because $|x|=150$ so $x^{150} = e$).
ii) If $0 < k < 10$ then $0 < 15k < 150$ and $(x^{15})^k = x^{15k} \ne e$ (because $|x|= 150$ so when $15k< 150$ we know $x^{15k}\ne e$).
So by definition $|x^{15}| = 10$.
So if $|x| =150$ is possible, then $|x^{15}| = 10$ is inevitable. And $|x|=150$ is clearly possible (for example: If $\phi = \mathbb Z_{150}$ and $x = 1$....)