Please if someone can help and can take 3 minutes I would be so so unbelievably happy because it is really important to me... Thank you :)
We assume we have a $m$-th root of unity $\zeta_m=e^{\frac{2\pi i}{m}}$.
Also obviously for $i=0,\ldots,m-1$ we have in total $m$ different $m$-th roots of unity in $\mathbb{C}$, namely:
$$\zeta_m^k=e^{\frac{2\pi i k}{m}}, k \in \{0\ldots,m-1\}.$$
Also we obviously have from the geometric sum formula $\zeta_m^{m-1}=e^{\frac{2\pi i (m-1)}{m}}= \sum \limits_{k=0}^{m-2} -\zeta_m^k$ because from the geometric sum formula we get $\sum \limits_{k=0}^{m-1} \zeta_m^k=0$. So for this reason we can transfer any sum like $x_0 \zeta_m^0+ \ldots+ x_{m-1} \zeta_{m}^{m-1}$ with $x_i \in \mathbb{Z}$ into a sum like $y_0 \zeta_m^0+ \ldots+ y_{m-2} \zeta_{m}^{m-2}$ with $y_i \in \mathbb{Z}$.
For this reason we can write any sum like this as a vector of dimension $1\times (m-1)$. Now we say two sums like that are congruent modulo $n$ if all the entries of the vector are congruent modulo $n$;
So now we can finally go to that lemma I don't understand:
The lemma says:
If $m$ and $n$ are natural numbers and $n \nmid m$ and assume $\zeta_m^j \equiv \zeta_m^k \pmod{n}$. Then we already get according to the lemma $\zeta_m^j = \zeta_m^k$.
What I don't understand: What is the sense of the lemma? Of course if I take $m=2$, then I could find with $n=2$ or $n=1$ an example where the elements are congruent but still not the same; Also if I take $m$ larger than $2$ and $n=1$;
But can anyone give me an example where $n \mid m$ and then the saying of the lemma would not be correct? (considering $m>2$ and $n$ is not $1$) Because the lemma seems to be so useless that I am afraid I didn't understand it properly..
Thank you again so much for helping.. :)
Simplest case: Let $m=4$, $n=2$. Then $\zeta_m=i$ and $i\equiv -i=i^3\pmod 2$, but of course $i\ne -i$. More generally, if $m=2k$ and $n=2$ then $\zeta_m^k=-1$ and hence $\zeta_m^k\equiv \zeta_m^0\pmod n$.