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I'm really not sure how to prove this. I know that $|\langle e^{2\pi i/n}\rangle| = | e^{2\pi i/n}|$ and that $|\langle e^{2\pi im/n}\rangle| = |e^{2\pi im/n}|$. clearly $| e^{2\pi i/n}| = n$ and I think since that $\gcd(m,n)=1$, then $m/n$ have no common factor so the order will also be $n$ since $n$ will be the smallest integer canceling $n$.and since theyre both cylic groups then, having the same order means they are the same group am I on the good track or am I completely wrong?

This duplicate doesn't answer my question, am I allowed to conclude they are the same group since they have the same order ?

user20194358
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  • Yes, I vote to reopen this question. – MathFail Apr 19 '23 at 23:02
  • Will this help? https://math.stackexchange.com/questions/3253788/prove-that-2-cyclic-subgroups-of-same-order-are-either-equal-or-contain-e-only – bluebril Apr 19 '23 at 23:24
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    @bluebril Yes, but now i just have no idea how to show that the two generators having the same order has to do with them generating the same group. $e^{2\pi i} = e^{2\pi im}$ = 1 = identity of the roots of unity group. so they both start on the X-axis. so $e^{2\pi i/n}$ and $e^{2\pi im/n}$ will both hit the same exact points in the unit circle logically.and they will come back to the intial point in the same time i just have no idea how to show this – user20194358 Apr 19 '23 at 23:36
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    As $(e^{2\pi i/n})^m=e^{2\pi i m/n}$, you have that $\langle e^{2\pi i m/}\rangle \subseteq \langle e^{2\pi i/n}\rangle$. So if they have the same order, then the group they generate have the same order. But as one is a subset of the other, they must be equal (as they are finite). – Severin Schraven Apr 20 '23 at 10:08
  • @SeverinSchraven i don't get it – user20194358 Apr 21 '23 at 01:37

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You are on the right track.

The question is: what is the order of an element $g$ of a group $G$? It's the smallest natural number $a$ satisfying $g^a=1_G$, where $1_G$ is the identity element of $G$.

Next, let's investigate $\langle e^{2\pi i/n} \rangle$.

What is the identity element of this group? Clearly it's $1 = e^0 = e^{k2\pi i} = e^{kn2\pi i/n}=(e^{2\pi i/n})^{kn}$ for $k\in\mathbb{Z}$. Then to minimize $kn$, we choose $k=1$, and thus the order of $e^{2\pi i/n}$ is $n$ (it seems like you got this so far).

What about $\langle e^{2\pi i m/n}\rangle $? Again, in the exponent of $e$, we want an integer multiple of $2\pi i$.

Here we may assume a couple things. For one, if $n=1$, then $\langle e^{2\pi i/n} \rangle = \langle 1 \rangle = \langle e^{2\pi i m/n}\rangle$, and your result trivially follows. So let's assume $n>1$. Also, we may assume that $m<n$, because if $m=n$, then $\gcd(m,n)=n>1$ (a contradiction), and if $m>n$, we have that $m=qn+r$ for some $q\in \mathbb{N}, 0<r<n$ (Division Algorithm), so $e^{2\pi i m/n} = e^{2\pi i (qn + r)/n} = e^{2q\pi i n/n + 2\pi i r/n} = e^{2q\pi i}e^{2\pi i r/n} = 1\cdot e^{2\pi i r/n} = e^{2\pi i r/n}$.

Now, we see that $(e^{2\pi i m/n})^a = e^{2\pi i am/n}$, so in particular, we are looking for the smallest natural number $a$ such that $\frac{am}{n}$ is an integer. In other words, we are looking for the smallest $a$ such that $am$ is a multiple of $n$. But then in this case, $am$ is exactly the definition of $\text{lcm}(m, n)$.

Using the identity $\text{lcm}(m,n)\cdot \gcd(m,n) = mn$, we see that $\gcd(m,n)=1$ implies that $\text{lcm}(m,n) = mn$. Therefore $a=n$, and hence the order of $e^{2\pi i m/n}$ (and hence the order of $\langle e^{2\pi i m/n} \rangle$) is $n$.

Lastly, we see that $e^{2\pi i m/n} = (e^{2\pi i/n})^m \in \langle e^{2\pi i/n}\rangle$, so $\langle e^{2\pi i m/n}\rangle \subseteq \langle e^{2\pi i /n}\rangle$. But since they are both finite and have the same number of elements, it is indeed the case that $\langle e^{2\pi i /n}\rangle = \langle e^{2\pi i m/n}\rangle$.

IAAW
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