$G$ is a cyclic group of order $n$ and there is an integer $m$ that divides $n$.Prove that there is a subgroup of $G$ of order $\frac{n}{m}$.

Question

$G$ is a cyclic group of order $n$ and there is an integer $m$ that divides $n$.
Prove that there is a subgroup of $G$ of order $\frac{n}{m}$.

Can I use Lagrange's theorem to help with this proof?

score 0 · Answer 1 · edited Dec 22 '19 at 19:34

0

Let $x$ be a generator of $G$. $x^m$ generates a subgroup of order $n/m$

edited Dec 22 '19 at 19:34

J. W. Tanner

answered Dec 22 '19 at 19:17

Tsemo Aristide

why does that imply the order is n/m? I get that x is an element that when raised to consecutive powers will create all elements in the group. – Lil Dec 22 '19 at 19:19
$(x^m)^{n/m}=x^n=e$, we deduce that the order$l$ of $x^m$ is inferior or equal to $n/m$, if $l<n/m, (x^m)^l=x^{lm}=e$ implies that the order of $x$ is $lm<n$ contradiction. – Tsemo Aristide Dec 22 '19 at 19:23

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