Let $(G,\cdot)$ be a group, $n\in \mathbb{N}, n\geq 2$ and $x_1,x_2,...,x_n\in G$ where any two elements commute. If the orders of these elements are finite and any two are relatively prime, then does this relation hold: $$ord(x_1^{k_1}x_2^{k_2}...x_n^{k_n})=\frac{ord(x_1)ord(x_2)...ord(x_n)}{(k_1,ord(x_1))(k_2,ord(x_2))...(k_n,ord(x_n))}?$$

Lemma: Let $(G,\cdot)$ be a group and $x\in G$, with $ord(x)=n\in\mathbb{N^{*}}$. Then, $\forall k\in \mathbb{Z}, ord(x^k)|n$. Moreover, $$\forall k\in \mathbb{Z}, ord(x^k)=\frac{n}{(k,n)}.$$ The proof goes as follows: for $k\in\mathbb{Z}$ we have $(x^k)^n=e$, where $e$ is the neutral element of the group, thus $ord(x^k)|n$. Let $d=(k,n)$ ($ (k,n)$ represents the greatest common divisor of $k$ and $n$) and $n_1,k_1\in\mathbb{Z}$ such that $(n_1,k_1)=1$ and $n=dn_1, k=dk_1$. Because $(x^k)^{n_1}=x^{nk_1}=e$, we have $ord(x^k)|n_1$. We have $x^{ord(x^k)k}=e$ and because $ord(x)=n$, we obtain $n|ord(x^k)k$, so $dn_1|dk_1ord(x^k)$, hence $n_1|k_1ord(x^k)$. Because $(n_1,k_1)=1$, we get $n_1|ord(x^k)$, but $ord(x^k)|n_1$, thus $ord(x^k)=n_1=\frac{n}{d}=\frac{n}{(k,n)}$. The proof is finished. Now we know that $ord(x_i^{k_i})=\frac{ord(x_i)}{(k_i,ord(x_i))}, \forall i\in${$1,2,...,n$}, hence $\frac{ord(x_1)ord(x_2)...ord(x_n)}{(k_1,ord(x_1))(k_2,ord(x_2))...(k_n,ord(x_n))}=ord(x_1^{k_1})ord(x_2^{k_2})...ord(x_n^{n_n})$. Therefore it suffices to prove that $ord(x_1^{k_1}x_2^{k_2}..x_n^{k_n})=ord(x_1^{k_1})ord(x_2^{k_2})...ord(x_n^{n_n})$. I don't think this relation is true. Thank you!

  • $\begingroup$ So, does your formula show that the product of two elements is the the product of the orders? ($k_i = 1$) $\endgroup$ – Thomas Nov 6 '15 at 13:56
  • $\begingroup$ But the orders are relatively prime and the elements commute. $\endgroup$ – Travi Nov 6 '15 at 14:15
  • $\begingroup$ The product of the orders is the order of the product (for $k_i=1$) $\endgroup$ – Travi Nov 6 '15 at 14:16
  • $\begingroup$ Yes it does, with your lemma and an induction, it suffices to show that if $order(x)$ and $order(y)$ are relatively prime then $order(xy)=order(x)order(y)$. Try to show that this is true. $\endgroup$ – Clément Guérin Nov 6 '15 at 15:17
  • $\begingroup$ @ClémentGuérin The thing is, we don't know that $ord(x_1^{k_1})$ is relatively prime to $ord(x_2^{k_2})$ $\endgroup$ – Travi Nov 6 '15 at 17:04

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