Let $n =p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}$, where $p_i$'s are odd primes. Then $$U(n) = \mathbb Z_{m_1} \times \cdots \times \mathbb Z_{m_r}$$ where $m_i = p_i^{k_i-1} (p_i-1)$.

I want to find a condition on $n$ such that any two elements of $U(n)$ of the form $(a_1,a_2, \dots, a_i, \dots, a_n)$ and $(a_1, a_2, \dots, ka_i, \dots, a_n)$, implies there is an integer $t$ such that either $$(a_1,a_2, \dots, a_i, \dots, a_n) = t(a_1, a_2, \dots, ka_i, \dots, a_n)$$ $$\text{ or }$$ $$(a_1, a_2, \dots, ka_i, \dots, a_n) = t(a_1,a_2, \dots, a_i, \dots, a_n).$$

Take $l -1 = \mathrm{lcm}\ \{ o(a_j) : \; j \neq i \}$ and if we take $t$ a multiple of $l$, then $t(a_1,a_2, \dots, a_i, \dots, a_n) = (a_1,a_2, \dots, ta_i, \dots, a_n)$, change only the $i^{th}$ cordinate.

I would be thankful for any kind of help.

  • $\begingroup$ The group you're discussing is the group of units in $\mathbb{Z_n}$. The unitary group $U(n) = \{g\in GL_n(\mathbb{C}):\, g^\dagger g = 1\}$. (The former is donated by $U(n)$ occasionally (though $\mathbb{Z}_n^\times$ is more common), but I've never heard it called 'the unitary group'.) $\endgroup$ – anomaly Oct 25 '17 at 6:50
  • $\begingroup$ I have seen this notation of this group in the book Contemorary Abstract algebra ( Joseph A. Gallian) . i am sorry about the name. I will edit it and thanks for the remind me. $\endgroup$ – Struggler Oct 25 '17 at 7:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.