Consider the cyclic group $\mathbb Z_n$. Let $d_1$ be a positive proper divisor of $n$. Let $K_{d_1}=\{x:\gcd(x,n)=d_1\}$.
If $a,b\in K_{d_1}$, is it true that $\langle a\rangle =\langle b\rangle$? Here $\langle a\rangle$ denotes the cylcic group generated by $a$.
My try: I need to show that $\langle a\rangle\subseteq\langle b\rangle$ and then $\langle b\rangle\subseteq\langle a\rangle$.
Let $x\in \langle a\rangle$, then $x=ma$. Also $\gcd(a,n)=d_1$ so $a=kd_1$, so we have $x=mkd_1$
I am stuck on how to show that $x\in \langle b\rangle$ from here. Please help someone.
Continuing your steps:
By Bézout's identity, we know that there exist $x_0,y_0, x_1, y_1 \in \mathbb{Z}$ such that $d_1 = ax_0 + ny_0 = bx_1+ny_1$ by hypothesis. So $$ x = mkd_1 = mk( bx_1 + ny_1) = b(mkx_1) + n(mky_1) = b(mkx_1) \mod n \in \langle b \rangle$$ so we have $\langle a \rangle \subseteq \langle b \rangle$
To show the other inclusion, comsider $x \in \langle b \rangle$ then there exist integers $m_1$ and $k_1$ such that $x = m_1k_1d_1$, similar to previous argument we have $x = a(m_1k_1x_0) \mod n \in \langle a \rangle$.