generator of cyclic group I've been reading this article (http://arxiv.org/PS_cache/arxiv/pdf/1109/1109.0024v2.pdf, page 7, paragraph 2) about a generalized Goursat lemma and in the article the author determines the cyclic subgroups of $A \times B$. In the proof he makes the following statement (sort of): 
Suppose we are given finite cyclic groups $\overline{G}_1, \overline{G}_2$ of orders $m=m_1d$ and $n=n_1d$, where $d=gcd(m, n)$. Let $G_1, G_2$ be subgroups of $\overline{G}_1, \overline{G}_2$ of coprime orders $m_1, n_1$. Suppose there exists an isomorphism $\theta: \overline{G}_1/G_1\rightarrow \overline{G}_2/G_2$. If $\alpha$ is a generator of $\overline{G}_1$ and $\beta G_2=\theta(\alpha G_1)$, then both $\alpha G_1$ and $\beta G_2$ are of order $d$, whence $\beta$ is of order $d|G_2|=dn_1=n$.
Why must $\beta$ be of order $n$?? All I can prove is that the order of $\beta$ is of the form $d\cdot r$, where $r|n_1$.  
 A: Admittedly this step is weak: without further explanation the claim may be false.
However, we can fix things. Note that the condition $\beta G_2=\theta(\alpha G_1)$ does not determine the element $\beta$ uniquely. Only its coset modulo $G_2$ is known. It turns out that we can choose $\beta$ within this coset in such a way that $\beta$ does have order $n$.
Let $\overline{G}_2$ be generated by an element $y$ of order $n_1d$, so $G_2=\langle y^d\rangle$.
The condition $\beta G_2=\theta(\alpha G_1)$ tells us that $\beta G_2$ generates the quotient group $\overline{G}_2/G_2$, so $\beta\in y^i G_2$ for some integer
$i, 0<i<d, (i,d)=1$. The choices that were made earlier ($\alpha$, $\theta$ and $y$) will determine $i$, but only modulo $d$. Let $n^*$ be the product of those prime factors of $n$ that are not factors of $d$. By the Chinese remainder theorem we can find an integer $\ell$ such that $\ell\equiv i\pmod d$ and $\ell\equiv 1\pmod {n^*}$. Then $(\ell,n)=1$, because otherwise $\ell$ would have common prime factors with either $d$ or $n^*$.
The former congruence says that $y^\ell G_2=y^i G_2=\theta(\alpha G_1)$, and the conclusion $(\ell,n)=1$ says that $y^{\ell}$ has order $n$. Therefore we can choose $\beta=y^\ell$.
