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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 $r|n_1$.

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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 $\beta=y^\ell$.

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IOW: The natural projection $p:\mathbf{Z}_n\rightarrow\mathbf{Z}_d$ of residue class rings remains surjective, when restricted to the group of units $p: \mathbf{Z}_n^*\rightarrow\mathbf{Z}_d^*,$ and by lifting correctly we can be sure that $\beta$ is a generator. –  Jyrki Lahtonen Sep 26 '11 at 19:33
    
I only glanced at the paper before I was told you had posted an answer, but it seems like on that particular page the authors are assuming that the group $G\lt A\times B$ is cyclic; $G_1$, $\overline{G_1}$, $G_2$, and $\overline{G_2}$ are the groups you get from Goursat's Lemma ($\overline{G_i}$ is the projection of $G$ onto the $i$th component, $G_i$ is the intersection of $G$ with the $i$th component of the product). So perhaps the assumption that the subgroup $G$ is cyclic comes into play (if I'm correct that this is an overarching assumption). –  Arturo Magidin Sep 26 '11 at 19:54
    
I'll try to take a look as soon as I have the time. Probably not just right now, though... –  Arturo Magidin Sep 27 '11 at 2:11
    
After looking at it more carefully, the paragraph in question is trying to establish that $G$ is cyclic (it is part of an "if and only if"). And I think you were spot on in terms of what the authors intended (namely, that by switching $\beta$ if necessary they could select it of the appropriate order). –  Arturo Magidin Sep 27 '11 at 3:49
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