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Let $G=(g^i)_{i\in\mathbb{N}},\ H=(h^i)_{i\in\mathbb{N}}$ be finite cyclic groups such that $(g,h)$ generates $G\times H$. I want to prove that $\gcd(|g|,|h|)=1$ (where $|g|=|G|$,etc)

Let $k\in\mathbb{Z}$ such that $(g,h)^k=(g,h^2)$. Then, $$k=1\mod|g|,\quad k=2\mod|h|$$ So, we can take $\alpha,\beta\in\mathbb{Z}$ such that $k=1+\alpha|g|=2+\beta|h|$. From this we get $\alpha|g|-\beta|h|=1$ and, then, by Bezout's lemma, we finally deduce that $|g|$ and $|h|$ are coprime.

Is this correct? A Yes or No is enough

Veridian Dynamics
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1 Answers1

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Yes, it's a perfect proof.

However, it took me some time to parse the formula of your very idea: how did that exponent $2$ appear, shouldn't it be $k$, what happens to $g$, etc. So, some preliminary words could help the reader (perhaps only an "in particular").

Alternatively, you could e.g. choose a $k$ such that $(g,h)^k=(1,h)$ to get the same conclusion.

Berci
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