Consider a finite group $G$. Let $g_1$ and $g_2$ two elements of $G$ that commute and have co-prime orders. Are $g_1$ and $g_2$ elements of $\langle g_1g_2\rangle$ ?
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3@DietrichBurde Please, read carefully the question before closing it. The OP asks "is it possible to generate $g_1$ and $g_2$ from $g_1g_2$?", I don't see how it relates to the problem you link. – Crostul Mar 01 '20 at 10:15
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@Crostul Sorry, I read that $g_1g_2$ then generates the whole group $G$, hence also $g_1$ and $g_2$. The assumption was, however, that $G$ is abelian. So the question is reopened. – Dietrich Burde Mar 01 '20 at 13:56
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1What have you tried? Why do you think this might be true? Do you know any structural theorems for finite abelian groups? – user1729 Mar 01 '20 at 13:59
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BTW this was a cross-post (1 hour later) from MO (https://mathoverflow.net/questions/353909/) where it's been closed and deleted. – YCor Mar 01 '20 at 21:41
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Let $o(g_1)=m$ and $o(g_2)=n$, so gcd$(m,n)=1$. By Bézout's Theorem we can find $k,l \in \mathbb{Z}$ with $1=km+ln$. It follows that $g_1=g_1^{ln}$ and $g_2=g_2^{km}$. Hence $(g_1g_2)^{ln}=g_1$ and $(g_1g_2)^{km}=g_2$.
Nicky Hekster
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1@RTK If you deem this answer the right one,please tick it as such, complying with the StackExchange rules, thanks. – Nicky Hekster Mar 01 '20 at 18:07