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Consider normal Sylow subgroups $G_1, \ldots, G_r$ of a finite abelian group $G$. Let's set up the homomorphism $\phi: G_1 \times \ldots \times G_r \to G$ s.t. $(g_1, \ldots, g_r) \mapsto g_1\ldots g_r$ as in this question. How to show that the homomorphism is injective, i.e., the kernel is trivial?

I understand that $g_1, \ldots, g_n$ raised to their respective orders is certainly in the kernel but I'm not sure how to show that no other $r$-tuple is in the kernel too. I think the fact that "in abelian groups, the order product of commuting elements with relatively prime orders is equal to the product of their orders" needs to be used somehow but I'm not sure how.

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  • $\begingroup$ What've you tried? $\endgroup$ – mrtaurho Apr 8 at 19:52
  • $\begingroup$ @mrtaurho I understand that $g_1, \ldots, g_n$ raised to their respective orders is certainly in the kernel but I'm not sure how to show that no other r-tuple is in the kernel too. $\endgroup$ – S.D. Apr 8 at 19:54
  • $\begingroup$ @mrtaurho I think the fact that "in abelian groups product of commuting elements with relatively prime orders is equal to the product of their orders" needs to be used somehow but I'm not sure how. $\endgroup$ – S.D. Apr 8 at 19:58
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    $\begingroup$ Don't tell this to me; tell this to the people who come across your question :D So, think about adding these thoughts with an edit to prevent your question from being downvoted and/or closed. $\endgroup$ – mrtaurho Apr 8 at 20:02
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Let $|G_{i}|=p_{i}^{n_{i}}$ for all $i$. Let $(g_{1},\ldots,g_{r})\in\ker\phi$. Then $g_{1}\cdots g_{r}=1$. If $r=1$, then we are done. If $r\geq 2$, for any $i$, we have $g_{i}=(g_{1}\cdots g_{i-1}g_{i+1}\cdots g_{r})^{-1}$. Since $g_{i}^{p_{i}^{n_{i}}}=1$, $|g_{i}|\mid p_{i}^{n_{i}}$. Since $$ g_{i}^{p_{1}^{n_{1}}\cdots p_{i-1}^{n_{i-1}}p_{i+1}^{n_{i+1}}\cdots p_{r}^{n_{r}}}=(g_{1}\cdots g_{i-1}g_{i+1}\cdots g_{r})^{-p_{1}^{n_{1}}\cdots p_{i-1}^{n_{i-1}}p_{i+1}^{n_{i+1}}\cdots p_{r}^{n_{r}}}=1, $$ $|g_{i}|\mid p_{1}^{n_{1}}\cdots p_{i-1}^{n_{i-1}}p_{i+1}^{n_{i+1}}\cdots p_{r}^{n_{r}}$. So $|g_{i}|\mid \gcd(p_{i}^{n_{i}},p_{1}^{n_{1}}\cdots p_{i-1}^{n_{i-1}}p_{i+1}^{n_{i+1}}\cdots p_{r}^{n_{r}})=1$. Then $g_{i}=1$. So $\phi$ is injective.

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