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Let $G_1,...,G_n$ be groups of finite order such that theirs orders $|G_i|$ are pairwise coprime. Show that $\xi:$$\prod_{i=1}^{n}\text{Aut}(G_i)\cong \text{Aut}\Big(\prod_{i=1}^{n}G_i\Big)$. Here I won't really consider the case for $n$. I would like to consider $n=2$ and write my intuition on this statement. I saw that there are some posts for this statement (for $n=2$), but I didn't really find a clear explication. If someone could comment, give a feedback on it or help to solve the problem, I would really appreciate it. Thank you in advance

So let's consider the case for $n=2$. Let $G_1,G_2$ be groups of finite order such that theirs orders $|G_1|, |G_2|$ are coprime. What I think is that to show that ${\rm Aut}(G_1\times G_2)\cong{\rm Aut}(G_1)\times {\rm Aut}(G_2)$, I can show that there is injection in both directions (and then conclude by Cantor Schroeder Bernstein). In my previous post I showed that there is an injection $\xi:{\rm Aut}(G_1)\times{\rm Aut}(G_2)\to{\rm Aut}(G_1\times G_2)$.

My problem, now, is how to define properly the injection $\eta:{\rm Aut}(G_1\times G_2)\to {\rm Aut}(G_1)\times{\rm Aut}(G_2)$.

What I think, is that I can define $\eta$ as:

$\eta(\phi)=(\phi_1,\phi_2)$ where $\phi_1 \in \text{Aut}(G_1\times \{e\})$ and $\phi_2\in \text{Aut}(\{e\}\times G_2)$.

But, once I send $\phi$ to ${\rm Aut}(G_1)\times{\rm Aut}(G_2)$ by $\eta$, I don't really see which elements to apply to the 2-tuple $(\phi_1,\phi_2) \in {\rm Aut}(G_1)\times{\rm Aut}(G_2)$.

For example if we take $\phi \in{\rm Aut}(G)$, then we can take $g\in G$ and send it by $\phi$ to $G$. But how does it work in ${\rm Aut}(G_1)\times{\rm Aut}(G_2)$?

Could I define it as the following?

$(\phi_1,\phi_2)(g_1,g_2)=(\phi_1(g_1),\phi_2(g_2))$.

Then, if yes, I don't understand where to apply the coprimeness of $|G_1|$ and $|G_2|$... With this definition of $(\phi_1,\phi_2)$, we can only notice that $\phi(g_1,g_2)=\phi_1(g_1)\cdot \phi_2(g_2)$.

Sorry in advance for kinda "vague" explication and thanks for help!

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    $\begingroup$ The usual way to say that any two are coprime in (American?) English is “pairwise coprime”. “two-by-two coprime” sounds like a literal translation, perhaps form the Spanish version of the phrase? (coprimos dos a dos) $\endgroup$ May 7 at 14:53
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    $\begingroup$ @ArturoMagidin From French x) Thank you for correction, I edit it now. $\endgroup$
    – Daniil
    May 7 at 14:54
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    $\begingroup$ The case $n=2$ is all you need - the general case follows by an easy induction. $\endgroup$ May 7 at 14:54
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    $\begingroup$ No, Schroeder Bernstein doesn't say the two functions are bijections, only that a bijection exists, and there is no reason to think the bijection resulting would be a homomorphism. $\endgroup$ May 7 at 15:08
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    $\begingroup$ Much weaker, however, is that all the groups are finite, so if there are injections both ways, then the injections are isomorphisms. But I think it is easier to prove this homomorphism is onto. $\endgroup$ May 7 at 15:13
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You want to find a homomorphism $$ \eta: \text{Aut}(G_1\times G_2)\to \text{Aut}(G_1)\times \text{Aut}(G_2). $$

The best guess is $\eta(\phi)=(\phi|_{G_1}, \phi|_{G_2})$: but that will only work if $\phi|_{G_i}(G_i)=G_i$. It is here that the coprimeness comes into play. It is easy to prove the following lemma; and with it we are sure that $\phi|_{G_i}(G_i)=G_i$. It's easy to check the injectivity, easy to check that $\eta$ is a homomorphism, and easy to check that $\eta$ is the inverse of your map $\text{Aut}(G_1)\times \text{Aut}(G_2)\to \text{Aut}(G_1\times G_2)$.

Lemma The only elements of order dividing $|G_i|$ in $G_1\times G_2$ lie in $G_i$.

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