# Show that $\prod_{i=1}^{n}\text{Aut}(G_i)\cong \text{Aut}\Big(\prod_{i=1}^{n}G_i\Big)$ (intuition on smaller $n$)

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!

• 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) May 7 at 14:53
• @ArturoMagidin From French x) Thank you for correction, I edit it now. May 7 at 14:54
• The case $n=2$ is all you need - the general case follows by an easy induction. May 7 at 14:54
• 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. May 7 at 15:08
• 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. May 7 at 15:13

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 $$G_i$$.