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Here's the proposition:

Let $G=G_1\times\cdots\times G_n.$

If factors $G_1,...,G_n$ are characteristic subgroups of $G$, then $${\rm Aut}(G)\cong{\rm Aut}(G_1)\times\cdots\times{\rm Aut}(G_n).$$

The proof goes like this:

If $\alpha_i$ is an automorphism of $G_i~(\text{for }i=1,...,n)$, then$$(g_1,...,g_n)^\alpha:=(g_1^{\alpha_1},...,g_n^{\alpha_n})$$ defines an automorphism of $G=G_1\times \cdots\times G_n$, and $$\varphi:{\rm Aut}(G_1)\times\cdots\times{\rm Aut}(G_n)\rightarrow{\rm Aut}(G)~\text{with } (\alpha_1,...,\alpha_2)\mapsto\alpha$$ is a monomorphism. Moreover, $\varphi$ is surjective if the factors $G_1,...,G_n$ are characteristic subgroups of $G$.

My Questions:

$1)$ What does it mean by "define"? "How does it "define a automorphism"?

$2)$ How can I show it's a monomorphism?

$3)$ How to show it's surjective?

PS:

[1] The questions above might be quite easy and obvious for you, but as a beginner, I get rather confused. How could I manage to fully understand this proposition, and how should I think? I'd really appreciate it if you could give me some help!

[2] It's 1.6.3 at the bottom of page 29 of my textbook, The Theory of Finite Groups, An Introduction.

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  • $\begingroup$ It just means that what is written there is an automorphism. $\endgroup$ Commented Mar 13, 2018 at 11:29
  • $\begingroup$ @TobiasKildetoft Thanks, but how could I manage to gain an insight into this proposition, how to understand it? $\endgroup$
    – user517681
    Commented Mar 13, 2018 at 11:33
  • $\begingroup$ I am really not sure what to tell you. I would usually say that if you have studied enough group theory to start considering characteristic subgroups, then this sort of thing should look completely natural. $\endgroup$ Commented Mar 13, 2018 at 11:35
  • $\begingroup$ @TobiasKildetoft $\ddot\smile$. Frankly, I think so... but I still want some hints to this small question. $\endgroup$
    – user517681
    Commented Mar 13, 2018 at 11:38

1 Answer 1

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You have to decompose the questions. Do you have any difficulty with any of the item below? For any of the item below in 1) or 2), it suffices to write the definition and to apply the hypothesis you are given.

1)

a) what is written there defines a map $\alpha$ from $G$ to $G$.

b) this map $\alpha$ is a morphism.

c) this morphism $\alpha$ is one-to-one.

d) this morphism $\alpha$ is onto.

2)

a) what is written there defines a map $\varphi$ from $Aut(G_1)\times \cdots\times Aut(G_n)\to Aut(G)$.

b) The map $\phi$ is a morphism.

c) The map $\phi$ is one-to-one.

3) Ok, You have an $\alpha\in Aut(G)$ and you want to prove that, given your hypothesis, $\alpha=\varphi(\alpha_1,\dots,\alpha_n)$.

When you have to prove this kind of things, it is usually a good idea to go backward to have an idea on how to construct $\alpha_1,\dots, \alpha_n$. Assume that $\alpha=\varphi(\alpha_1,\dots,\alpha_n)$, how do you recover $\alpha_1$? Well $\alpha_1(g_1)$ is equal, by definition, to the first coordinate of $\alpha((g_1,1_{G_2},\dots, 1_{G_n}))$ isn't it?

Given $\alpha\in Aut(G)$ you simply have to define $\alpha_1:=\pi_1\circ \alpha$ where $\pi_1$ is the natural projection of $G=G_1\times \cdots\times G_n$ to $G_1$. Of course do the same thing for $\alpha_i:=\pi_i\circ \alpha$.

Finally, you want to prove that $\alpha=\varphi(\alpha_1,\dots,\alpha_n)$ (Hint: use the hypothesis).

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