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Can anyone help with the following proof?

Let H,G be final groups.

$(|G|,|H|) = 1$, prove that $Aut(G×H) ≅ Aut(G) ×Aut(H)$

I found this question Show that if $ \gcd(|G|,|H|) = 1 $, then $ \text{Aut}(G \times H) \cong \text{Aut}(G) \times \text{Aut}(H) $. but didnt really understand the solution, would appreciate if anyone could help with that.

thank you

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Bar Dubovski
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2 Answers2

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HINT: There's two things you need to show.

  • $(i)$ $\vert Aut(G)\times Aut(H)\vert\le \vert Aut(G\times H)\vert$, and

  • $(ii)$ $\vert Aut(G)\times Aut(H)\vert\ge \vert Aut(G\times H)\vert$.

Let's start with $(i)$. Suppose I have an automorphism $\alpha$ of $G$, and an automorphism $\beta$ of $H$. Do you see how to build a "product automorphism" $\alpha\times \beta$ of $G\times H$? Do you see how to show that this is injective - that is, $\alpha_0\times\beta_0=\alpha_1\times\beta_1\implies \alpha_0=\alpha_1, \beta_0=\beta_1$?

As for $(ii)$, this is harder. We basically want to show that every automorphism of $G\times H$ is of the form $\alpha\times \beta$ for some $\alpha\in Aut(G)$ and $\beta\in Aut(H)$. This is where the condition on the gcd of the orders of $G$ and $H$ comes in. HINT: suppose the orders of $G$ and $H$ are relatively prime. Can $G$ and $H$ have nonidentity elements $a$ and $b$ with the same order?

Noah Schweber
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    for (i) , I guess that in order to build the "product automorphism" all i have to do is take $\alpha \in Aut(G)$ and $\beta \in Aut(H)$ and just do the product, no? I would really appreciate if you could show how to prove that it is injective – Bar Dubovski Jan 09 '16 at 17:13
  • @BarDubovski What does "just do the product" mean? If I have an element $(g, h)\in G\times H$, what should $(\alpha\times\beta)(g, h)$ be? – Noah Schweber Jan 09 '16 at 17:14
  • it would be $(\alpha(g),\beta(h))$ – Bar Dubovski Jan 09 '16 at 17:16
  • @BarDubovski OK, good. So now suppose $\alpha_0\times\beta_0=\alpha_1\times\beta_1$. Do you see how to prove $\alpha_0=\alpha_1$? (HINT: $\alpha_0=\alpha_1$ iff for every $g$, $\alpha_0(g)=\alpha_1(g)$ . . .) – Noah Schweber Jan 09 '16 at 17:18
  • I understand what i need to prove but but isn't that immediate? I mean if $\alpha_0 = \alpha_1$ than its obvious that for every $g$, $\alpha_0(g)=\alpha_1(g)$. I feel like im missing something.. – Bar Dubovski Jan 09 '16 at 17:24
  • @BarDubovski Yes, but it's the converse you need here. Suppose $\alpha_0\times\beta_0=\alpha_1\times\beta_1$ - that is, suppose $(\alpha_0\times\beta_0)(g, h)=(\alpha_1\times\beta_1)(g, h)$ for every $g, h$; how do I conclude that $\alpha_0=\alpha_1$? (HINT: If $\alpha_0\not=\alpha_1$, then for some $g$ we have $\alpha_0(g)\not=\alpha_1(g)$. What can you say about $(\alpha_0\times\beta_0)(g, e_H)$ vs. $(\alpha_1\times\beta_1(g, e_H)$? Here "$e_H$" is $H$'s identity element.) The same argument will show $\beta_0=\beta_1$ of course. – Noah Schweber Jan 09 '16 at 17:28
  • well for the case you mentioned I can say that $(\alpha_0\times\beta_0)(g, e_H) = (\alpha_1\times\beta_1)(g, e_H) \implies (\alpha_0(g),\beta_0(e_H) = (\alpha_1(g),\beta_1(e_H) \implies (\alpha_0(g),e_H = (\alpha_1(g),e_H) \implies \alpha_0(g) = \alpha_1(g)$ and that would be a contradiction so we conclude that $\alpha_0(g) = \alpha_1(g)$ for every $g$, is that correct? – Bar Dubovski Jan 09 '16 at 17:40
  • Yup! So that's the proof of injectivity, so you've got $\vert Aut(G)\times Aut(H)\vert\le \vert Aut(G\times H)\vert$. The other part is a bit more interesting - HINT: if $g$ is an element of $G$, what is the relationship between the order of $g$ in $G$ and $\vert G\vert$? – Noah Schweber Jan 09 '16 at 17:43
  • $g^|G|=1$, you meant this relationship? – Bar Dubovski Jan 09 '16 at 17:47
  • I'm not sure what that means. A concrete example: if $G$ has an odd number of elements, can any element of $G$ have even order? – Noah Schweber Jan 09 '16 at 17:49
  • well no, because every order of any element have to divide $|G|$, that was the relationship you meant? – Bar Dubovski Jan 09 '16 at 17:54
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    @BarDubovski Yes. Now you can apply this to show that any automorphism $\gamma$ of $G\times H$ is of the form $\alpha\times\beta$. HINT: Look at the "projections" $\gamma(g, e_H)$ and $\gamma(e_G, h)$ - show that e.g. $\gamma(g, e_H)=(g', e_H)$ for some $g'$. This is where you use the order assumption on $G$ and $H$. – Noah Schweber Jan 09 '16 at 18:02
  • Im sorry , I understand what I have to prove here, but don't quite know how. I still don't understand how the order assumption helps me if I know that $\gamma(g,e_H) = (g',e_H)$ for some $g'$ , any chance that you could give me another hint? – Bar Dubovski Jan 09 '16 at 18:11
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    @BarDubovski Sure. Suppose $\gamma(g, e_H)=(a, b)$ where $b\not=e_H$. Then what can you say about the order of $(a, b)$ (think about the order of $a$ in $G$ and of $b$ in $H$ first)? What can you say about the order of $(g, e_H)$? Why is this a problem? – Noah Schweber Jan 09 '16 at 18:18
  • the order of $(g,e_H)$ is $order(g)$. I don't really know what is the order of $(a,b)$ because I don't know the order of b.. I can say that the order is the $lcm(order(a),order(b))$. is this what you mean? or you meant something else...? – Bar Dubovski Jan 09 '16 at 18:24
  • @BarDubovski That's exactly what I mean. Now, the order of $(g, e_H)$ is (as you say) the order of $g$ in $G$, which is a factor of $\vert G\vert$. And the order of $(a, b)$ is the lcm of the order of $a$ in $G$ and the order of $b$ in $H$ - so can this be a factor of $\vert G\vert$? (Remember $b\not=e_H$.) – Noah Schweber Jan 09 '16 at 18:26
  • the lcm can't be a factor of $|G|$ because of the gcd assumption. – Bar Dubovski Jan 09 '16 at 18:31
  • @BarDubovski Right, so the order of $(a, b)$ cannot be the same as the order of $(g, e_H)$. Now all you need to show is automorphisms preserve order. (I'm going to sleep now btw; good luck!) – Noah Schweber Jan 09 '16 at 18:39
  • Thank you, I'm still not quite sure what to do from here, the fact that automorphisms preserve order is straightforward (I guess), let $g \in G , O(g) = m$, we need to prove that $O(f(g)) = m$ where $f(g) \in Aut(G)$. lets assume that the order of $ Aut(G)$ is $t \lt m$ then , $(f(g))^t = e_G = f(g)...f(g){[t-times]}= f(gg...*g{[t-times]}) \implies g^t=e_G$ , contradiction to the fact that m is the order$. After showing that automorphisms preserve order I'm a bit stuck (again) – Bar Dubovski Jan 09 '16 at 19:14
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    @BarDubovski So you've shown that $(i)$ the order of $(a, b)$ cannot equal the order of $(g, e_H)$ unless $b=e_H$, and that automorphisms preserve order. So for every $g\in G$, we have $\gamma(g, e_H)=(g', e_H)$ for some $g'\in G$! Similarly, for every $h\in H$ we'll have $\gamma(e_G, h)=(e_G, h')$ for some $h'\in H$. This gives us two "projection maps" $\gamma_G$ and $\gamma_H$ given by $\gamma_G(g)=g'$, $\gamma_H(h)=h'$ - that is, $\gamma(g, e_H)=(\gamma_G(g), e_H)$ and $\gamma(e_G, h)=(e_G, \gamma_H(h))$. (cont'd) – Noah Schweber Jan 10 '16 at 11:53
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    Now just show $\gamma=\gamma_G\times \gamma_H$. HINT: $\gamma(g, h)=\gamma(g, e_H)*\gamma(e_G, h)$ . . . – Noah Schweber Jan 10 '16 at 11:54
  • @NoahSchweber in (i), when you wrote $|Aut(G_{1}) \times Aut (G_{2})| \leq |Aut(G_{1} \times G_{2})|$, did you mean $|Aut(G_{1})| \cdot |Aut(G_{2})| \leq |Aut(G_{1} \times G_{2})| $? As in, is that supposed to be regular multiplication of real numbers on the left hand side? –  Nov 06 '16 at 05:57
  • @NoahSchweber why does showing that $\alpha \times \beta$ is injective give us that $|Aut(G_{1}) \times Aut(G_{2})| \leq |Aut(G_{1})| \times |Aut(G_{2})|$? I'm trying very hard to solve this problem on my own, but I don't understand how showing that our "product map" is injective gives us the inequality we want there. Also, do we need to show that $\alpha \times \beta $ is a homomorphism in order to even begin to consider whether it's injective or surjective? –  Nov 06 '16 at 15:44
  • @NoahSchweber: One has to show not only that there is a bijection between $\Aut(G \times H)$ and $\Aut(G) \times \Aut(H)$, but that there is a bijective homomorphism. The point that is still missing here is to check if $(\alpha,\beta)\mapsto \alpha\times\beta$ is a homomorphism. – Thrash Feb 10 '22 at 14:52
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Consider the map $\psi\colon \operatorname{Aut}(G) \times \operatorname{Aut}(H) \to \operatorname{Aut}(G\times H)$ given by $(\alpha,\beta)\mapsto \alpha\times\beta$ where $\alpha\times\beta\colon G\times H \to G\times H$ is given by $(g,h)\mapsto(\alpha(g),\beta(h))$.

I skip the step of showing that $\psi$ is well-defined, which means that $\alpha\times\beta$ is indeed an automorphism of $G\times H$. This can be easily checked.

  1. We show that $\psi$ is a group homomorphism. First we have $$ ((\alpha_1\circ\alpha_2)\times(\beta_1\circ\beta_2))(g,h)=(\alpha_1(\alpha_2(g)),\beta_1(\beta_2(h))) = ((\alpha_1\times\beta_1)\circ(\alpha_2\times\beta_2))(g,h), $$ which means $(\alpha_1\circ\alpha_2)\times(\beta_1\circ\beta_2) = (\alpha_1\times\beta_1)\circ(\alpha_2\times\beta_2)$. Thus we get \begin{align*} \psi((\alpha_1,\beta_1)(\alpha_2,\beta_2)) &= \psi(\alpha_1\circ\alpha_2,\beta_1\circ\beta_2)\\&=(\alpha_1\circ\alpha_2)\times(\beta_1\circ\beta_2) \\&= (\alpha_1\times\beta_1)\circ(\alpha_2\times\beta_2) \\&= \psi(\alpha_1,\beta_1)\circ\psi(\alpha_2,\beta_2). \end{align*}

  2. Too see that $\psi$ is injective, we determine $\ker(\psi)$: \begin{align*} (\alpha,\beta)\in\ker(\psi) &\iff \alpha\times\beta=\operatorname{id}_{G\times H} \\&\iff \forall g\in G, h\in H\colon (\alpha(g),\beta(h))=(g,h) \\&\iff \forall g\in G, h\in H\colon \alpha(g)=g \ \land \ \beta(h)=h \\&\iff \alpha=\operatorname{id}_G \ \land \ \beta=\operatorname{id}_H. \end{align*} Thus the kernel is trivial, which means that $\psi$ is injective.

  3. Too see that $\psi$ is surjective, consider $\gamma\in\operatorname{Aut}(G\times H)$ and $g\in G$. Assume $\gamma(g,e_H)=(x,y)$ with $y\ne e_H$. Thus we have $n:=\operatorname{ord}(y)>1$, where $n\mid|H|$. If $g=e_G$, then $y(g,e_H)=(e_G,e_H)$, so $g\ne e_G$. Thus we have $m:=\operatorname{ord}(g)>1$ as well, where $m\mid|G|$. From $$ (e_G,e_H)=\gamma(e_G,e_H)=\gamma(g^m,e_H)=\gamma(g,e_H)^m = (x,y)^m = (x^m,y^m) $$ it follows $n\mid m$ and hence $n\mid|G|$. So $n>1$ is common divisor of $|G|$ and $|H|$, which is a contradiction to our condition $\gcd(|G|,|H|)=1$. This means $\gamma(g,e_H)$ has the form $(x,e_H)$, and analogously, $\gamma(e_G,h)$ has the form $(e_G,y)$. Let $\pi_G$ denote the projection homomorphism $G\times H \to G$ given by $(g,h)\mapsto g$. Thus the map $\alpha\colon G\to G$ given by $g\mapsto\pi_G(\gamma(g,e_H))$ is an automorphism (with inverse $g\mapsto\pi_G(\gamma^{-1}(g,e_H))$). Analogously, the map $\beta\colon H\to H$ given by $h\mapsto\pi_H(\gamma(e_G,h))$ is an automorphism as well. In other words: $\gamma(g,e_H)=(\alpha(g),e_H)$ and $\gamma(e_G,h)=(e_G,\beta(h))$. It remains to check $\gamma = \alpha\times\beta$, which can be easily done. Thus we have $\gamma=\alpha\times\beta=\psi(\alpha,\beta)$, which means that $\psi$ is surjective.

Summary: $\psi$ is a group isomorphism.

Thrash
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