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Background:

Proposition 1: Let $L$ be a group and $G$ and $H$ be two subgroups. Then, the following statements are equivalent:

  1. $L,G$ and $H$ satisfy the following relations: $L=GH, G\cap H=\{1\},$ and $|G,H|=\{1\}.$

  2. $L$ is the direct product of $G$ and $H,$ i.e. $L\cong G\times H.$

Proof: Write

$$G=\{(g,1),g\in G\}$$

and

$$H=\{(1,h),h\in H\}$$

Then, obviously $L=GH$ and $G\cap H=\{1\}.$ For $[G,H]=\{1\},$ we have

$$(g,1)(1,h)(g,1)^{-1}(1,h)^{-1}=(1,1).$$

Questions:

I just have a quick question about: $(g,1)(1,h)(g,1)^{-1}(1,h)^{-1}=(1,1).$, specifically in $(g,1)^{-1}(1,h)^{-1}$, does $(g,1)^{-1}=(1,g^{-1})$ and similarly $(1,h)^{-1}=(h^{-1}, 1)$?

Thank you in advance

Arturo Magidin
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Seth
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  • No, in a direct product $A\times B$, $(x,y)^{-1}$ is $(x^{-1},y^{-1})$. – Gerry Myerson Mar 24 '24 at 02:17
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    @GerryMyerson ah kk. thank you. – Seth Mar 24 '24 at 02:38
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    Stop using mathjax to format text. Why do we need to tell you this hundreds of times, and why do you stubbornly insist on doing it? – Arturo Magidin Mar 24 '24 at 02:43
  • @ArturoMagidin I thought I am using Latex code, and the corrections you made are html markup code – Seth Mar 24 '24 at 03:14
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    You are using mathjax in order to use bad LaTeX code (in LaTeX you do not use math mode for text formatting either). I've explained this to you multiple times: you should not use mathjax for text formatting, you should use mark up. I have also explained to you why. But clearly, you are not listening. Why that is, I don't know. But I am not going to explain it to you yet again. I'm just going to call you out and downvote you until you stop misusing mathjax. – Arturo Magidin Mar 24 '24 at 03:21
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    And I am not the only one who has told you this. I told you about forcing black color a month ago. And about sizing problems you cause. And you are not even doing good $\LaTeX$. You are doing terrible $\LaTeX$ with the mathjax. – Arturo Magidin Mar 24 '24 at 03:32

1 Answers1

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Let $(g,h)\in G\times H$. Consider $(g^{-1}, h^{-1})$. We have

$$\begin{align} (g,h)(g^{-1}, h^{-1})&=(gg^{-1}, hh^{-1})\\ &=(e_G, e_H)=e_{G\times H}\\ &=(g^{-1}g,h^{-1}h)\\ &=(g^{-1}, h^{-1})(g,h).\\ \end{align}$$

Therefore, by uniqueness of inverses, we have $(g,h)^{-1}=(g^{-1}, h^{-1}).$

Shaun
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