This might be a stupid question but if I have a subgroup $H$ of $G$, and I have $h\in H$ and $g\notin H$, then $gh\notin H$ for if $gh\in H$ then $g\in Hh^{-1}=H$ is a contradiction. But what if I have $x\notin H$, what is my argument for why $gx\notin H$?
Product of two elements not in subgroup is not in subgroup
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abstract-algebra
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0Let your $G = \mathbb{Z}$, and let $H = 4\mathbb{Z}$. Let $g = x = 2$. Then $g$ and $x$ are both not in $H$, but what's $g + x$? (I love being able to reduce a question to this identity.) – 2012-06-14
1 Answers
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It is not true that:
$x\notin H$ and $g\notin H$ $\implies$ $xg\notin H$.
(If I understand your question, you wanted to justify this claim.)
Counterexample:
Just take $H=\{e_G\}$ and $x=g^{-1}$ for some $g\ne e_G$.
E.g. $H=\{0\}$ and $g=1$, $x=-1$ in $(\mathbb Z,+)$.
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2then that is why I can't prove it. I just hoped it was so.. But my real problem is the following: I have that $H\cap H^g = 1$ for all $g\in G\backslash H$, and I want to show that $H^x\cap H^g = 1$ for different conjugates of $H$. So I wanted to show that $gx^{-1}\notin H$ so that $H\cap H^{gx^{-1}} = 1$ and then $H^x\cap H^g = (H\cap H^{gx^{-1}})^x = 1$. How do I do that then? – 2012-06-14
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1@Helen Perhaps you could edit the question? Just from skimming it I think you want to be careful about the statement. If $g$ and $x$ are in the same coset then what you're saying cannot be true. – 2012-06-14
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0@Helen: I would not edit the question, I would ask a new one. – 2012-06-14
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0@Helen: If $H^x \neq H^g$, then $H \neq H^{gx^{-1}}$ and so $gx^{-1} \notin H$. – 2012-06-14
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0How do I prove that $H=H^{gx^{-1}} \Longleftrightarrow gx^{-1} \in H$? – 2012-06-14
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0I can see it if I am talking about cosets: $Hg = Hx \Longleftrightarrow gx^{-1} \in H$. Is it sort of the same thing? – 2012-06-14
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1@Helen: you only need to prove the easy direction. $H^h=H$. The other direction happens to be true, but this is because $N_G(H)=H$ in your case. – 2012-06-14
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0@JackSchmidt: Of course! thank you! – 2012-06-14