CAF

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	490 reputation	bio	website		visits	member for	7 months
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May 9	comment	Proving rigorously the supremum of a set. Do you mean that < becomes $\leq$?
May 9	comment	Proving rigorously the supremum of a set. Maybe I could also use the density of rationals: There exists a rational in the interval $2 - \frac{1}{n} < q < 2$, where $q$ is the possible 'lower' bound < $2$. Take the limit and we get $2 < q < 2$, a contradiction? I think a similar way is done to prove the case if this proposed lower bound was irrational.
May 9	comment	Proving rigorously the supremum of a set. Is there anywhere I can make a slight change that would make it work?
May 9	comment	Proving rigorously the supremum of a set. I see, that is unfortunate, it was going well up to then.
May 9	comment	Proving rigorously the supremum of a set. That is quicker indeed - but is my proof okay? I suppose I could have deleted the bit about Archimedean since I didn't have to use such an $n$.
May 9	revised	Proving rigorously the supremum of a set. added 43 characters in body
May 9	comment	Proving rigorously the supremum of a set. Okay, I will edit it. Thanks. Is my proof correct?
May 9	asked	Proving rigorously the supremum of a set.
May 3	accepted	Isomorphism between $\operatorname{Stab_G(x)}$ and $S_{n-1}$
May 3	comment	Isomorphism between $\operatorname{Stab_G(x)}$ and $S_{n-1}$ Ok thanks for clarifying that - I think I was wrong too - I can't say the stabiliser acts trivially on $X$, since it only fixes one element and none of the others. Is this right? (Sorry for the deviation from the thread - I just want to ensure I understand terminology correctly).
May 3	comment	Isomorphism between $\operatorname{Stab_G(x)}$ and $S_{n-1}$ Why would the stabilizer act transitively? I.e if we consider $Stab_G(x)$ then all it's elements fix x, so there is no element g in this stabilizer that maps x to another element $y \in X$?
May 3	comment	Isomorphism between $\operatorname{Stab_G(x)}$ and $S_{n-1}$ Thanks, can I also construct a homomorphism from what I wrote in my post starting from 'Is it correct..' I don't know how to show there that the symmetric group is explicitly $S_{n-1}$.
May 3	comment	Isomorphism between $\operatorname{Stab_G(x)}$ and $S_{n-1}$ How would the Orbit Stabilizer help? Sure, I can see why it helps to see that the two groups are indeed the same size, but how else does it help?
May 3	comment	Isomorphism between $\operatorname{Stab_G(x)}$ and $S_{n-1}$ Do you mean like, if in $S_3$ for example $1 \rightarrow 1,\,2 \rightarrow 3,\,3 \rightarrow 2$, then in cyclic notation we have $(23)$. But surely this is still a permutation in $S_3$? The absence of $1$ is just implicitly saying that $1$ is mapped to $1$.
May 3	comment	Isomorphism between $\operatorname{Stab_G(x)}$ and $S_{n-1}$ How would you prove that this is a homomorphism though? Also, in the edit, what is $\eta$? Thank you.
May 3	asked	Isomorphism between $\operatorname{Stab_G(x)}$ and $S_{n-1}$
May 2	revised	(Non) Faithful Group Action question edited title
May 2	comment	(Non) Faithful Group Action question $g^3$ and $g^6$ both fix each respective triangle within the $9-$gon. Since $G = D_9$ acts on the three triangles by permuting the vertices, the elements $g^3$ and $g^6$ permute the vertices such that the triangles remain in place. Hence they are elements of the kernel of the action. Is this correct analysis?
May 2	comment	(Non) Faithful Group Action question Yes, of course - $D_9$ has order $18$ and $X$ has order $3$, so there is no way for there to be an injection between the two sets, i.e the action is not faithful. So when I said that $D_9$ permutes the vertices, this is not the case here, correct? How else could it act on the triangles?
May 2	comment	(Non) Faithful Group Action question Thanks for your response - do you have any comments on my method - I can't see where I made my error.