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Dec
10
awarded  Caucus
Dec
2
comment Find the number of elements of each order in a simple group of order $60$
@DerekHolt, but you could imaginably still have elements of order 6.
Dec
2
comment Find the number of elements of each order in a simple group of order $60$
@DerekHolt, thanks for the heads-up, and it seems that if I show that claim, then it follows that all the elements in the Sylow 2-subgroups must be of order $2$, correct? So I guess that would take care of part 2) of my question, and I'd still need to establish $n_2 = 5$.
Dec
2
comment Find the number of elements of each order in a simple group of order $60$
You say "Since $H$ is not cyclic", but I think that's the hardest thing to show, and that's what my question was about. I.e. how do I show that?
Dec
2
comment Find the number of elements of each order in a simple group of order $60$
@DerekHolt, I was aware of Cayley's Theorem, I just wasn't sure what exactly you were referring to with the image of the regular representation. But why would the image be 15 $4$-cycles? And when you say "intersecting with the alternating group", intersecting what with the alternating group? The representation of $G$? If so, I don't see why the intersection of this subgroup of $S_{60}$ would necessarily have to have index $2$. I assume the odd permutation has to come into play somehow, but how? I mean, it's just one of the elements.
Dec
2
comment Find the number of elements of each order in a simple group of order $60$
@DerekHolt, can you be more explicit? I don't really follow. I haven't learned about regular representations, so how would I consider this as a subgroup of $S_{60}$? Is it just multiplication on the left? But even if so, why would it be an odd permutation? Ugh, and I also don't see the connection between odd permutations and simplicity (or normal groups).
Dec
2
asked Find the number of elements of each order in a simple group of order $60$
Nov
2
revised If an analytic function $f: \mathbb{R}^2 \to \mathbb{R}^2$ is locally invertible at $(x_0, y_0)$, then $Df(x_0,y_0) \not = 0$.
added 1 character in body
Nov
2
comment If an analytic function $f: \mathbb{R}^2 \to \mathbb{R}^2$ is locally invertible at $(x_0, y_0)$, then $Df(x_0,y_0) \not = 0$.
@guest, I edited the original post and stated that then. And, yes, as I said, I cannot use any notions from complex analysis whatsoever.
Nov
2
revised If an analytic function $f: \mathbb{R}^2 \to \mathbb{R}^2$ is locally invertible at $(x_0, y_0)$, then $Df(x_0,y_0) \not = 0$.
added 131 characters in body
Nov
2
comment If an analytic function $f: \mathbb{R}^2 \to \mathbb{R}^2$ is locally invertible at $(x_0, y_0)$, then $Df(x_0,y_0) \not = 0$.
I am not working in $\mathbb{C}$, and this problem was not part of complex analysis, so would you mind explaining why the "multiplicity" would be greater than $1$?
Nov
2
revised If an analytic function $f: \mathbb{R}^2 \to \mathbb{R}^2$ is locally invertible at $(x_0, y_0)$, then $Df(x_0,y_0) \not = 0$.
added 29 characters in body; edited title
Nov
2
asked If an analytic function $f: \mathbb{R}^2 \to \mathbb{R}^2$ is locally invertible at $(x_0, y_0)$, then $Df(x_0,y_0) \not = 0$.
Oct
20
comment Prove that the evaluation map $E_{x_0}: C(K) \to \mathbb{R}$ is differentiable
@PhoemueX, great, thanks aplenty!
Oct
20
comment Prove that the evaluation map $E_{x_0}: C(K) \to \mathbb{R}$ is differentiable
@Dorebell, can you help me solve this problem then? I mean, if $E_{x_0}$ is a linear transformation, then it'd make sense for the derivative to just be the representation of $E_{x_0}$ itself, wouldn't it? But again, I fail to see the significance of $K$ being compact, and it seems that this would then be almost too easy. So I'm thinking it might not be quite right to just say $L = E_{x_0}$, making the numerator $0$.
Oct
20
asked Prove that the evaluation map $E_{x_0}: C(K) \to \mathbb{R}$ is differentiable
Oct
5
comment Mean Value Theorem for Multiple Variables
Actually, $A$ does have to be open, because you can only apply the chain rule to interior points of a set.
Oct
1
comment To prove that these matrices are invertible
Could you elaborate on this a bit? Intuitively, it makes sense that the geometric series would converge to that, but how does that follow rigorously? And if you just expand that and multiply on both sides with $AB$, what's the significance of $||K|| \leq 1$. Why would it not work without that? Is it because then we could find a vector whose norm would get larger and larger as we $i$ increases, and so that the sum would not converge?
Sep
29
accepted Show $|\sin(y)y - \sin(x)x| \leq C|y - x|$ for some $C > 0$
Sep
29
comment Show $|\sin(y)y - \sin(x)x| \leq C|y - x|$ for some $C > 0$
@gnasher729, $y-x = \frac{\pi}{2}$, I guess, but I see you guys are absolutely right. This isn't bounded this way, ugh.