I agree with you, (assuming its counts multiplicity) should be a completely additive arithmetic function.(Original post by Oxmath)
Guys after reading this thread, I went and got Algebra and Geometry by Beardon. Its been interesting so far (although I've just begun chapter 2).I was reading the page on "The Principle of Induction II", and I didn't understand something.When he says: "If not, then we can write and so (obviously) ".Shouldn't that be ?Edit: How the **** do I put paragraphs in?
Anywho, surely the example given is completely asinine. It folds quickly to contradiction. i.e: assume that the number of prime factors is , then since every prime factor is we get a number which is easily shown to be bigger than for all (via induction, if you wish)  so contradiction, hence .
A Summer of Maths (ASoM) 2016
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 02072016 20:11
Last edited by Zacken; 02072016 at 20:15. 
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 02072016 21:47
(Original post by EnglishMuon)
For Q 4 (ii) page 15 of the Beardon book is this supposed to be an obvious solution? From i I just showed any cycle
so should I be reapplying the identity to this expression to get terms of the form but this seems fiddly.
Every permutation in can be written as the product of disjoint cycles. Every cycle of length of at least 2 can be written as a product of transpositions. Every transposition (a b) can be written as the product of 3 transpositions of the form (1 a) (1 b) (1 a), so any element in can written in the required form.Last edited by Zacken; 02072016 at 21:53. 
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 02072016 21:52
(Original post by Zacken)
It is meant to be obvious:
Every permutation in can be written as the product of disjoint cycles. Every cycle of length of at least 2 can be written as a product of transpositions. Every transposition (a b) can be written as the product of 3 transpositions of the form (1 a) (1 b) (1 a), so any element in can written in the required form. 
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 02072016 21:58
(Original post by EnglishMuon)
huh? yeah thats wat I did for the first bit. So wat about the 2nd?
(1 2) is of both forms, so that's okay. Assume (1 k) can be written as a product of transpositions of the form (m m+1) then the identity (1 k)(k, k+1)(1 k) = (1 k+1) completes the induction. 
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 02072016 22:03
(Original post by Zacken)
Oh ffs, I didn't even read your question properly. (Sorry!!) Try induction on to show that transpositions of the form (1 a) can be written as product of transpositions of the form (k, k+1).
(1 2) is of both forms, so that's okay. Assume (1 k) can be written as a product of transpositions of the form (m m+1) then the identity (1 k)(k, k+1)(1 k) = (1 k+1) completes the induction. 
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 03072016 01:58
Babshshshabsmsm
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 03072016 02:23
How are you doing? Things going OK? 
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 03072016 02:34
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 03072016 02:43
(Original post by EricPiphany)
OK, good on ya.
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 03072016 02:45
(Original post by drandy76)
Got s group related question, will ask when I remember mate
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 03072016 02:47
(Original post by EricPiphany)
K, I doubt I'll be able to do it though, coz I haven't put pen to paper w.r.t. GT yet. Ask me anyways
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 03072016 02:51
(Original post by drandy76)
I do mental notes anyway so neither have I tbh, it was either dihedral groups of homomorphism/iso
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 03072016 15:56
(Original post by EricPiphany)
Cool
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 03072016 16:27
(Original post by drandy76)
Turns out it was both, I'm not fully convinced of the fact that S3 group is an isomorphism of the D3 group even after looking at the Cayley table, or rather I'm not entirely sure what I'm meant to be noticing between the permutations between the two groups that allows me to identify them as isomorphic
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With S(3), it turns out that this spacing is always conserved with any of the 3! permutations of {1,2,3}; 1 is always between 2 and 3, 2 always between 1 and 3, and 3 always between 1 and 2. This is precisely (isomorphic to) D(3)!
Another interesting observation, that can be made from a very highlevel point of view, is that if m divides n, then D(m), the group of symmetries of the regular mpolygon, is (isomorphic to) a subgroup of D(n). For example the symmetries of a triangle are contained within the symmetries of a hexagon. Why? (Draw a hexagon and find a triangle within it.)
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 03072016 16:33
(Original post by Ecasx)
Each symmetry in D(n), the symmetry group of the regular nsided polygon, is just a permutation of the vertices numbered {1,2,...,n} where the relative 'spacing' between the vertices is conserved. For example, 2 must be in between 1 and 3, and 1 must be in between 2 and n. From this you can immediately conclude that D(n) is isomorphic to a subgroup of S(n); in particular, D(n) is isomorphic to the subgroup of S(n) in which this relative 'spacing' is conserved.
With S(3), it turns out that this spacing is always conserved with any of the 3! permutations of {1,2,3}; 1 is always between 2 and 3, 2 always between 1 and 3, and 3 always between 1 and 2. This is precisely (isomorphic to) D(3)!
Another interesting observation, that can be made from a very highlevel point of view, is that if m divides n, then D(m), the group of symmetries of the regular mpolygon, is (isomorphic to) a subgroup of D(n). For example the symmetries of a triangle are contained within the symmetries of a hexagon. Why? (Draw a hexagon and find a triangle within it.)
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 03072016 16:43
(Original post by drandy76)
Ahh I see thanks, was missing the idea of the conservation of relative spacing, would I be correct in saying that this isomorphism only holds when n=3, or is there more overlap in cases where the order of Dn doesn't equal 2n( not sure if this is an identity or it's something that happens most of the time)
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You can confirm this by first understanding that each 'symmetry' of the npolygon (a rotation, reflection, or combination of both) conserves this relative spacing. Vertex 1 can be moved to any of n places, and then vertex 2 has two possible places adjacent to vertex 1. After this, vertex 3 is fixed (it is in the other place adjacent to vertex 1), and then every other vertex is fixed. So we have an upper bound of 2n when counting the number of symmetries of the npolygon.
Your next task should be to confirm that all possible 2n symmetries (again, symmetries are just permutations of vertices which conserve this relative spacing) are indeed achievable via rotations and reflections. Once you do this you prove that the order of D(n) is 2n.
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 03072016 17:05
(Original post by Ecasx)
Each symmetry in D(n), the symmetry group of the regular nsided polygon, is just a permutation of the vertices numbered {1,2,...,n} where the relative 'spacing' between the vertices is conserved. For example, 2 must be in between 1 and 3, and 1 must be in between 2 and n. From this you can immediately conclude that D(n) is isomorphic to a subgroup of S(n); in particular, D(n) is isomorphic to the subgroup of S(n) in which this relative 'spacing' is conserved.
With S(3), it turns out that this spacing is always conserved with any of the 3! permutations of {1,2,3}; 1 is always between 2 and 3, 2 always between 1 and 3, and 3 always between 1 and 2. This is precisely (isomorphic to) D(3)!
Another interesting observation, that can be made from a very highlevel point of view, is that if m divides n, then D(m), the group of symmetries of the regular mpolygon, is (isomorphic to) a subgroup of D(n). For example the symmetries of a triangle are contained within the symmetries of a hexagon. Why? (Draw a hexagon and find a triangle within it.)
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2006 Q6 IMO. 
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 04072016 14:45
Is there any nice way to deal with the factorisation of the determinant of D? The textbook method uses EROs to change the 1s to 0s in the rightmost two columns:
But this method isn't really obvious to me, I tried just working with the determinant as it comes so I end up with:
But can't figure out how to factorise that, at least not without seeing the answer already.
Is there a good technique for spotting factorisations like that? physicsmaths
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Updated: September 28, 2016
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