A Summer of Maths (ASoM) 2016

huh? yeah thats wat I did for the first bit. So wat about the 2nd?

Oh ffs, I didn't even read your question properly. (Sorry!!) Try induction on $n$ 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.

Babshshshabsmsm

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I know...

How are you doing? Things going OK?

Bit drunk tbh


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OK, good on ya.

Got s group related question, will ask when I remember mate


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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

I do mental notes anyway so neither have I tbh, it was either dihedral groups of homomorphism/iso


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Cool

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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Each symmetry in D(n), the symmetry group of the regular n-sided 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 high-level point of view, is that if m divides n, then D(m), the group of symmetries of the regular m-polygon, 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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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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Each symmetry in D(n), the symmetry group of the regular n-sided 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 high-level point of view, is that if m divides n, then D(m), the group of symmetries of the regular m-polygon, 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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But can't figure out how to factorise that, at least not without seeing the answer already.

Well, you certainly took the painful way through the problem, row reduction works better. But anywho, it's not hard, looking at the polynomial, to guess that $b=c$ is a solution, and indeed:

$c^3 - c^3 - ac^2 + ca^2 + ac^2 - ca^2 = 0$, so by symmetry $c=a$ and $b=a$ are also solutions, given that the polynomial has a highest power of $a^2, b^2, c^2$ then $(b-a)(b-c)(c-a) = 0$ is the only possible factorisation (since the coefficients are all 1).

In fact, a rather harder version of this sort of technique comes up in STEP II 2016 Q2.

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:

$det(D) = \left| \begin{array}{ccc} 1 & 0 & 0 \\a & b - a & c - a & a^2 & b^2 - a^2 & c^2 - a^2 \end{array} \right| = (b - a)(c - a)\left| \begin{array}{ccc} 1 & 0 & 0 \\a & 1 & 1 & a^2 & b + a & c + a \end{array} \right|$

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