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2014-06-28 20:15:17 -0500 marked best answer Permutation group: (1234)=(12)(13)(14)

How do I show in sage that

 (1234)=(12)(13)(14)

I tried make use of:

 PermutationGroup([1,2]) * PermutationGroup([1,3])

but it doesn't do what is needed.

2014-06-28 20:15:17 -0500 marked best answer S3's elements

Here's elements of Symmetric group of 6th order: S3:

The book says

I want to get the same in Sage. So I do:

sage: G = SymmetricGroup(3)
sage: G.list()
[(), (2,3), (1,2), (1,2,3), (1,3,2), (1,3)]

Now I can't find (1,3,2) element in the book. As far as I understand:

P1 -> ()
P2 -> (1,2,3)
P3 -> (2,3)
P4 -> (1,2)
P5 -> (1,3)
P6 -> ??? also (1,2,3) ???

So my question is to set the correct map from sage to my book...

2014-06-28 20:15:17 -0500 marked best answer What can I delete?

So I've compiled sage from sources. The sage folder is now 3Gb. I guess I've no need for some files... What can I delete? Is there any reason I should not delete these things?

2014-06-28 20:14:50 -0500 marked best answer Symbolic product in Sage?

Suppose I'd like to compute

prod(1/x^4, x, 1, oo)

How can this be done?

I found an old thread, but with no answers.

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2013-06-16 00:38:35 -0500 marked best answer Symbolic product in Sage?

Take the natural logarithm of your product and you get a sum which can be evaluated:

$$\ln\left( \prod_{x=1}^k \frac{1}{x^4} \right) = \sum_{x=1}^k \ln\left(\frac{1}{x^4}\right)$$

... now take the limit as $k \to \infty$:

sage: sum(ln(1/x^4), x, 1, oo)
-Infinity
sage: e^sum(ln(1/x^4), x, 1, oo)
0
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2012-06-18 23:39:28 -0500 commented question Python thing that doesn't work in Sage, works in pure Python

See [here](http://www.sagemath.org/doc/faq/faq-usage.html#i-have-type-issues-using-scipy-cvxopt-or-numpy-from-sage).

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2012-06-14 04:23:20 -0500 commented answer Symbolic product in Sage?

Am... Great, but why post it here?

2012-06-13 03:37:57 -0500 marked best answer Check that P3*P6=P4

The problem is that you are comparing lists, not Sage objects. achrzesz gives an answer which compares two Sage group elements, so they are equal. The lists you give are simple Python ordered lists, and as ordered things, certainly aren't the same. Why didn't you just compare P[3]*P[6] and P[4]?

By the way, they're not the same. But that's a different issue. They're using the notation of the second row of your notation in your original question, not cycle notation.

sage: P[3] * P[6]
(1,3)
sage: P[4]
(1,2)
2012-06-12 07:45:50 -0500 commented answer Check that P3*P6=P4

But mine are also the same. [3, 2, 1] is same as [2, 1, 3], isn't it?

2012-06-12 07:18:09 -0500 asked a question Check that P3*P6=P4

Here's elements of Symmetric group of 6th order: S3:

The book says

I want to check that P3*P6=P4.

G = SymmetricGroup(3)

BookNumbers = [1, 4, 2, 3, 6, 5]

P = [0]
for i in BookNumbers:
    P.append(sorted(G.list())[i-1])

print (P[3] * P[6]).list(), P[4].list()
print (P[3] * P[6]) == P[4]

it gives:

[3, 2, 1] [2, 1, 3]
False

so they are the same actually. But how do I make sage say True?.

2012-06-12 05:52:41 -0500 marked best answer S3's elements

The results of

sage: G = SymmetricGroup(3)
sage: G.list()
[(), (2,3), (1,2), (1,2,3), (1,3,2), (1,3)]

are given in cycle notation. So the book's equivalent of (1,3,2) is one where 1 becomes 3, 3 becomes 2, and 2 becomes 1, which is P6, if I'm reading correctly.

If you want to match the bottom three elements of your matrix, you can simply convert each to a list, or maybe a dict would make the mapping more explicit:

sage: for g in G:
....:     print g, g.list(), g.dict()
....:     
() [1, 2, 3] {1: 1, 2: 2, 3: 3}
(2,3) [1, 3, 2] {1: 1, 2: 3, 3: 2}
(1,2) [2, 1, 3] {1: 2, 2: 1, 3: 3}
(1,2,3) [2, 3, 1] {1: 2, 2: 3, 3: 1}
(1,3,2) [3, 1, 2] {1: 3, 2: 1, 3: 2}
(1,3) [3, 2, 1] {1: 3, 2: 2, 3: 1}
2012-06-12 05:52:40 -0500 commented answer S3's elements

That's very cool. I did this: `for e in sorted(G): print '{0:>8s}{1:>10s}{2:>19s}'.format(e, e.list(), e.dict())` (I type more then, but it inserts >...)

2012-06-12 05:47:52 -0500 commented answer Polynomial representation of GF(7)?

Oh I see - so it's just because `n=1`. Thank You for Your answer.

2012-06-12 05:47:31 -0500 marked best answer Polynomial representation of GF(7)?

If $p$ is a prime, then GF(p^n,'x') is obtained by computing $F_p[x] / (f(x))$ where $f$ is a monic, irreducible polynomial of degree $n$ in $F_p[x]$. For $n=1$, you just get $F_p[x] / (x) \cong F_p$.

So, for any prime $p$, GF(p,'x') is [0,1,2,...,p-1].

2012-06-12 03:49:54 -0500 commented question separation of variables / parameterizing equations in sage math

I'd've recommend You to read 'symbolic expression' of reference manual. Take a look at `collect`, lhs(), rhs() there.

2012-06-12 03:45:31 -0500 marked best answer Permutation group: (1234)=(12)(13)(14)

Sage's built-in help is useful here.

sage: PermutationGroup?
Definition:     PermutationGroup(gens=None, gap_group=None, domain=None, canonicalize=True, category=None)
Docstring:
       Return the permutation group associated to x (typically a list of
       generators).

       INPUT:

       * "gens" - list of generators (default: "None")

       * "gap_group" - a gap permutation group (default: "None")

       * "canonicalize" - bool (default: "True"); if "True", sort
         generators and remove duplicates

       OUTPUT:

       * A permutation group.

So indeed you get a permutation group, with these things as generators. You can't multiply groups.

A similar look at help will show you that Permutation does not give an element of a group, though you can convert an element of such a group to one.

But you can get group elements from your group. Try this.

sage: G = SymmetricGroup(4)
sage: G([(1,2)])
(1,2)
sage: G([(1,2)])*G([(1,3)])*G([(1,4)])
(1,2,3,4)
2012-06-12 03:38:30 -0500 asked a question Polynomial representation of GF(7)?

Why sage would give me polynomial representation of GF(8), but not GF(7)?

sage: G = GF(8, 'x')
sage: G.list()
[0, x, x^2, x + 1, x^2 + x, x^2 + x + 1, x^2 + 1, 1]
sage: G = GF(7, 'x')
sage: G.list()
[0, 1, 2, 3, 4, 5, 6]

Maybe there's no such thing as polynomial represenation of GF(7)?