Wikipedia:Reference desk/Archives/Mathematics/2010 September 19

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September 19

Expression of a rational number in powers of p

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Let p be a fixed prime. Let x be a non zero rational number. Express x as: $x={\frac {p^{j}A}{B}},j,A\in Z,B\in N$ and gcd(A,B)=1, and p dividing neither A nor B. Let A/B = a (mod p). Now my book says that x can be expressed as $x=ap^{j}$ + higher order terms. I see that in the case when x is an integer, but fail to understand how it follows for rationals. Can someone explain this please. Thanks-Shahab (talk) 15:45, 19 September 2010 (UTC)

Higher order terms in what ? Does this mean higher powers of p ? And do the coefficients have to be integers in the range [0, p − 1], like a is ? If so, it sounds as if your book is describing the p-adic expansion of x. Gandalf61 (talk) 19:54, 19 September 2010 (UTC)

By higher order terms I assume it means powers of p which are greater then j, (j may be negative), because that is used later. About where the coefficients lie, I am not sure, but a is assumed non zero later, one more thing I have been puzzling about. -Shahab (talk) 03:12, 20 September 2010 (UTC)

Then, yes, this sounds like the p-adic expansion of x, which is

$x=p^{j}\sum _{k=0}^{\infty }a_{k}p^{k}$

where a₀ is an integer in the range [1, p-1] and the other a_k are integers in the range [0, p-1] - so the integers a_k can be thought of as the "digits" of the expansion. This infinite series does not converge under the standard metric on the real numbers. However, it does converge under the p-adic metric induced by the p-adic absolute value. This is all explained in more detail in our article on p-adic numbers. Gandalf61 (talk) 09:19, 20 September 2010 (UTC)

Thanks-Shahab (talk) 16:52, 20 September 2010 (UTC)

Is every permutation group decomposable into a group direct product of cyclic subgroups

The fundamental theorem of finite abelian groups states that every finite abelian group can be expressed as the direct sum of cyclic subgroups of prime-power order. However, this theorem does not apply to permutation groups, since they are not abelian. Is it true that every permutation group is decomposable into a group direct product of cyclic subgroups (of any order)? Hints to a proof or a counter-example would be great. Thanks! --Masatran (talk) 15:52, 19 September 2010 (UTC)

No. Any direct product of cyclic groups is abelian. Algebraist 15:53, 19 September 2010 (UTC)

Thanks for the answer. I found the Jordan-Hölder theorem, which applies in certain cases. --Masatran (talk) 19:56, 19 September 2010 (UTC)

It may be worthwhile to mention the related semi-direct product construction. While no permutation group on more than 2 letters can be the direct product of cyclic (or even abelian) groups (as Algebraist explains), in low-order cases, they do occur as semi-direct products of "well-behaved groups". For example, S₃, the permutation group on three letters, is the semidirect product of the cyclic group of order 2 acting on the cyclic group of order 3. Similarly, S₄, the permutation group on four letters, is the semi-direct product of a cyclic group of order 2 acting on the alternating group on 4 letters. This generalizes: the permutation group on n letters is the semidirect product of the cyclic group of order 2 acting on the alternating group on n letters, for all $n\geq 3$.

In fact, since S_n contains only one non-trivial proper normal subgroup for $n\geq 5$ (the alternating group on n letters, A_n), for $n\geq 5$, one cannot write S_n as a non-trivial semidirect product in a way distinct (up to isomorphism of the groups involved) to the one described in the previous paragraph. (Note that A_n is a simple group for all $n\geq 5$.)

There are other useful constructions in group theory other than the direct product and the semidirect product - for example, the wreath product is a particularly interesting example. To give one interesting application of the wreath product construction (in case you are interested), I quote the following remarkable theorem due to Yoshida:

“ Let G be a finite group, let P be a Sylow p-subgroup of G, and let N denote N_G(P). (Here N denotes the normalizer of P in G.) Then N controls p-transfer in G unless P has a homomorphic image isomorphic to the wreath product of two cyclic groups of prime order. ”

It is often useful to know when p-transfer is controlled and this is the reason Yoshida's theorem is so powerful. Hope that helps! PST 23:41, 19 September 2010 (UTC)

I'd like to modify my question. Is it true that every permutation group is decomposable into a group direct product of (not-necessarily cyclic) subgroups (of any order), each of which cannot be decomposed further this way? And what is the appropriate theoretical framework for doing this? --Masatran (talk) 14:49, 23 September 2010 (UTC)

braids, knots, and links

I'm thinking about a simple braid with three strands. The leftmost strand crosses over the middle strand, then the rightmost strand crosses over what is now the middle strand, then repeat,.... and repeat,.... and repeat....

The length of the braid is the number of such crossings. Now take the end of the braid and glue it strand-by-strand to the beginning of the braid. One end of a strand may get glued to its other end, or may get glued to the other end of a different strand; this depends on the number of crossings. The total number of strands after gluing—I'll call them "components"—is the number of cycles of the corresponding permutation of three elements. Each component will be either knotted or unknotted, and the several components will be either linked or unlinked, both depending on the number of crossings. And the particular type of knot or link will depend on the number of crossings. So I drew some crude—but I hope accurate—illustrations, and I get this:

0 crossings, 3 components, unlinked, unknotted.

1 crossing, 2 components, unlinked, unknotted.

2 crossings, 1 component, unknotted.

3 crossings, 2 components, linked, unknotted.

4 crossings, 1 component, knotted.

5 crossings, 2 components, linked, unknotted.

6 crossings, 3 components, Borromean links, unknotted.

I haven't (yet) gone beyond 6. The number of components must clearly be a periodic function of the number of crossings, but I don't know about the knotted or linked status or the types of knots or links; I'm guessing these could grow in complexity. Only in the case of 4 crossings did I see a knot, and it's not a simple overhand knot, but it's one I think I've seen a picture of within Wikipedia. The cases with 3 and 5 crossings gave identical results—in particular, the links with the simplest possible. This suggests you should be able to rearrange the strands from one of these cases to the other without cutting any strands.

What is known about this problem? Michael Hardy (talk) 18:29, 19 September 2010 (UTC)

...and now I see we have a (short) section on closed braids. So probably someone's thought this through. Michael Hardy (talk) 18:34, 19 September 2010 (UTC)

You may be interested in Wikipedia:WikiProject Knots. -- Wavelength (talk) 19:15, 19 September 2010 (UTC)

Braid group is the most appropriate article, you seem to be looking at the braid group on three strings B₃ which is isomorphic to the knot group of the trefoil knot – in particular, it is an infinite non-abelian group. This group has the presentation $\langle x,y\mid x^{2}=y^{3}\rangle$ or $\langle a,b\mid aba=bab\rangle$. --Salix (talk): 19:46, 19 September 2010 (UTC)

The knot for 4 crossings is the figure-eight knot if that is helpful. Rckrone (talk) 19:56, 19 September 2010 (UTC)

Definitely the figure-eight knot matches the picture I drew. I said 3 and 5 are the same link, but I now think I should have said those are different but 5 and 7 are the same link. Michael Hardy (talk) 22:12, 19 September 2010 (UTC)

I think 5 and 7 are different as well. 3 is the Hopf link, 5 is the Whitehead link and 7 is this thing [1]. Rckrone (talk) 22:07, 20 September 2010 (UTC)