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April 10

x^y=y^x

is it possible to express y explicitly from the equation
x^y=y^x ?
ie is it possible to write y as a function of x from this equation? Rkr1991 (talk) 04:16, 10 April 2009 (UTC)[reply]

y = -x W(-ln(x)/x)/ln(x) where W is the Lambert W function. McKay (talk) 04:29, 10 April 2009 (UTC)[reply]

In LaTeX you can write it more clearly as

y={\frac {W(-\ln x/x)}{-\ln x/x}}

--CiaPan (talk) 05:33, 10 April 2009 (UTC)[reply]

And there is y=x, also. Any pair x>0, y>0 verifying x^y=y^x is of either this form or the other. --pma (talk) 08:58, 10 April 2009 (UTC)[reply]

In fact, y=x is a special case of the Lambert W solution, for appropriately chosen branches. Fredrik Johansson 23:56, 10 April 2009 (UTC)[reply]

Nontrivial Normal Abelian Subgroups of a Solvable Group

I'm working through Dummit&Footes book on algebra and cannot seem to figure out one of the exercises. They are asking for a proof that if G is solvable and has a nontrivial normal subgroup, then there is a nontrivial abelian subgroup A of H so that A is also a normal subgroup of G. I can demonstrate this in the case that G is finitely generated, or in the case that H has a nontrivial center; but I can't seem to show this in general. My best guess is that I'm over complicating things and that the solution involves some clever use of the fact that there is H' so H/H' is abelian and an application of one of the isomorphism theorems, but I'm not sure. Any help would be greatly appreciated; I'd actually be happier with a decent hint:) Thanks Phoenix1177 (talk) 09:35, 10 April 2009 (UTC)[reply]

one word hint: centres. 129.67.186.200 (talk) 09:51, 10 April 2009 (UTC)[reply]

Thank you. I'm assuming it's something along the lines H is solvable implies Z(H) is nontrivial, Z(H) char H implies Z(H) is a normal subgroup of G. I just can't seem to see why Z(H) needs to be nontrivial...though I'm sure I'll get it eventually. 66.202.66.78 (talk) 10:13, 10 April 2009 (UTC)[reply]

The symmetric groups S₃ and S₄ are solvable and have trivial centres, so this can't quite work. — Emil J. 10:35, 10 April 2009 (UTC)[reply]

I was thinking something might be wrong with the approach since it wasn't going anywhere fruitiful. The methods of approach I can think of would be to either use the fact that if 1 < A <...H < B <...G is a normal series with abelian factors for G, then A is a subgroup of H that is abelian, but this seems a long shot; or to use induction over the minimal length normal series with abelian factors, but there doesn't seem to be any guarentee that the series containing H will be minimal. I imagine that there is a rather obvious way to proving this, and I'm just not seeing it. Any further help would be greatly appreciated, I'm rather stuck:) 76.125.236.118 (talk) 13:09, 10 April 2009 (UTC)[reply]

What is H? If H = G, the result is easy (so in this case, given a solvable group G, you want to find a non-trivial Abelian normal subgroup of G given that G is not simple). Consider the minimal normal subgroup of G (this proof does not even require non-simplicity). Otherwise, what is H supposed to be? --PS T 13:28, 10 April 2009 (UTC)[reply]

By the way, are you familiar with the result that for solvable groups, all chief factors are Abelian p-groups? --PS T 13:37, 10 April 2009 (UTC)[reply]

As I understand it, H denotes the nontrivial normal subgroup of G assumed to exist in the beginning of the second sentence of the original post. That is, the problem is: given a solvable group G and its nontrivial normal subgroup H, find a nontrivial abelian subgroup of H which is normal in G. — Emil J. 13:43, 10 April 2009 (UTC)[reply]

In that case, consider my previous hint - consider the minimal normal subgroup of G (this subgroup is solvable as a group so what can you say about its commutator subgroup?). Furthermore, what is the relation between this subgroup and H? --PS T 22:11, 10 April 2009 (UTC)[reply]

The terms of the derived series of H are characteristic in H, so normal in G. The last non-identity term of the derived series is abelian. Some people use nontrivial to also mean "proper", but with this usage there need not be any such subgroup: take G to be a finite solvable group of composite order and H to be a minimal normal subgroup, then H contains no non-identity proper subgroups that are normal in G. JackSchmidt (talk) 14:24, 10 April 2009 (UTC)[reply]

If "non-trivial subgroups" are precisely those that are neither the trivial group or the whole group, then non-simplicity (which is one of the hypothesis mentioned) should do it (along with solvability, of course). Just consider the minimal normal subgroup of G (which is not G because of non-simplicity). But basically, I am still not totally sure as to what the OP is asking because of his undefined terminology so I can't be sure whether this answers his/her question. --PS T 04:15, 11 April 2009 (UTC)[reply]

The solvable group G = Z has no minimal normal subgroup. The OP has already handled the case of finitely generated solvable groups. Also the requirement is that the abelian G-normal subgroup is contained in a given normal subgroup H. JackSchmidt (talk) 15:57, 11 April 2009 (UTC)[reply]

Apologies - I was thinking of the case for finite groups. Thanks for correcting me. But for finite solvable groups that are not simple, it is always true that there exists a proper Abelian non-trivial normal subgroup (the minimal normal subgroup). --PS T 02:53, 12 April 2009 (UTC)[reply]

April 10

xy=yx

Nontrivial Normal Abelian Subgroups of a Solvable Group

x^y=y^x