My Algebra Questions: Answers & Updates

[JasonRox]

My Algebra Questions (continuously updated)

Note: Continuously updated because I will post new questions in the thread. I will bold them so you spot them.

Ok, this isn't really a problem solving question. It's basically me asking "what is this?"

The paragraphs starts as follows...

The first mathematicians who studied group-theoretic problems, e.g., Lagrange, were concerned with the question: What happens to the polynomial g(x_1,...,x_n) if one permutes the variables?

My question is...

What is g(x_1,...,x_n)?

How to I write 6x^2 + 4x +1 as g(x_1,...,x_n)?

Permuting what variables?

 

[ircdan]

g(x_1, ..., x_n) is a polynomial in x_1, ..., x_n, it's an element of R[x_1, ..., x_n], where R is some ring, maybe a field, etc. So it's a polynomial in n indeterminates(ie "variables") x_1, x_2,..., x_n with coefficients in some ring.

To say we permute the variables means just that, we interchange them. Sometimes polynomials don't change when we do this, like say g(x_1, x_2) = x_1 + x_2, this is the same as g(x_2,x_1) = x_2 + x_1. These are called symmetric polynomials. This doesn't always work though of course, if g(x_1, x_2) = x_1 - x_2, then g(x_2, x_1) = x_2 - x_1 != g(x_1, x_2). In your example you only have 1 variable, 1 indeterminate, and it's x.These are studied in Galois Theory, I'll let others say more on this since I don't know enough about this topic.

 

[JasonRox]

So, g(x_1,x_2) can be 6x_1^2 + 4x_2 + 1?

 

[theperthvan]

Nope. Think of x_1 = x ... x_2 = y ... x_3 = z etcetera

So yours is just g(x), but for a polynomial g(x_1,x_2)=g(x,y) and could be like x^2-3y-14x^3+37 y^99. Then you switch x and y so give g(y,x)= y^2-3x-14y^3+37 x^99
which is different to g(x,y) so they aren't symmetric.

 

[matt grime]

i don't get perthvan's objection. Jason didn't ask about symmetric polys.

 

[JasonRox]

OH! I get it now.

 

[JasonRox]

The next part is...

Note: Summation and Product is from i=0 to n, where S is the summation and P is the product (the capital Pi looking thing).

If a polynomial f(x) = S a_ix^i has roots r_1,...,r_n, then each of the coeficients a_i of f(x) = a_n P (x - r_i) is a symmetric function of r_1,...,r_n.

Does this mean whichever order we choose to put r_1,...,r_n, we still get f(x) = g(x) (if g(x) is where we interchange say r_1 with r_2)?

Or is it, no matter how we interchange the coefficients, the roots don't change?

 

[JasonRox]

Now, here is a new question...

Prove that A_4 (the alternating group) is centerless.

Note: Centerless is when the center of the group only contains the identity element.

I have shown that for S_n, n>=3, it is centerless.

I can't think of a direct way to show that A_4 is centerless without using brute force. Going by brute force wouldn't take that long, but I'm curious to know if there is another way.

I'm pretty sure that A_n for n>=5 is centerless. (A_n, n>=5 is simple.)

 

[ircdan]

Suppose the center of A_4 is not trivial, set H = Z(A_4), then by lagrange the possible orders of H are 1, 2, 3, 4, 6, 12. If the order is 1, then H is the trivial group, contradiction. If the order is 12 then H = A_4 so A_4 is abelian, which is ridiculous.

So the possible orders of H are 2, 3, 4, and 6.

Note H is normal in G so we can play with the quotient group.

Now if |H| = 6(the same argument I'm about to do works if |H| = 3)

Consider A_4/H, this has order 2. Now let f be 3-cycle in A_4(A_4 has 8 3-cycles). Then (Hf)^3 = Hf^3 = H, so o(Hf) | 3, but by lagrange the order of Hf divides 2 also, so we must have o(Hf) = 1, so Hf = H, so f is in H, so H contains at least all 8 3 cycles + the identity, so H contains at least 9 elements, a contradiction.

You'll have to do the cases when |H| = 2 and |H| = 4 differently, shouldn't be too bad to do them directly. I have an idea but I will let you do it.
Note that if |H| = 4, then H = {1, (12)(34), (13)(24),(14)(23)} =~ klein 4 =~ C_2 x C_2.Also note the reason the I knew how to do the case when |H| = 6, is because A_4 is the smallest subgroup for which the converse of lagranges fails, ie, what I did above is one way to show A_4 has no subgroup of order 6, for if it did, call it H, then [A_4: H] = 2, so H is normal, and then repeat the above.

 

[teleport]

Actually for any non-abelian group G, the factor group G/Z(G) is non-cyclic, i.e. it definitely cannot have prime order. So the given |H| cannot be either 6 or 4, just by that statement.

 

[JasonRox]

I got a faster proof, but it relies on a theorem we used later.

It's as follows...

Theorem - If H is normal in A_n, and H contains a 3-cycle, then H = A_n.

Therefore, Z(G) cannot contain a 3-cycle because then that means H = A_n.

Therefore, |H| can only be 2 and 4.

Then by what teleport said, we are left with only 4. That is all the elements in A_4 that is not a 3-cycle. That group is the 4-V group. This group can easily be checked to be centerless.

Ok, but the problem with the above is that the proof and teleport's statement comes after that question. The next question is to prove teleport's statement, which I know I can do. (I think I already did. I did so many I forgot.)

Note: 4-V is the Klein 4. I didn't know that.

Note: That was a nice note at the bottom, ircdan.

Thanks for the help guys!

 

[JasonRox]

Of course I can still use what ircdan said, but then it still comes down to it being almost equivalent of just doing it by brute force.

 

[teleport]

For the cases |H| = 2, 4, all the elements except e, are product of two 2-cycles. Take one, and denote it by (ab)(cd). Then take a 3-cycle from A_4 - H, say (abc). Now check both do not commute when multiplied.

 

[teleport]

Actually it is much more simple. Because no product of 2-cycles commute with all the other elements of A_4 (in particular see my previous post) no 2-cycles will be in H. The same can be said of 3-cycles using the same previous sample calculation. That's it.

 

[JasonRox]

That's what I was thinking about, but it still seems equivalent to doing it by brute force because it simply comes down to solve like 6-7 products.

The next part is...

Note: Summation and Product is from i=0 to n, where S is the summation and P is the product (the capital Pi looking thing).

If a polynomial f(x) = S a_ix^i has roots r_1,...,r_n, then each of the coeficients a_i of f(x) = a_n P (x - r_i) is a symmetric function of r_1,...,r_n.

Does this mean whichever order we choose to put r_1,...,r_n, we still get f(x) = g(x) (if g(x) is where we interchange say r_1 with r_2)?

Or is it, no matter how we interchange the coefficients, the roots don't change?

Any help on this part though?

 

[teleport]

"That's what I was thinking about, but it still seems equivalent to doing it by brute force because it simply comes down to solve like 6-7 products."

No. Like I said, take any 2-cycles (ab)(cd). In A_4, (abc) exists. Then by showing (just these two) they do not commute, you have shown no 2-cycles are in H. Then you do a similar thing with 3-cycles. Take any (abc). In A_4, (ab)(cd) exists...

 

[teleport]

For the other question I have no clue. We still haven't covered rings, etc. in my class yet. Sorry :)

 

[JasonRox]

It's in my Group Theory textbook!

I want to skip it but I like understanding everything.

 

[JasonRox]

You're awesome help by the way.

 

[Kummer]

You can show that A_4 has trivial center by looking at is conjugacy classes. Two classes of permutations are in the same conjugacy class if and only if they have the same # of disjoint cycles and length. Now we need to look at the even permutations.

 

[Kummer]

Here is a nice group theory problem to practice with. Let G be a finite group not divisible by three and so that (ab)^3=a^3b^3 prove that it is abelian.

 

[JasonRox]

You can show that A_4 has trivial center by looking at is conjugacy classes. Two classes of permutations are in the same conjugacy class if and only if they have the same # of disjoint cycles and length. Now we need to look at the even permutations.

You'll find that A_4 has 3 conjugacy classes, the 3-cycle's (8 of them), the two 2-cycle's (3 of them), and the one identity.

Since there are no conjugacy classes with cardinality 1, we conclude that no element commutes with all other elements in A_4.

Now that is an answer!

Thanks for pointing that out.

 

[JasonRox]

Solving all these problems takes a long freaking time!

 

[Hurkyl]

The next part is...

Note: Summation and Product is from i=0 to n, where S is the summation and P is the product (the capital Pi looking thing).

If a polynomial f(x) = S a_ix^i has roots r_1,...,r_n, then each of the coeficients a_i of f(x) = a_n P (x - r_i) is a symmetric function of r_1,...,r_n.

Does this mean whichever order we choose to put r_1,...,r_n, we still get f(x) = g(x) (if g(x) is where we interchange say r_1 with r_2)?

Or is it, no matter how we interchange the coefficients, the roots don't change?

Define the polynomial-valued expression of n + 1 variables
F(a_n, r_1, \cdots, r_n)(x) := a_n \prod_{i=1}^n (x - r_i)

The "k-th coordinate of F" function can easily be seen to be a polynomial in a_n, r_1, \cdots, r_n.

F is symmetric in its last n arguments. e.g. if n = 2, then F(a, r, s) = F(a, s, r).


Things become interesting when you use a_n, r_1, \cdots, r_n as indeterminates! If my ring of coefficients is R, then I might want to work in the polynomial ring

R[a_n, r_1, \cdots, r_n]

 

[JasonRox]

Ok, that definitely helps!

Now, after reading the section, the study of polynomials sounds more interesting! I hope it stays that way.

I hated polynomials last year during some Ring Theory. I didn't learn a thing. I remember extension fields and all that jazz, but I just knew enough to get by. Hopefully that can change.

 

[Hurkyl]

Here's a cute little application of how some of these things fit together.

Suppose you want to factor the polynomial

f(x) = 2x^4 + 7x^3 + 7x^2 + 7x + 2.​

The rational root theorem fails to find any rational solutions, but what about two quadratics?

Notice that f has an obvious symmetry: it's its own reverse, and so

x^4 f(1/x) = f(x).​

In particular, this means that inversion is a permutation of its roots.


So, we proceed as follows.

Let E be the splitting field of f -- it's the field generated by Q and the four roots of f.

Let S be the symmetry of E given by inverting the roots of f. I.E.
S(a_1 r_1 + a_2 r_2 + a_3 r_3 + a_4 r_4) = <br /> \frac{a_1}{r_1} + \frac{a_2}{r_2} + \frac{a_3}{r_3} + \frac{a_4}{r_4}

The subfield of E fixed by S should be interesting. Clearly, r_1 is not an invariant of S, but f(r_1) is. Can you think of any other expressions involving r_1 that are invariant under S?


(pausing while you think)


Let s = r_1 + 1 / r_1. This is an invariant under S. This suggests that we should make the substitution y = x + 1/x. With a little bit of work, you can compute
x^{-2} f(x) = 2y^2 + 7 y + 3
which factors, as expected.

<br /> \begin{equation*}\begin{split}<br /> 2x^4 + 7x^3 + 7x^2 + 7x + 2 &amp;= x^2(2y^2 + 7 y + 3) = x^2(2y + 1)(y + 3)<br /> \\&amp;= (2x^2 + x + 2) (x^2 + 3x + 1)<br /> \end{split}<br /> \end{equation*}<br />

 

[matt grime]

Jason, some times it really helps to do some examples. But you know this. So when you ask 'is it that the roots don't change if I swap coefficients', you should think 'do I really mean to ask if x+2 and 2x+1 have the same roots?'

Obviously swapping the roots of a polynomial doesn't alter the polynomial - you're multiplying the same factors together just in a different order.

I only say this because it seems like there's too much maths in this thread rather than just examining what's going on and thinking about it.

 

[JasonRox]

Obviously swapping the roots of a polynomial doesn't alter the polynomial - you're multiplying the same factors together just in a different order.

Yes, I thought about that of course.

 

[teleport]

You're awesome help by the way.

Probably untrue, but awesome feedback. Thanks.

PF gives me a lot. Whenever I can, I'll give back.

 

[JasonRox]

I have a new question now. Be free to discuss previous problems if you like, or add extra notes regarding anything. I read all the posts and think about each one.

The question is...

Let X be a G-set, let x, y, e X, and let y = g*x (the group action) for some g e G. Prove that G_y = g G_x g^-1 (stabilizer); conclude that |G_y| = |G_x|.

I proved the second part without even using the first part. It's just a matter of using...

|O(x)| = [G:G_x] , where O(x) is the orbit of x, and showing O(x) = O(y).

Any help on starting the first part?