My Algebra Questions

1. Nov 12, 2007

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?

Last edited: Nov 12, 2007
2. Nov 12, 2007

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.

Last edited: Nov 12, 2007
3. Nov 12, 2007

JasonRox

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

4. Nov 12, 2007

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.

5. Nov 12, 2007

matt grime

6. Nov 12, 2007

JasonRox

OH! I get it now.

7. Nov 12, 2007

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?

Last edited: Nov 12, 2007
8. Nov 12, 2007

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.)

9. Nov 12, 2007

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 arguement 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.

Last edited: Nov 12, 2007
10. Nov 12, 2007

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.

11. Nov 12, 2007

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!

12. Nov 12, 2007

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.

13. Nov 12, 2007

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.

14. Nov 12, 2007

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.

15. Nov 12, 2007

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.

Any help on this part though?

16. Nov 12, 2007

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...

17. Nov 12, 2007

teleport

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

18. Nov 12, 2007

JasonRox

It's in my Group Theory textbook!

I want to skip it but I like understanding everything.

19. Nov 12, 2007

JasonRox

You're awesome help by the way.

20. Nov 12, 2007

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.