305 – Homework V

This is the last homework set of the term. It is due Friday, April 27, 2012, at the beginning of lecture, but I am fine collecting it during dead week, if that works better.

Recall that, given a group G, the commutator of x,y\in G is {}[x,y]=xyx^{-1}y^{-1}. The derived group G' is the subgroup of G generated by the commutators,

G'=\langle [x,y]\mid x,y\in G\rangle.

  • Suppose that G=S_n. In class we saw that G'\le A_n, and that equality holds for n\le 7. Determine (with a proof or a counterexample) whether equality holds for all n.

[Here is one possible approach: Can you write any product of two transpositions as a commutator? You may want to consider two cases, depending on whether the two transpositions are disjoint or not. Do these products generate A_n, or what else is needed?]

  • The book lists for all n\le 25 all groups (up to isomorphism) of order n, see pages 210-213. Pick a group G with 20\le |G|\le 25, and show that

    G'=\{[x,y]\mid x,y\in G\}.

(Note the difference with the definition above. Here we are saying that every element of G' is a commutator, rather than simply saying that it is a product of commutators, as in the definition.)

It is not true that for every G, the elements of G' are commutators. The smallest counterexample has order 96, so they are not that easy to identify.

  • As an extra credit problem, show that if G=S_n, then any element of G' is a commutator.

Here I sketch an argument (due to Phyllis Joan Cassidy) showing an explicit (infinite) counterexample. For additional references, you may want to see this MathOverflow question, or this Math.StackExchange question.

Suppose that f is a polynomial in x, that g is a polynomial in y, and that h is a polynomial in x and y — we are allowing h to only depend on one of the two variables, or to be constant. Similarly with f and g.

Define m(f,g,h) to be the matrix

\left[\begin{array}{ccc} 1&f(x)&h(x,y)\\ 0&1&g(y)\\ 0&0&1 \end{array}\right].

Let G be the set of all these matrices.

  • Show that the matrix inverse of m(f,g,h) is m(-f,-g,-h+fg).
  • Show that the product of m(f_1,g_1,h_1) and m(f_2,g_2,h_2) is the matrix m(f_1+f_2,g_1+g_2,h_1+h_2+f_1g_2), and conclude that G is a group (with matrix multiplication as the operation).
  • Show that the commutator [m(f_1,g_1,h_1),m(f_2,g_2,h_2)] is m(0,0,f_1g_2-f_2g_1).
  • Define H to be the (abelian) group of matrices of the form m(0,0,h) with h a polynomial in x and y. Note that the operation is particularly simple here: m(0,0,h_1)\cdot m(0,0,h_2)=m(0,0,h_1+h_2). Conclude that H is isomorphic to the (additive) group of polynomials in x,y. Also, show that G'\le H.
  • In fact, G'=H. To see this, we must show that, for any polynomial h in x,y, we have that m(0,0,h) is a product of commutators. Prove this by showing that, if \displaystyle h=\sum a_{ij} x^i y^j, then

    m(0,0,h)=\prod[m(a_{ij}x^i,0,0),m(0,y^j,0)].

    (As usual, you may want to try a few examples first to make sure you understand this formula.)

Now we get to the crucial part of the argument: We want to show that not every m(0,0,h) is a commutator. In fact, let’s show that, for any n, there is an h such that m(0,0,h) is not a product of n commutators:

  • Let \displaystyle h(x,y)=\sum_{i=0}^{2n+1}x^i y^i. From the formulas above, show that what we need to prove is that we cannothave an equality of the form (*)

    h(x,y)=\sum_{j=1}^n(f_j(x)g_j(y)-h_j(x)k_j(y)),

    for some polynomials f_j,g_j,h_j,k_j, with 1\le j\le n.

To prove this, argue by contradiction: Suppose we have an equation of this form. Consider both sides of the equation as polynomials in x whose “coefficients” are polynomials in y:

Write \displaystyle f_j(x)=\sum_i a_{ij}x^i and \displaystyle h_j(x)=\sum_i b_{ij}x^i, and compare the coefficients of 1,x,\dots,x^{2n+1} on the left and right of (*).

  • Check that we get, for each i=1,\dots,2n+1, an equality of the form \displaystyle \sum_{j=1}^n(a_{ij}g_j(y)-b_{ij}k_j(y))=y^i.
  • Now, we invoke some linear algebra to finish: The polynomials 1,y,\dots,y^{2n+1} are linearly independent. However, the equations above show that they belong to the space generated by the polynomials g_j(y),k_j(y), j=1,\dots, n. Explain why this is a contradiction.
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This entry was posted on Monday, April 9th, 2012 at 2:33 pm and is filed under 305: Abstract Algebra I. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

2 Responses to 305 – Homework V

  1. 305 – Derived subgroups of symmetric groups « Teaching blog says:
    April 11, 2012 at 2:04 am

    […] of the problems in the last homework set is to study the derived group of the symmetric group […]

    Reply
  2. 305 – Derived subgroups of symmetric groups « A kind of library says:
    August 26, 2012 at 10:42 pm

    […] of the problems in the last homework set is to study the derived group of the symmetric group […]

    Reply

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