## 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.