## SL(2,IR) is the commutator subgroup of GL(2,IR)

Here is a proof of the above fact.

Let $N$ be the commutator subgroup of the general linear group $GL(2,\mathbb R)$; i.e.,

$N=\langle ABA^{-1}B^{-1}:A,B\in GL(2,\mathbb R)\rangle$.

First, it is clear that $N$ is contained in the special linear group $SL(2,\mathbb R)$, since $\det(ABA^{-1}B^{-1})=1$ for any $A,B\in GL(2,\mathbb R)$. Next, we claim that $N$ contains all matrices

$\begin{pmatrix} 1 & b\\ 0 & 1\end{pmatrix}$.

This follows from noting that

$\begin{pmatrix} 1 & b\\ 0 & 1\end{pmatrix}=\begin{pmatrix} 1 & b\\ 0 & b\end{pmatrix}\begin{pmatrix} 1 & 1\\ 0 & 1\end{pmatrix}\begin{pmatrix} 1 & b\\ 0 & b\end{pmatrix}^{-1}\begin{pmatrix} 1 & 1\\ 0 & 1\end{pmatrix}^{-1}$.

By taking transposes, it also follows that $N$ contains all matrices

$\begin{pmatrix} 1 & 0\\ c & 1\end{pmatrix}$.

Further, $N$ contains all matrices

$\begin{pmatrix} a & 0\\ 0 & 1/a\end{pmatrix}$

since

$\begin{pmatrix} a & 0\\ 0 & 1/a\end{pmatrix}=\begin{pmatrix} a & 0\\ 0 & 1\end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}\begin{pmatrix} a & 0\\ 0 & 1\end{pmatrix}^{-1}\begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}^{-1}$

for any $a\neq 0$.

Now let

$\begin{pmatrix} a & b\\ c & d\end{pmatrix}\in SL(2,\mathbb R)$.

Then $ad-bc=1$. Using the above results,

$\begin{pmatrix} a & b\\ c & d\end{pmatrix}=\begin{pmatrix} 1 & 0\\ c/a & 1\end{pmatrix}\begin{pmatrix} 1 & ab\\ 0 & 1\end{pmatrix}\begin{pmatrix} a & 0\\ 0 & 1/a\end{pmatrix}\in N$

if $a\neq 0$, and

$\begin{pmatrix} a & b\\ c & d\end{pmatrix}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}1&-\frac{d}{b}\\ 0&1\end{pmatrix}\begin{pmatrix}1&0\\ ab&1\end{pmatrix}\begin{pmatrix}1/b&0\\ 0&b\end{pmatrix}\in N$

if $b\neq 0$, and the latter since

\begin{aligned}\begin{pmatrix} 0 & -1\\ 1 & 0\end{pmatrix}=&\begin{pmatrix}x&y\\0&-x-y\end{pmatrix}\begin{pmatrix}-x-y&0\\ x&y\end{pmatrix}\begin{pmatrix}x&y\\0&-x-y\end{pmatrix}^{-1}\begin{pmatrix}-x-y&0\\ x&y\end{pmatrix}^{-1}\\ \in &N\end{aligned}

for any $x,y,x+y\neq 0$. Thus $SL(2,\mathbb R)\subseteq N$, i.e., $N=SL(2,\mathbb R)$.

Filed under Linear algebra

## Minkowski’s criterion

Here is a related post that I find interesting.

In linear algebra, Minkowski‘s criterion states the following.

Theorem (Minkowski’s Criterion). Let $A$ be an $n\times n$ matrix with real entries such that the diagonal entries are all positive, off diagonal entries are all negative, and the row sums are all positive. Then $\det(A)\neq 0$.

This is a nice criterion and is not very difficult to prove, but for a random matrix it is asking too much. To decide whether a matrix is singular one usually looks for a row/column consisting of zeros or adding up to zero. The following result gives sufficient conditions for this to work. Unfortunately, it does not generalise Minkowski’s result.

Theorem. Let $A$ be a $n\times n$ matrix with real entries such that its row sums are all $\ge 0$, its lower diagonal entries are $\ge 0$ and its upper diagonal entries are $\le 0$. Then $\det(A)=0$ if and only if $A$ has either a row consisting entirely of zeros or all the row sums equal to zero.

Proof. Suppose that $Ab=0$, where $b=(b_1,\dots,b_n)^T\neq 0$. Assume that $b_1\ge\cdots\ge b_n$. Then there exists $1\le m such that $a_{1,1}',\dots,a_{1,m}'\ge 0$ and $a_{1,m+1}',\dots,a_{1,n}'\le 0$. Hence

\begin{aligned}0=\sum_{j=1}^na_{1,j}'b_j&\ge b_m\sum_{j=1}^ma_{1,j}'+b_{m+1}\sum_{j=m+1}^na_{1,j}\\&\ge (b_m-b_{m+1})\sum_{j=1}^ma_{1,j}'\ge 0.\end{aligned}

So we must have (i) $b_1=\cdots=b_m$, (ii) $b_{m+1}=\cdots=b_n$, (iii) $a_{1,1}'+\cdots+a_{1,n}'=0$ and (iv) either $b_m=b_{m+1}$ or $a_{1,1}'+\cdots+a_{1,m}'=0$. These boil down to having either $b_1=\cdots=b_n$ or $a_{1,1}'=\cdots=a_{1,n}'=0$. Apply this argument to each row of $A$ to obtain the desired conclusion. $\square$

Filed under Linear algebra

## Simple cases of Jacobson’s theorem

A celebrated theorem of Jacobson states that

Theorem. Let $R$ be a ring, not necessarily containing $1$. If, for each $a\in R$ there exists a positive integer $n$ such that $a^n=a$, then $R$ is commutative.

This is a very strong and difficult result (although not very useful in practice). However, we can obtain some special cases via elementary means.

Proposition 1. Let $R$ be a ring such that for each $a\in R$ we have $a^2=a$. Then $R$ is commutative.

Proof. Let $a,b\in R$. Then $a+b=(a+b)^2=a^2+ab+ba+b^2=a+ab+ba+b$, i.e., $ab=-ba$. Again, $a-b=(a-b)^2=a^2-ab-ba+b^2=a-ab-ba+b$, i.e., $ab=-ba+2b$. Thus $2b=0$, i.e., $b=-b$ for each $b\in R$. Thus $ab=-ba=ba$, as desired. $\square$

The next case is already considerably harder.

Proposition 2. Let $R$ be a ring such that for each $a\in R$ we have $a^3=a$. Then $R$ is commutative.

Proof. Let $a,b\in R$. Then $a+b=(a+b)^3$ shows that

$(*)\qquad\qquad\qquad a^2b+aba+ba^2+ab^2+bab+b^2a=0$,

and $a-b=(a-b)^3$ shows that

$a^2b+aba+ba^2=ab^2+bab+b^2a$.

Hence

$(**)\qquad\qquad\qquad\qquad 2(a^2b+aba+ba^2)=0$

for all $a,b\in R$.

Plugging $a=b$ into $(**)$ gives $6a=0$, i.e., $3a=-3a$ for each $a\in R$.

Plugging $b=a^2$ into $(*)$ gives $3(a^2+a)=0$, i.e., $3a^2=3a$ for each $a\in R$. Replacing $a$ by $a+b$ gives $3(ab+ba)=0$, i.e., $3(ab-ba)=0$.

Also, multiplying $(**)$ by $a$ first on the left and then on the right and then subtracting the two gives $2(ab-ba)=0$.

From the last two paragraphs we conclude that $ab-ba=0$ for all $a,b\in R$. $\square$

Corollary. Let $R$ be a ring such that for each $a\in R$ we have $a^n=a$ for some $n\le 3$. Then $R$ is commutative.

Proof. Note that if $a^n=a$ for some $n\le 3$ then $a^3=a$. Hence the result follows by Proposition 2. $\square$

Filed under Algebra

## A nice group theory result

While working on some group theory problems today a friend and I came up with the following result.

Lemma. Let $H$ be a normal subgroup of a finite group $G$ such that $\gcd(|H|,|G/H|)=1$. If the order of $g\in G$ divides $|H|$, then $g\in H$.

Proof. Let $d$ be the order of $g$. Then the order $d'$ of $gH$ in $G/H$ divides both $d$ and $|G/H|$. But $\gcd(d,|G/H|)=1$. Hence $d'=1$, i.e., $gH=H$, i.e., $g\in H$. $\square$

Corollary 1. Let $H$ be a normal subgroup of a finite group $G$ such that $\gcd(|H|,|G/H|)=1$. If $K\le G$ such that $|K|$ divides $|H|$, then $K\le H$.

Proof. Apply the lemma to the elements of $K$. $\square$

Corollary 2. Let $H$ be a normal subgroup of a finite group $G$ such that $\gcd(|H|,|G/H|)=1$. Then $H$ is the unique subgroup of $G$ of order $|H|$.

Proof. Use Corollary 1. $\square$

Here is an example of the lemma in action.

Problem. Show that $S_4$ has no normal subgroup of order $8$ or $3$.

Solution. If $H$ is a normal subgroup of $S_4$ of order $8$, then $\gcd(|H|,|S_4/H|)=1$. Hence every element of order $2$ or $4$ in $S_4$ must lie in $H$. In particular, $(1\ 2),(1\ 2\ 3\ 4)\in H$. By a result in the previous post, $H=S_4$, a contradiction.

Likewise, if $H$ is a normal subgroup of $S_4$ of order $3$, then $H$ must contain every 3-cycle; in particular, $(1\ 2\ 3),(2\ 3\ 4)\in H$. Hence $(1\ 2\ 3)(2\ 3\ 4)=(1\ 2)(3\ 4)\in H$. But this has order 2, and $2\nmid 3$, a contradiction. $\square$

More generally, we can prove the following.

Corollary 3. $S_n$ for $n\ge 4$ has no non-trivial proper normal subgroup $H$ with $\gcd(|H|,|S_n/H|)=1$.

Proof. Suppose otherwise and let $d$ divide $|H|$. Then $H$ must contain all $d$-cycles. So if $|H|$ is even then taking $d=2$ gives $H=S_n$. If $|H|$ is odd, it contains the cycles $\sigma=(1\ \cdots\ d)$ and $\rho=(d\ \cdots\ 2\ n)$ for some $3\le d. Then $\sigma\rho=(1\ 2\ n)\in H$ has order 3. So $|H|$ contains all 3-cycles, i.e., $A_n\le H$. Since $A_n\le S_n$ is maximal, either $H=A_n$ or $H=S_n$, a contradiction. $\square$

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## Generating the symmetric group

It’s a fairly well-known fact that the symmetric group $S_n$ can be generated by the transposition $(1\ 2)$ and the $n$cycle $(1\ \cdots\ n)$. One way to prove it is as follows.

1. Show that the transpositions $(a\ b)$ for $a,b\in\{1,\dots,n\}$ generate $S_n$.
2. Show that any transposition $(a\ b)$ can be obtained from $(1\ 2)$ and $(1\ \cdots\ n)$.

We also need the following key lemma the proof of which is routine.

Lemma. $\rho (a\ b)\rho^{-1}=(\rho(a)\ \rho(b))$ for any $\rho\in S_n$ and $a,b\in\{1,\dots,n\}$.

(1) is easily proven by the observation that any permutation of $1,\dots,n$ can be obtained by swapping two elements at a time. (2) is a bit more interesting.

We first use the lemma to observe that any transposition of the form $(a\ a+1)$ can be obtained from $(1\ 2)$ upon repeated conjugation by $(1\ \cdots\ n)$. Now, since $(a\ b)=(b\ a)$, WLOG let $a. Using the lemma, conjugating $(a\ a+1)$ by $(a+1\ a+2)$ gives $(a\ a+2)$, conjugating $(a\ a+2)$ by $(a+2\ a+3)$ gives $(a\ a+3)$, etc. In this way we can eventually get $(a\ b)$. So we are done by (1).

This argument shows that $S_n$ can in fact be generated by $(a\ a+1)$ and $(1\ \cdots\ n)$ for any $a$.

Now let’s consider an arbitrary transposition $\tau$ and an $n$-cycle $\sigma$ in $S_n$. By relabeling $1,\dots,n$, we can assume that $\sigma=(1\ \cdots\ n)$ and $\tau=(a\ b)$ for $a. Note that $\sigma^{-a+1}\tau\sigma^{a-1}=(1\ c)$ where $c=b-a+1$, so WLOG $\tau=(1\ c)$. Then $\sigma^k\tau\sigma^{-k}=(1+k\ c+k)$ for each $k$. In particular, taking $k=c-1$ gives $\sigma^{c-1}\tau\sigma^{-c+1}=(c\ 2c-1)=:\rho$. Then $\rho\tau\rho^{-1}=(1\ 2c-1)$. Repeating this procedure produces $(1\ c+k(c-1))$ for $k=0,1,2,\dots$. Now $\{c+k(c-1):k=0,1,\dots,n-1\}$ is a complete set of residues mod $n$ if and only if $\gcd(c-1,n)=1$, i.e., $\gcd(b-a,n)=1$. So we’ve shown that

Theorem. Let $\tau=(a\ b)$ be a transposition and $\sigma=(c_1\ \cdots\ c_n)$ be an $n$-cycle in $S_n$. Then $\tau$ and $\sigma$ generate $S_n$ if and only if $\gcd(c_b-c_a,n)=1$.

In particular, $(a\ b)$ and $(1\ \cdots\ n)$ generate $S_n$ if and only if $\gcd(b-a,n)=1$.

Corollary. $S_p$ is generated by any transposition and any $p$-cycle for $p$ prime.

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