# Transitive subgroups of the symmetric group

Let $S_n$ denote the symmetric group on $A=\{1,2,\ldots n\}$. A subgroup $G\subseteq S_n$ is said to be a transitive subgroup if the natural action on the set $A$ is transitive.

Q : Find all isomorphism classes of transitive subgroups of size n of $S_n$.

Let us try to find one subgroup .., well, $\mathbb{Z}_n$ generated by the $n$-cycle $(1 2 \ldots n)$. Clearly it is transitive, and in fact, more is true, namely, the stabilizer of any element is the trivial permutation.

Can we think of more ?

Let us look at $S_{2n}$ and look at $p=(1, 2, \ldots n)(n+1, \ldots, 2n)$ [this is the cycle notation of permutations]. $p$ has order $n$. Now if we can find a $q$ which somehow can make 1 go to n+1, then the subgroup generated by $p,q$ would be transitive.

In fact, we can find $q$ such that $qp=pq$ or rather $qpq^{-1}=p$. Just look at $q=(1, n+1)(2 , n+2)(3 , n+3)\ldots (n , 2n)$. Then $qpq^{-1}$ = $(1, 2 ,\ldots, n)(n+1, \ldots, 2n)=p$

The subgroup generated by $p,q$ is isomorphic to $\mathbb{Z}_n\times \mathbb{Z}_2$, so we see that $\mathbb{Z}_{n}\times \mathbb{Z}_{2}$ is also a transitive subgroup of size 2n in $S_{2n}$

In fact, more is true !

Claim : Let $G$ be ANY subgroup of size $n$. Then it can be embedded as a transitive subgroup of $S_n$.

Proof : Let $G=\{g_1,\ldots,g_n\}$ and let $S_n$ be the symmetric group on a set of size n, which we can assume to be G ! We have the following (usual) embedding of $G$ into $S_n$ got by the action of $G$ on itself by a modified left translation, namely

$f:G\to S_n$ which sends $g\to [g_i\to g^{-1}g_i]$, where $S_n$ is thought of as set bijections of G and itself.

1. The map $[g_i\to g^{-1}g_i]$ is a set bijection because $g^{-1}g_i=g^{-1}g_j\iff g_i=g_j$
2. $f$ is a group homomorphism – easy check. [This is why we need $g_i\to g^{-1}g_i$ instead of $g_i\to gg_i$
3. $f$ is a 1-1 function (or an embedding) because the map $[g_i\to g^{-1}g_i]$ is identity if and only if $g^{-1}g_i=g_i\forall g_i$ which happens if and only if $g=e$

So G can be identified as a subgroup of $S_n$ via $f$. Now we check that it is a transitive subgroup !

Any $g_i$ can be moved to $g_j$ by the element $a=g_ig_j^{-1}$ because $a\star g_i = a^{-1}g_i = g_jg_i^{-1}g_i=g_j$.

Infact, what is the stabilizer of any $g_i$ under this action of G ? It is the identity ! So this action of G  is in fact regular (or transtive+free = simply transitive, namely given $g_i$ and $g_j$, there is a unique element in G which takes one to the other)

# Zero divisors vs right and left cancellations

Let $R$ be a ring (not necessarily commutative) with $1$.

Left zero divisor

$a\in R$ is said to be a left zero divisor (LZD) if there is a $b\neq 0$ in $R$ such that $ab=0$.

Right zero divisor

$a\in R$ is said to be a right zero divisor (RZD) if there is a $b\neq 0$ in $R$ such that $ba=0$.

Zero divisor

$a\in R$ is said to be a zero divisor (ZD) if it is both a left and a right zero divisor.

Left cancellation :     $a\neq 0, ab=ac\implies b=c$.

Right cancellation :     $a\neq 0, ba=ca\implies b=c$.

Prove that R has no non-trivial zero divisors if and only if both right and left cancellations hold.

Proof : Right cancellation holds will imply there are no nontrivial right zero divisors ($ba=0\implies ba=0a \implies b=0.$). Similarly left cancellation holds will imply no nontrivial left zero divisors. So clearly no element can be a zero divisor except 0 (for it can’t even be a LZD or an RZD)

The other direction is more complicated.

1. Assume $x$ is a nontrivial LZD and not a ZD (so x is not 0).
2. So there is a $y\neq 0$ such that $xy=0$.
3. If $yx=0$ also, then $x$ is an RZD and hence a ZD, a contradiction
4. So $yx\neq 0$.
5. $(yx)y=yxy=y(xy)=0$ and $y\neq 0$,  so yx is an LZD
6. $(x)(yx)=(xy)x=0$ and $x\neq 0$, so yx is an RZD
7. So $yx$ is a nontrivial ($yx\neq 0$ by Step 4) ZD.

So we have shown that $x,y\neq 0, xy=0, x\neq ZD\implies yx$ is a nontrivial ZD.

Let us show left cancellation holds assuming no nontrivial ZD in R. If $a\neq 0, ab=ac$, this implies $a(b-c)=0$. If $b-c=0$, we are done, else put $x=a, y=b-c$. Thus $yx=(b-c)a$ is a nontrivial ZD, a contradiction.

Let us show right cancellation holds assuming no nontrivial ZD in R. If $a\neq 0, ba=ca$, this implies $(b-c)a=0$. If $b-c=0$, we are done, else put $y=a, x=b-c$ and they are both not ZD as R has no nontrivial ZD. Thus $yx=a(b-c)$ is a nontrivial ZD, a contradiction.

# Can an irreducible curve in a surface be the divisor of a function

Question : Let S be a projective surface and C, an irreducible curve on it. Show that C can’t be the divisor of a function.

Answer : A surface is a two dimensional scheme. I guess we assume S is defined over a field $k$. So S being projective means that the natural map $S\to Spec(k)$ is a projective morphism, in that it factors through the following closed immersion $i$:

$S\to^{i} \mathbb{P}^n_k\to Spec(k)$

Here $\mathbb{P}^n_k =Proj k[x_0,x_1,\ldots,x_n]$ and the map from this to $Spec(k)$ is given by gluing maps from the usual affine cover $D_{+}(x_i)\forall x_i\to Spec(k)$ where

$D_+(f)=\{P\in Proj k[x_0,\ldots,x_n]| f\not\in P\} \cong Spec$ degree 0 homogeneous elements of $k[x_0,\ldots,x_n]_{f}$

To check : Thus $D_{+}(x_i)\cong Spec k[\frac{x_0}{x_i},\ldots, \frac{x_n}{x_i}]$

which has a natural map to $Spec(k)$.

Now the ring of functions regular on S is $k^*$ because $S$ is projective. (I guess that for this you need that S is projective over a field, or atleast a nice enough affine scheme below, like an integral domain ??)

What is the divisor of a function ?

If $f\in k(S)^*$, the function field of $S$, then the divisor of $f$ is $div(f)=\sum v_{P}(f)$ where P runs over all codimension 1 points of $S$. This is the construction of a principal weil divisor.

In more detail, if $S$ is a nice enough scheme (noetherian, separated, all local rings of dimension one are regular), then at any codimension 1 point P (which will correspond to a height one prime ideal of the ring of any affine open subset containing it), the local ring will be a noetherian regular one dimensional local ring and hence a DVR. This DVR defines a valuation on $k(S)$ called $v_P$.

Coming back to our problem, let $C$ be an irreducible curve. If possible, let it be $div(f)$. Thus we get $div(f)=nP$ (for an irreducible curve can’t have more than 1 generic point in it. Note that since S is two dimensional, closure of any codimension 1 point is 1-dimensional)

We will show that this is not possible.

Note that we can assume wlog that $n>0$ (if $n<0$, just look at $\frac{1}{f}$. Of course if $n=0$, then it is not a curve !)

So this means $f\in m_{P,S}$, the maximal ideal of the local ring of S at P and in $O_{Q,S}$, the local ring of S at every other codimension 1 point Q.

Probably a stupid question : Is an integral projective scheme normal ? Ie, are the local rings integrally closed ?

Let us assume for a moment that our scheme S is normal. That is, all the local rings (whether at codimension 1 points or not are integrally closed in the function field). Let us look at any affine open set $Spec(R)$.

Claim : $R$ is also integrally closed.

Proof : $R=\cap R_M$ where M runs over maximal ideals of $R$

(Look at $a\in \cap R_M$ and look at $I=\{x\in R|xa\in R\}$, this is an ideal of R. If it is all of R, then $1\in I$ and hence $a.1=a\in R$. If it is a proper ideal, then let it be inside maximal ideal $N$. Since $a\in R_N$, $a=b/c$ for $c\not\in N$ and hence $c\not\in I$. But $ca=b\in R$ and hence $c$ has to be in $I$, a contradiction.)

We have assumed every local ring is normal, so $R_M$ are all integrally closed in $k(S)=K$. If $b\in K$ such that it is integral over $R$, then it is clearly integral over each $R_M$ (a bigger ring than $R$) and hence because of normality,  in each $R_M$. Thus $b\in \cap R_M=R$.

Finally we have one more theorem :

Thm : If R is a noetherian integrally closed domain, then $R=\cap R_P$ where P runs over height one prime ideals only.

Thus, our $f$ will be in $R$ since it is in each $R_P$.

This means that $f$ is regular on each affine open set of S, and hence it is a regular function on S, and hence a constant in $k$. In which case, it can’t have positive valuation at any codimension 1 point (for it is a unit in every local ring which are all k-algebras)

Hence we are done, modulo the question about whether S is a normal surface or not.

Counter example

If we drop the assumption that S is projective, then this is false. For example, take $S=\mathbb{A}^2=Spec k[x,y]$.

What are the height one primes here ?

Well  every prime ideal has an irreducible element in it. And the ideal generated by any irreducible element is a prime ideal because $k[x,y]$ is a UFD.

But the Krull dimension of $k[x,y]$ is 2, hence every height one prime looks like $(t)$ where $t$ is an irreducible polynomial in $k[x,y]$.

Let us look at $div(x)$. Since $x$ is irreducible, $x\in (t)$ implies $(t)=(x)$, thus $x$ is a unit in $k[x,y]_P$ for every height one prime ideal $P$ except $Q=(x)$. At $Q$, x has valuation 1.

Thus $div(x)=Q$.

# Finding zeros (or not) of quadratic forms over number fields

Q : Find a quadratic extension $K$ of $\mathbb{Q}$ where $u^2-3v^2-3w^2+9z^2$ has no nontrivial zero.

Non-solutions :

1. Over $\mathbb{Q}(\sqrt{2})$, $(0,\sqrt{2},1,1)$ is a solution.
2. Over $\mathbb{Q}(\sqrt{3})$, $(0,\sqrt{3},0,1)$ is a solution.

Solution : $\mathbb{Q}(\sqrt{7})$

Sketch of how to go about it :

• Finding a zero of $<1,-3,-3,9>$ is equivalent to finding a zero of $<1,-3,-3>$ . In fact, more generally, finding zero of $<1,-a,-b,ab>$ is equivalent to finding a zero of $<1,-a,-b>$ over any field $K$. A nice explanation of this fact is via the theory of quaternions. Our form is a really nice form, it is the norm form of the quaternions called $(a,b)$ which has generators : $i, j$ and relations : $i^2 = 3 =a , j^2 = 3 = b , ij=-ji= k$.
• A quaternion algebra is either a skew field (all elements have inverses) or it is $M_2(K)$, the 2×2 matrices.
• The norm is multiplicative, so it is a good test to check whether our quaternion is a skew field or not. The norm form has a nontrivial zero if and only if the quaternion is a matrix algebra.
• Another if and only if test for whether the quaternion $(a,b)$ is a matrix algebra over $K$ is to check that the form $<1,-a,-b>$ has a nontrivial zero.
• So from now on, we can assume our form is $<1,-3,-3>$.
• Hasse-Minkowski’s theorem : If q is a quadratic form over K, a number field, then q has a nontrivial zero if and only if it has a nontrivial zero over $K_v$, all the completions of K.
• For example, $\mathbb{Q}_p$ is the field of $p$-adic numbers, which is the completion of $Q$ at prime $p$. In fact, here we only need the trivial direction of this theorem, namely the fact that if there is a rational solution, there is a local solution.
1.  So we try to find a $K=\mathbb{Q}(\sqrt{d})$ and a completion $K_v$ such that our form $q=<1,-3,-3>$ has no solution over $K_v$.
2. In fact, let us try to imitate the proof of why this form has no solution over $\mathbb{Q}$ : If it does, it should have a nontrivial solution over $\mathbb{Q}_3$ (the 3 adics), so by clearing denominators, the solution will be in $\mathbb{Z}_3$ (the 3-adic integers).
3. Let $a^2 - 3b^2 - 3c^2 =0$ where $a,b,c$ are in $\mathbb{Z}_3$ and not all 0 simultaneously. Since $\mathbb{Z}_3$ is a DVR with parameter 3, this says that $3|a$, so divide by 3, the whole equation to get $3x^2 - b^2 - c^2 =0$ . Continue this process till you can assume there is some $u,v,w\in \mathbb{Z}_3$ such that $3u^2 - v^2 - w^2 =0 \in \mathbb{Z}_3$and 3 does not divide one of v and w.
4. $F_3$ is the field of 3 elements which is also $\mathbb{Z}_3$ modulo its maximal ideal $(3)$, so the previous solution gives a nontrivial solution $v^2+w^2=0 \in F_3$, which is a contradiction (you can check manually in $F_3$)
5. So if we find a $K= Q(\sqrt{d})$ and ring of integers $O_K$ and a prime $P$ of $O_K$ such that the residue field of $K_P$ (completion of K at P) is $F_3$, and 3 is not ramified at P, the same explanation as above shows that $u^2 - 3v^2 - 3 w^2$ has no solution over K.
6. For $K = Q(\sqrt{d})$ where d is 2 or 3 mod 4, $O_K = Z[\sqrt{d}]$, and I think we can directly see how any prime p in $\mathbb{Z}$ splits in K [look at factorization of $x^2-d$ in $F_p[x]$]
7. For $Q(\sqrt{7})$ , $3 = PQ$ where P and Q are distinct primes in $O_K$ (because $x^2-7 = x^2 - 1 = (x+1)(x-1) \in F_3[x]$ In fact $P=(3,\sqrt{7}+1), Q=(3,\sqrt{7}-1)$). So 3 does not get ramified in $K_P$. Also we see that residue field of $K_P = \mathbb{Z}[\sqrt{d}]/P = F_3$
8. In fact, there is a formula to check whether an odd prime $p$ is ramified, split or inert in $\mathbb{Q}(\sqrt{d})$ depending on whether $d$ is a square modulo $p$ or not.

# Corestriction of Brauer group of local field is an isomorphism

Brauer group of any field is the set of finite dimensional central simple algebras over that field modulo an equivalence relation, called the Brauer equivalence. It turns out to be an abelian group under the operation $\otimes$.  Every equivalence class has a unique central division algebra as a representative. So if we can classify all central division algebras over the field, we know its Brauer group.

Another way to describe the Brauer group of a field is via the profinite group cohomology of the Galois group of $K$:

$H^2(K, \mathbb{Q}/\mathbb{Z})$

Thus for (finite) field extensions $L/K$, we have the following morphisms :

1. Restriction : $Res : Br(K)\to Br(L)$, which at the central simple algebra level is just a base change  from K to L
2. Corestriction : $Cor : Br(L)\to Br(K)$

If K is an algebraically closed field, then there are no nontrivial finite dimensional central division algebras over it (If D is one such, then it must contain a nontrivial field extension of finite degree over K, which is not possible). Thus Br(K) = (0).

If K is a finite field, then there are still no nontrivial finite dimensional central division algebras over it (This is not as simple as the case of an algebraically closed field though, but the proof is still very elementary. One elegant way of showing this is to show K is a $C_1$ field. A $C_1$ field is any field such that any homogeneous polynomial with degree strictly less than the number of variables used has a nontrivial zero. And it is easy to see that there are no nontrivial central division algebras over such a field by simply observing that the reduced norm polynomial is a homogeneous polynomial with degree = $\sqrt{[D:K]}$ in $[D:K]$ number of variables where $D$ is the central division algebra over $C_1$ field $K$) Thus Br(K)=(0) here also.

The field $\mathbb{R}$ has exactly one nontrivial central division algebra over it. It is the famous Hamiltonian quaternions. Thus $Br(\mathbb{R})\cong \mathbb{Z}/{2\mathbb{Z}}$. I have forgotten how to prove this ..

We now come to $K$, a local field (examples are $\mathbb{Q}_p$ and $k((t))$ where $k$ is a finite field). Any finite extension of a local field is also a local field. Local class field theory tells us that $Br(K)\cong \mathbb{Q}/{\mathbb{Z}}$. In fact, it gives us the following very useful commutative diagram

$\begin{array}{ccc} Br(K) & \to^{Res} & Br(L) \\ \downarrow & . &\downarrow \\ \frac{\mathbb{Q}}{\mathbb{Z}} & \to ^{[L:K]} & \frac{\mathbb{Q}}{\mathbb{Z}} \end{array}$

The vertical arrows are isomorphisms. These maps which give you the isomorphism of the Brauer groups of local fields with $\frac{\mathbb{Q}}{\mathbb{Z}}$ are called the Hasse-invariants or simply $inv$.

To see the usefulness of this diagram, here is a nice application :

If $K$ is a local field and central division algebra $D/K$ has exponent $r$ (that is $r[D]=0\in BR(K)$), then any field $L/K$ of degree $r$ splits $D$.

Thus checking whether a field is a splitting field of a division algebra becomes as trivial as checking the degree of the field extension. The proof easily follows from the diagram :

$\begin{array}{ccc} D & \to^{Res} & D\otimes L \\ \downarrow & . &\downarrow \\ a & \to ^{[L:K]} & {ra=0} \end{array}$

Thus the restriction map $Res : Br(K)\to Br(L)$ can be simply described as multiplication map $\frac{\mathbb{Q}}{\mathbb{Z}} \to \frac{\mathbb{Q}}{\mathbb{Z}}$, sending $p/q \to np/q$ where $n=[L:K]$. This is clearly surjective.

Thus restriction map between Brauer groups of local fields is a surjection

What about the corestriction $Cor : Br(L)\to Br(K)$ ? It corresponds to the identity map of $\frac{\mathbb{Q}}{\mathbb{Z}}$ and in fact fits into the following commutative diagram :

$\begin{array}{ccc} Br(L) & \to^{Cor} & Br(K) \\ \downarrow & &\downarrow \\ \frac{\mathbb{Q}}{\mathbb{Z}} & \to ^{id} & \frac{\mathbb{Q}}{\mathbb{Z}} \end{array}$

Here is a proof :

Once we show that the above diagram commutes, then since the vertical maps are isomorphisms and so is the bottom map, $Cor$ will be an isomorphism.

Let $[D]\in Br(L)$. Since retsriction map is surjective, it comes from $[C]\in BR(K)$. The following is a well-known and very useful fact from finite group cohomology, namely if $H$ is a subgroup of $G$ of index $n$, then $Cor\circ Res : H^i(G,-)\to H^i(H,-)\to H^i(G,-)$ is simply multiplication by $n$.

Thus $Cor([D]) = Cor(Res([C])) = n[C]$ where $n=[L:K]$. Let $inv([C])=a$ and $inv([D])=b$. The following diagram will make things clear !

$\begin{array}{ccccc} C & \to^{Res} & D & \to^{Cor} & nC\\| & . & | & . & |\\a & \to & b & \to & na \end{array}$

# Both a semi and a direct product

We all know what a direct product of two groups A and B is. It is isomorphic to the group defined as follows : As a set it is  $A \times B$ and has the group operation  $(a,b)*(c,d)=(ac,bd)$ where ac and bd make sense inside A and B respectively. And clearly you identify $A\times\{1\}$ with A and $\{1\}\times B$ with B and both these are normal subgroups inside $A\times B$.

To make a semidirect product of two groups A and B, we need more information. Namely, we need a homomorphism of  $f: B\to Aut(A)$ which just means we need an action of B on A. This is what links the groups A and B together.

Then we define $A\rtimes_f B$ to be the semidirect product of A and B via f as follows :

• As a set, it is still $A\times B$
• $(a,b)*(c,d) = (a f(b)(c), bd)$

Notice that in the first coordinate, the map f is involved. A direct product $A\times B$ is just $A\rtimes_f B$ where $f:B\to Aut(A)$ sends every $b\to id_A$. Thus $f(b)=id_A$ and hence $(a,b)*(c,d)=(ac,bd)$.

It is generally taught in any group theory course that you can find an isomorphic copy of A as a normal subgroup in $A\rtimes_f B$ and an isomorphic copy of B as a subgroup (not neccesarily normal) in $A\rtimes B$.

Now here is a question :

Can a group be isomorphic to a nontrivial semidirect product of A and B (that is $f:B\to Aut(A)$ is not the trivial map) and also simultaneously be isomorphic to the direct product of A and B. In symbols, can you think of an example of A, B and f where

$A\rtimes_f B\equiv A\times B$

Note that we don’t demand that the isomorphism preserves A or B or anything like that.

OK, here’s an answer. Take A=B to be some nonabelian finite group and let $f: B\to Aut(A), b\to [a\to bab^{-1}]$. f is not the trivial map because there is some a, b such that $bab^{-1}\neq a$ because A=B is nonabelian.

$\alpha : T\rtimes_f A\to A\times A, (a,b)\leadsto (ab,b)$.

1. T is a group homomorphism because
1. $T((a,b)*(c,d))=T((abcb^{-1}, bd))= (abcb^{-1}bd, bd) = (abcd,bd)$
2. $T((a,b))T((c,d))=(ab,b)(cd,d)=(abcd,bd)$

2. T is 1-1 because $(ab,b)=(1,1)$ implies $b=1$ and $ab=1$ and hence $(a,b)=(1,1)$.
3. Since the domain and codomain are finite and have the same size, T is onto and hence is a group isomorphism.

Here’s the “picture” behind the proof [I am not sure whether there really is a mathematical analogy, but this picture helped me formalize the example]. The real plane is the span of the x and the y-axis. It is also the span of the line y=x and the x-axis… With this as an intuition, can we try to symbolize everything into groups ?

Well, look at $A \times A$. It has $A\times\{1\}=C$ as a normal subgroup. It has also the “diagonal” $D=\{(a,a)|a\in A\}$ as a subgroup which is not normal. Clearly C and D are isomorphic to A and even more clearly $C\cap D=\{(1,1)\}$. Also $CD=A\times A$ for any $(a,b)=(ab^{-1},1)(b,b)$. Thus $A\times A$ is the semidirect product (internal, if you will) of subgroups C and D ..