## Category Archives: Group Theory

### Sherman-Morrison Formula

Suppose $\eta\in \mathbb{R}^n$ is a column vector and $M_{n\times n}$ is an invertible matrix.  Set $A=M+\eta \eta^T$, then

$A^{-1}=M^{-1}-\frac{M^{-1}\eta\eta^TM^{-1}}{1+\eta^T M^{-1}\eta}$

This formula has more general forms.

### Polynomial of prime degree and solvable transitive subgroup

$L$ is the group of transformations of $\mathbb{Z}_p$ of the form $x\to ax+b$, $a\neq 0$.
$H$ is all the translations $x\to x+b$.

$\mathbf{Lemma 1:}$ $|L|=p(p-1)$ and $|H|=p$, $H\lhd L$.

$\mathbf{Lemma 2:}$ $G$ is a group of transformation in $\mathbb{Z}_p$ containing the group $H$ of translations as normal subgroup. Show that $G$ is a subgroup of $L$, this means $L$ is the normalizer of $H$.
$\mathbf{Proof:}$ Suppose $\tau:x\to x+1$ and let $\eta\in G$, $\eta \tau\eta^{-1}$ has the form $x\to x+k$. Hence $\eta(x+1)=\eta\tau(x)=\eta(x)+k$. If $\eta(0)=b$, then $\eta(x)=kx+b$.
$\text{Q.E.D}\hfill \square$

$\mathbf{Lemma 3:}$ Any solvable transitive subgroup of $S_p$, $p$ a prime, is equivalent to a subgroup of $L$ containing the group of translations.

$\mathbf{Proof:}$ Suppose $G$ is this group. Then $G$ has a composition series
$\displaystyle G\rhd H_1\rhd H_2\rhd \cdots \rhd H_{n-1}\rhd H_n=1$
whose factors are all cyclic groups of prime order.
By lemma 4, we know that $H_{n-1}$ is also a solvable transitive subgroup of $S_p$. Since $H_{n-1}$ is cyclic of prime order, then it must be generated by some p-cycle, otherwise it can not be transitive.
Apply some conjugation to $G$ in $S_p$, we can assume $H_{n-1}$ is generated by $(12\cdots p)$. So $H_{n-1}$ is a translation group. We can identify the element of $G$ as the transformation of $\mathbb{Z}_p$. By lemma 2, we know that $G$ is equivalent to some subgroup of $L$.
$\text{Q.E.D}\hfill \square$
$\mathbf{Lemma 4:}$ Let $H$ be a normal nontrivial subgroup of a transitive subgroup $G$ of $S_n$ of transformations of $\{1,2,\cdots,n\}$. Show that all $H-$orbits have the same cardinality. Hence show that if $n=p$ is a prime, then $H$ is transitive.

$\mathbf{Theorem(Galois):}$ Let $f(x)\in F[x]$ be irreducible of prime degree over $F$ of characteristic 0, $E$ a splitting field over $F$ of $f(x)$. Show that $f(x)$ is solvable by radicals over $F$ if and only if $E=F(r_i,r_j)$ for any two roots $r_i,r_j$ of $f(x)$.

$\mathbf{Proof:}$ Since $F$ has characteristic 0, $f(x)$ is solvable by radicals if and only if $G=\text{Gal }E/F$ is a solvable subgroup of symmetric group $S_p$. $f(x)$ is irreducible if and only if $G$ is transitive.
(1) if $f$ is solvable by radicals, then $G$ is a solvable transitive subgroup of $S_p$.
By lemma 3, we know $G$ is equivalent to a subgroup of $L$ containing $H$. WLOG, assume $H\leq G\leq L$. The action of $G$ on roots looks like $r_i\to r_{ai+b}$, $a\neq 0$.
Consider $\text{Gal }E/F(r_i,r_j)$, $r_i\neq r_j$. $\forall\,\eta\in \text{Gal }E/F(r_i,r_j)$, $\exists\, a,b$ unique such that $i=ai+b$ and $j=aj+b$. Then $(a-1)(i-j)=0$. Since $i\neq j$, we get $a=1$ and $b=0$, which means $\eta=id$. So $E=F(r_i,r_j)$.
(2) If $E=F(r_i,r_j)$, then $p| |\text{Gal }E/F(r_i,r_j)|$. This Galois group contains a normal Sylow p-subgorup which is isomorphic to $H$. By lemma 2, we know $G$ is equivalent to a subgroup of $L$ which is solvable. So $G$ is solvable, $f(x)$ is solvable by radicals.
$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ Jacobson p262.