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.