Modern Algebra: Groups, Rings, Modules, Fields

Let \(S=G/H\) and note that \(|S|=p\text{.}\) Let \(K\) denote the kernel of the Permutation Representation generated by \(G\) acting on \(S\) by left multiplication.

The First Isomorphism Theorem tells us that \(G/K\cong\im(\rho)\leq S_p\text{.}\) Thus \(|K|\big|p!\) by Lagrange’s Theorem. Let \(k\in K\text{.}\) Then \(kgH=gH\) for all \(g\in G\text{,}\) making \(k\in gHg\inv\) for all \(g\in G,\) including \(e\text{.}\) Thus \(K\nsg H\text{.}\)

This yields \([G:K]=[G:H][H:K]\) by Theorem 3.14. Let \(h=[H:K]\text{,}\) giving us \([G:K]=ph\text{.}\) As \(|K|\big|p!\) we have \(h|(p-1)!\text{,}\) so \(h<p\text{.}\) But \(p\) is the smallest prime dividing the order of \(G\text{,}\) and thus \(h=1\text{,}\) making \(H=K\) and \(H\nsg G\text{.}\)

Let \(G\) be a group of order \(385 = 5 · 7 · 11\text{.}\) By Sylow’s Theorems we know the \(n_{11}|35\) and \(n_11\equiv 1\mod{11}\text{,}\) and so \(n_{11}=1\text{,}\) making \(P\text{,}\) the unique Sylow \(11\)-subgroup of \(G\text{,}\) normal in \(G\) by Corollary 6.14. We also know \(n_7|55\) and \(n_7\equiv 1\mod{7}\text{,}\) so \(n_7=1\) as well.

Let \(P\) denote the unique Sylow \(7\)-subgroup, and let \(G\) act on \(P\) by conjugation. Thus \(\rho:G\to S_7\text{.}\) The First Isomorphism Theorem tells us that \(G/K\cong\im(\rho)\leq\Aut(P)\text{,}\) where \(K\) is the kernel of \(\rho\text{.}\) However, by [provisional cross-reference: cite] we know \(\Aut(P)=\Z_{7}^\times=\Z_6\text{,}\) meaning that the order of \(K\) must divide both \(385\) and \(6\) by [provisional cross-reference: cite], CITEX which cannot happen. Thus \(K\) must be trivial, meaning that \(gxg\inv=x\) for every \(g\in G\) and \(x\in P\text{,}\) making \(P\) a subgroup of \(Z(G)\) of order \(7\text{.}\)

Let \(G\) be a group and \(H\nsg G\text{.}\) Let \(x\in C_H(G)\text{,}\) and consider \(gxg\inv\text{.}\) Let \(h\in H\text{.}\) As \(H\nsg G\) we have \(h=gh'g\inv\) for some \(h\in H\text{,}\) and thus that 1. \(g\inv h=h'g\inv\) and 2. \(hg=gh'\text{.}\) Consider \(gxg\inv h\text{.}\) By (1), we see \(gxg\inv h=gxh'g\inv\text{.}\) As \(x\) commutes with everything in \(H\) we have \(gh'xg\inv\text{,}\) and by (2) we have \(hgxg\inv\text{.}\) Thus \(C_H(G)\nsg G\text{.}\)

Let \(G\) act on the left cosets of \(C_G(H)\) by left multiplication, giving rise to the permutation representation homomorphism \(\rho:G/C_G(H)\to\Aut(H)\text{.}\) By the First Isomorphism Theorem we see that \(G/C_G(H)\) is isomorphic to a subgroup of the automorphism group of \(H\text{.}\)

Let \(E/F\) be a finite Galois extension and let \(G\) be the Galois group of \(E/F\text{.}\) Suppose that \(E = F (\a)\) and let \(f(x)\) be the minimal polynomial of \(\a\) over \(F\text{.}\) Thus \(G\) acts on the roots of \(f\) faithfully. Additionally, as \(f\) is the minimal polynomial of \(\a\) it is irreducible, making the action transitive as well.

As \(E\) is Galois over \(F\) we know that \(f\) splits into linear factors, each of the form \(x-\b\text{,}\) where \(\b\) is a root of \(f\text{.}\) As our action is transitive, for every root \(\b\) there exists a \(\s\in G\) such that \(\s\cdot\a=b\text{,}\) or \(\s(\a)=\b\text{.}\)

Let \(F\) be a field and \(G\) a finite subgroup of order \(n\) the multiplicative group \(F^\times\text{.}\) Let \(k\) denote the LCM of all orders of elements in \(G\text{.}\) Notice that as \(g^n=1\) for all \(g\in G\) we have \(k\leq n\text{.}\)

However, as \(g^k=1\) for all \(g\in G\) we know that \(g\) is the root of the polynomial \(f(x)=x^k-1\) in \(F[x]\text{.}\) By the Factor Theorem there are thus at most \(k\) roots of \(f\text{,}\) but there are \(n\) distinct roots. Thus \(k=n\text{.}\)

Thus the LCM of all orders of elements in \(G\) is \(n\text{.}\) Notice that as \(G\) is abelian every Sylow \(p\)-subgroup of \(G\) is normal, and thus there is only one of each. This means that \(G\) can be written as a direct product of cyclic groups of relatively prime order; hence \(G\) is itself cyclic.

Let \(L\) be the algebraic closure of \(F\text{.}\) Thus \(A\) has a Jordan Canonical Form in \(L\text{.}\) For each Jordan block \(J_i\) in the JCF of \(A\text{,}\) let \(B_i\) denote the transpose of the identity matrix, and notice that \(B_iJ_iB_i\inv=J_i^T\text{.}\) As this is the case for every Jordan block, we see that the JCF of \(A\text{,}\) \(J\text{,}\) is similar to its transpose. As the \(A\) is similar to \(J\text{,}\) \(A^T\) is similar to \(J^T\text{,}\) and \(J\) is similar to \(J^T\text{,}\) we see that \(A\sim A^T\) in \(L\) by transitivity.

Suppose \(A\) and \(B\) are similar in \(\Mat_{n \times n}(L)\text{.}\) As \(A\) and \(B\) have entries in \(F\text{,}\) then they are both in \(\Mat_{n \times n}(F)\text{.}\) Thus there exist matrices \(C,D\in\Mat_{n \times n}(F)\) in RCF such that \(A\) is similar to \(C\) and that \(B\) is similar to \(D\text{.}\) However, \(A\) is similar to \(C\) and that \(B\) is similar to \(D\) in \(\Mat_{n \times n}(L)\) as well. Notice \(C\) and \(D\) are still in RCF. However, as the RCF is unique, this means that \(C=D\) in \(\Mat_{n \times n}(L)\text{,}\) making them equal in \(\Mat_{n \times n}(F)\) as well. Thus \(A\) is similar to \(B\text{,}\) as similarity is transitive.

This yields \(A\sim A^T\) in \(F\text{.}\)

First, note that \(\cp_A(x)=\det(A-xI)\text{,}\) and thus \(\cp_A(x)=\begin{bmatrix} -1-x & 3 & 0\\0 & 2-x & 0\\2 & 1 & -1-x\end{bmatrix}.\)

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