Modern Algebra: Groups, Rings, Modules, Fields

Let \(G\) be a group of order \(2^m \cdot 5\) with \(m \leq 1\text{,}\) and suppose by way of contradiction that \(G\) is simple. By Sylow’s Theorems we know the following: - \(n_{5}|2^m\) and \(n_5\equiv1\mod{5}\text{,}\) and - \(n_{2}|5\) and \(n_{2}\equiv1\mod{2}\text{,}\) so our options are \(3\) and \(5\text{.}\) Suppose its \(3\text{.}\) Let \(G\) act on \(\Syl_2(P)\) by conjugation, yielding the permutation representation homomorphism \(\rho:G\to S_{3}\text{.}\) The kernel of this homomorphism cannot be trivial as the conjugation action on Sylow subgroups is transitive by Part (2) of Sylow’s Theorems. Notice \(|G|\) does not divide the order of \(S_{3}\text{,}\) so our kernel cannot be all of \(G\text{,}\) making \(\ker(\rho)\) a nontrivial normal subgroup of \(G\text{.}\)

If \(n_2=5\text{,}\) then \(G\) must divide \(5!\text{,}\) meaning that \(m=1\text{.}\) However, by the cyclic subgroup generated by an element of order \(5\) has index \(2\) in \(G\text{,}\) making it normal.

Thus \(G\) cannot be simple.

Let \(p\) be any positive prime integer.

First, suppose \(p=2\text{.}\) Thus \(G\) is a group of order \(pq\text{,}\) making it abelian. So the only groups of order \(6\) are \(\Z_2\times\Z_3\) and \(\Z_6\text{.}\) The same applies when \(p=2\text{,}\) where the groups are \(\Z_3\times\Z_3\) and \(\Z_9\text{.}\)

Let \(P\) be a Sylow \(p\)-subgroup of \(G\text{,}\) and note that \([G:P]=3\text{,}\) the smallest prime dividing the order of \(G\text{,}\) making \(P\nsg G\text{.}\) Let \(Q\) denote a Sylow \(3\)-subgroup of \(G\text{.}\) As \(P\) and \(Q\) are groups of relatively prime order we have \(PQ=G\) and thus \(G\cong P\sdp_\rho Q\text{,}\) where \(\rho:Q\to\Aut(P)\text{.}\) Notice that since \(|P|=p\text{,}\) we have \(\Aut(P)\cong\Z^{\times}_p\cong\Z_{p-1}\text{.}\) Thus, by the First Isomorphism Theorem \(Q/\ker(\rho)\cong\im(\rho)\leq\Z_{p-1}.\) As \(Q\) has three elements, the kernel of \(\rho\) must be either all of \(Q\) or trivial. However, the order of the image must divide \(\Z_{p-1}\text{,}\) which is only possible when \(p\equiv 1 \mod 3\text{.}\) Thus when this is the case there are two groups of order \(3p\text{,}\) otherwise the kernel is always trivial and we have \(G=P\times Q\) as the only group.

Let \(F \subseteq L\) be a finite Galois field extension of degree \(45\text{.}\)

Note that as \(45=5\cdot 9\text{,}\) we see that the number of Sylow-\(5\) subgroups of \(G\) divides \(9\) and is congruent to \(1\mod{5}\text{.}\) Thus there is exactly one Sylow-\(5\) subgroup of \(G\text{,}\) which we denote \(H\text{.}\) By the Fundamental Theorem of Galois Theory CITEX, \(H\) corresponds to an intermediate field extension \(E\text{.}\) Note that as \(H\) has order 5, we see that \([G:H]=9\text{,}\) and thus \([L:E]=9\) as well. By the The Degree Formula we see that \([L:F]=[L:E][E:F]\text{.}\) As \([L:F]=45\) and \([L:E]=9\text{,}\) we see that \([E:F]=5\text{,}\) as desired. As \(E\) corresponds to the unique subgroup of \(G\) or order 5, we see that this extension must be unique as well.

Let \(L\) be the splitting field of the polynomial \(f(x)=x^7-18\) over \(\Q\text{.}\) First, note that \(f\) is irreducible in \(\Q[x]\) by Eisenstein’s Criterion (\(p=2\)). The roots of \(f\) are the following: - \(a_1=\sqrt[7]{18}e^{\frac{2\pi i}{7}}\text{,}\) - \(a_2=\sqrt[7]{18}e^{\frac{4\pi i}{7}}\text{,}\) - \(a_3=\sqrt[7]{18}e^{\frac{6\pi i}{7}}\text{,}\) - \(a_4=\sqrt[7]{18}e^{\frac{8\pi i}{7}}\text{,}\) - \(a_5=\sqrt[7]{18}e^{\frac{10\pi i}{7}}\text{,}\) - \(a_6=\sqrt[7]{18}e^{\frac{12\pi i}{7}}\text{,}\) and - \(a_7=\sqrt[7]{18}e^{\frac{14\pi i}{7}}=\sqrt[7]{18}\text{.}\) Let \(\z=e^{\frac{2\pi i}{7}}\text{,}\) which is the root of the \(7\th\) cyclotomic polynomial (which has degree \(6,\) and thus irreducible in \(\Q[x]\text{.}\) Let \(F=\Q(\z)\) and \(E=\Q(\sqrt[7]{18})\text{.}\) Thus we have \([F:\Q]=6\) and \([E:\Q]=7\text{.}\) By the \(\gcd\) irreducible argument we see that \([L:\Q]=42\text{.}\)

Let \(G=\Gal(L/\Q)\text{.}\) From Sylow’s Theorems we have the following: \(n_7=1\text{,}\) as \(7\not\mid6\text{,}\) \(n_3|14\) and \(n_3\equiv1\mod3\text{,}\) so \(n_3\) is either \(1\) or \(7\text{.}\)

Let \(F \sse L\) is a finite extension of fields and that the characteristic of \(F\) is \(p\text{,}\) where \(p\) is a prime, and suppose there exists an element \(\a\in L\setminus F\) such that \(\a^p \in F\text{.}\)

Consider the polynomial \(f(x)=x^{p}-\a^p\in F[x]\text{,}\) and notice that \(f'(x)=px^{p-1}=0\text{,}\) as we are in a field of characteristic \(p\text{.}\) However, this characteristic also yields \(x^p-\a^p=(x-\a)^p\text{.}\) As \(F\) is a field we have \(F[x]\) as a UFD, and thus as \(x-\a\) is irreducible in \(F[x]\) it is also prime. Therefore this is the unique factorization of \(f\) up to associates. If \(f\) was reducible \(F\) it would thus have to be divisible into power as \(x-\a\text{,}\) which will never be reducible as \(\a\not\in F\text{.}\) Thus \(f\) is irreducible in \(F[x]\text{,}\) making it the minimal polynomial of \(\a\text{.}\) However, if \(|\Aut(L/F )|= [L : F ]\) this would make \(L\) the splitting field of \(f\) over \(F\text{,}\) which it is not, given \(\a\not\in F\text{.}\) Thus \(|\Aut(L/F )| < [L : F ]\text{.}\)

Let \(R\) be a PID, \(S\) a multiplicatively closed subset of \(R\) such that \(0 \not\in S\text{,}\) and \(I\) an ideal in \(S\inv R\text{.}\) Consider \(I \cap R\text{,}\) which is an ideal in \(R\) and is thus generated by some \(x\in R\text{.}\)

First, notice that \(x^3+x+1\) is irreducible in \(\Z[x]\) by the Rational Root Test. CITEX Thus the possibilities are the following: - \(I=(3,x^3+x+1)\text{,}\) - \(\Z[x]\text{.}\) Let \(J\) be an ideal containing \(I\text{.}\)

\((\Rightarrow)\) Suppose that \(g\) is diagonalizable. Thus there exists a change of basis matrix \(A\) such that \(A\) is diagonal. As it is diagonal, its diagonal entries are the eigenvalues of \(g\text{,}\) and thus the roots of the minimal polynomial of \(g\text{.}\) Using row and column operations we can rearrange \(A\) so that all repeated linear factors are next to each other in the diagonal, for convenience.

We know that \(\mp_A(x)\) is the smallest monic polynomial that sends \(A\) to 0. Take all the distinct eigenvalues \(\lambda_i\) and consider \(q(x)=\prod(x-\lambda_i)\text{.}\)

We examine \(q(A)\text{.}\) It will be a product of matrices, one for each \(\lambda_i\text{.}\) First, take the first matrix in this product, \((A-\lambda_1I)\text{,}\) and note that it sends all \(\lambda_1\) in \(A\) to 0. Thus all of the rows and columns that contained a \(\lambda_1\) are now 0, and thus all these rows and columns will be 0 in the final product \(q(A)\text{.}\) As this is we set \(q(x)=\prod(x-\lambda_i)\) for all \(l\lambda_i\in A\text{,}\) we see that for each row and column in \(A\) there will exist a matrix \(A-\lambda_iI\) in the product \(q(A)\) such that the row and column will be 0. Thus the entire matrix will be 0, and \(q(A)=0\text{.}\)

Note that if any \(\lambda_i\) were excluded from \(q(x)\) there would exist a non-zero row and column for every matrix in the product, and thus \(q\) would not send \(A\) to 0. Thus \(q(x)\) is indeed the minimal polynomial of \(A\text{.}\) As \(q(x)=\prod(x-\lambda_i)\text{,}\) we see it does indeed factor into distinct linear terms.

\((\Leftarrow)\) Suppose the minimum polynomial of \(g\) factors completely into distinct linear factors, each of which has the form \(x-\lambda_i\) for some \(\lambda_i\in F\text{.}\) As each \(\lambda_i\) is distinct, each elementary divisor is of the form \((x-\lambda_i)^1\text{.}\)

We construct the Jordan Canonical Form of \(g\text{.}\) As the elementary divisors are linear the Jordan blocks are \(1\times1\) matrices, making the \(JCF\) a diagonal matrix. As the JCF is itself a change of basis matrix, we see that \(g\) is diagonalizable.