Suppose every subgroup of \(G\) is normal. Prove that given any positive divisor \(d\) of \(|G|\) there exists a subgroup of \(G\) of order \(d\text{.}\)
Give an example, with justification, of a finite group \(G\) and a positive divisor \(d\) of \(|G|\) such that \(G\) has no subgroup of order \(d\text{.}\)
Solution.
Let \(G\) be a finite group.
Suppose every subgroup of \(G\) is normal. Let \(d\) be a positive divisor of \(|G|\text{.}\) For every prime \(p\) dividing the order of \(G\) there is exactly one Sylow \(p\)-subgroup of \(G\text{,}\) given that Sylow \(p\)-subgroups are normal if and only if they are unique. Thus \(G\) can be written as a direct product of its Sylow \(p\)-subgroups. Given the prime factorization of \(d\text{,}\) the direct product of the Sylow \(p\)-subgroups such that \(p|d\) is 1. A subgroup of \(G\) as each Sylow \(p\)-subgroup is normal, and 2. A direct product of cyclic groups of relatively prime order, yielding an element of order \(d\text{.}\) The subgroup generated by this element is a subgroup of \(G\) of order \(d\text{.}\)
We take a peak at \(G=A_4\) with \(d=6\text{.}\) Suppose \(A_4\) had a subgroup of order \(6\text{,}\)\(H\text{.}\) First, note that \([G:H]=2\text{,}\) the smallest prime dividing the order of \(G\text{,}\) making \(H\) normal in \(G\text{.}\)
As there are eight \(3\)-cycles in \(G\text{,}\) there exists some \(3\)-cycle, \(\sigma\text{,}\) such that \(\s\not\in H\text{.}\) Consider then \(H, \s H,\) and \(\s^2 H\) in \(G/H\text{.}\) Since \(|G/H|=\frac{|G|}{|H|}=2\text{,}\) it must be the case that either \(\s^2H=\s H\) or \(\s^2H=H\text{.}\)
If \(\s^2 H=H\) then \(\s^2\in H\text{.}\) As \(|\s|=3\) we have \(\s^2=\s\inv\text{,}\) but as \(H\) is a subgroup this would mean \((\s\inv)\inv=\s\in H\text{,}\) which is not the case.
If \(\s^2 H=\s H\) then \(\s\inv\s^2\in H\text{,}\) but \(\s\inv\s^2=\s\text{,}\) and so we have a contradiction. Thus \(H\) cannot exist, and \(G\) has no subgroup of order \(6\text{.}\)
ActivityC.65.Problem 2.
Let \(G\) be a group of order \(p^n\) for some prime \(p\) acting on a finite set \(X\text{.}\)
Suppose \(p\) does not divide \(|X|\text{.}\) Prove that there exists some element of \(X\) fixed by all elements of \(G\text{.}\)
Suppose \(G\) acts faithfully on \(X\text{.}\) Prove that \(|X| \geq np\text{.}\)
Solution.
Let \(G\) be a group of order \(p^n\text{,}\) for some prime \(p\text{,}\) acting on a finite set \(X\text{.}\)
Suppose there is no element in \(X\) that is fixed by all elements of \(G\text{.}\)
By [cross-reference to target(s) "cor-orbit-stabilizer" missing or not unique] we know that \(|\Orb_G(x)|\big||G|\) for all \(x\in X\text{.}\) Thus every orbit under this action has an order dividing \(p^n\text{,}\) so either \(1\) or some positive power of \(p\text{.}\) However, as no element of \(X\) is fixed by every element of \(G\text{,}\) there exists no stabilizer which is all of \(G\text{,}\) and thus there exist no orbits that have order \(1\text{.}\)
Recall that the orbits of this action partition \(X\text{,}\) and thus \(|X|=\sum_{x\in X}|\Orb_G(x)|\text{.}\) As every orbit is divisible by \(p\text{,}\) so too must be \(X\text{.}\)
Suppose \(G\) acts faithfully on \(X\text{.}\) Thus the permutation representation homomorphism \(\rho:G\to\Perm(X)\) is injective. Let \(k\) denote the order of \(X\text{.}\) Then \(\Perm(X)\cong S_k\text{,}\) which as order \(k!\text{.}\) As \(G\) is injective, we see that \(|G|=|\im\rho|\leq S_k\text{,}\) and thus \(p^n|k!\text{.}\)
Thus \(p\) must show up in the factorization of \(k!\) at least \(n\) times, meaning that \(|X| \geq np\text{.}\)
ActivityC.66.Problem 3.
Prove that any group of order \(3^2\cdot 11\cdot 17\) is abelian.
Solution.
Let \(G\) be a group of order \(3^2\cdot 11\cdot 17\text{.}\) By Sylow’s Theorems we see the following:
\(n_{11}|153\) and \(n_{11}\equiv 1\mod{11}\text{,}\) and so \(n_{11}=1\text{.}\)
\(n_{17}|99\) and \(n_{17}\equiv 1\mod{17}\text{,}\) and so \(n_{17}=1\) as well. - \(n_3|187\) and \(n_3\cong 1\mod{3}\text{,}\) so actually \(n_3=1\) too. Thus the unique Sylow \(11\)-subgroup and Sylow \(17\)-subgroup, denoted \(P\) and \(Q\text{,}\) respectively, are normal in \(G\text{.}\)
As \(P\) and \(Q\) are normal in \(G\) and intersect trivially, we see that \(PQ\leq G\text{.}\) Let \(g\in G\) and consider \(gPQg\inv\text{.}\) Let \(pq\in PQ\) and notice \(g(pq)g\inv=gpg\inv gqg\inv\text{.}\) As \(p\in P\nsg G\) and \(q\in Q\nsg G\) we see \(gpg\inv\in P\) and \(gqg\inv\in Q\text{,}\) thus \(gpqg\inv\in PQ\text{,}\) making \(PQ\nsg G\text{.}\)
Let \(R\) be the unique Sylow \(3\)-subgroup, which has order \(9\text{.}\) As \(R\nsg G\) and intersects with \(PQ\) trivially, we see \(G=PQ\times R\text{,}\) a direct product of cyclic groups of relatively prime order, making \(G\) abelian.
SubsectionSection II: Rings, Modules, and Linear Algebra
ActivityC.67.Problem 4.
Let \(R\) be a commutative domain.
State the definition for \(R\) to be a Euclidean domain.
Prove that in a Euclidean domain every ideal is principal.
Let \(I\) be an ideal in \(R\text{,}\) and let \(S\) denote the set of norms for a nonzero generating set of \(I\text{.}\) If none exist then \(I=(0)\) then in it is principally generated, so we can assume there is some nonzero \(x\in I\text{,}\) meaning our set \(S\) is nonempty.
As \(S\) is a set of natural numbers bounded below, by the Least Upper Bound Axiom there exists a smallest element. Let \(b\) denote the nonzero element in \(S\) with smallest norm.
Suppose \(a\) and \(b\) are nonzero elements of the generating set for \(I\text{.}\) As \(R\) is an ED there exist \(q,r\in R\) such that \(a=qb+r\) with either \(r=0\) or \(N(r)<N(b)\text{.}\)
Notice that \(r=a-qb\text{,}\) where \(a\in I\) and \(-qb\in I\) by absorption. Thus \(r\in I\text{,}\) meaning that \(N(r)\geq N(b)\text{.}\) Thus \(r=0\text{.}\) Then \(a=qb\text{,}\) meaning that \(a\in (b)\text{.}\) Thus every element in the generating set for \(I\) is contained in \((b)\text{,}\) and thus \(I=(b)\text{,}\) making \(I\) principally generated.
ActivityC.68.Problem 5.
Let \(R\) be a commutative ring in which every element \(x\) satisfies \(x^n = x\) for some \(n > 1\text{.}\) Show that every prime ideal in R is maximal.
Solution.
Let \(R\) be a commutative ring in which every element \(x\) satisfies \(x^n = x\) for some \(n > 1\text{.}\) Let \(P\) be a prime ideal in \(R\text{.}\)
Let \(rP\in R/P\text{,}\) meaning \(r\not\in P\text{.}\) Then \((rP)^n=r^nP=rP\) for some \(n>1\text{.}\) As \(R/P\) is a domain we can cancel an \(r\) to see that \(r^{n-1}P=P\text{,}\) and thus \(rP\cdot r^{n-2}P=P\text{,}\) making \(rP\) a unit in \(R/P\text{.}\) Thus \(R/P\) is a field, making \(P\) maximal in \(R\text{.}\)
ActivityC.69.Problem 6.
Let \(n\) be a positive integer and let \(J = J_{2n}(0)\) be the Jordan block matrix of size \(2n \times 2n\) with eigenvalue 0 in \(M_{2n\times2n}(\C)\text{.}\)
Find the minimal polynomials for \(J\) and for \(J^2\text{,}\) with justification.
Find the Jordan canonical form of \(J^2\text{,}\) with justification.
Hint.
Consider the kernel of \(J^2\text{.}\)
Solution.
Let \(n\) be a positive integer and let \(J = J_{2n}(0)\) be the Jordan block matrix of size \(2n \times 2n\) with eigenvalue 0 in \(M_{2n\times2n}(\C)\text{.}\)
Notice that \(J\) is a triangular matrix with \(0\)’s along the diagonal. Thus \(\cp_J(x)=x^{2n}\text{,}\) the product of the diagonal entries of the matrix \(J-xI\text{.}\)
Squaring a triangular matrix moves everything one down (Proof?), and so \(x^{2n-1}\) should do it. Squaring \(J\) just moves us one closer, so \(x^{2n-2}\)
As \(\cp_{J^2}(x)=x^{2n}\text{,}\) the only roots of it are \(0\text{,}\) and thus these are the only elementary divisors of \(J^2\text{.}\)
SubsectionSection III: Fields and Galois Theory
ActivityC.70.Problem 7.
Let \(L/F\) be a field extension and let \(K := \{a\in L | a \text{ is algebraic over }F \}\text{.}\) Show that if \(L\) is algebraically closed, then \(K\) is algebraically closed.
Solution.
Let \(L/F\) be a field extension let \(K := \{a\in L | a \text{ is algebraic over }F \}\text{,}\) and suppose that \(L\) is algebraically closed. Thus every polynomial in \(L[x]\) has a root in \(L\text{.}\)
Let \(f\in K[x]\) and let \(\a\) be a root of \(f\text{.}\) Thus \(f=a_nx^n + \cdots + a_1 x + a_0\text{,}\) where each \(a_i\in K\text{.}\) By definition of \(K\text{,}\) each \(a_i\) is algebraic over \(F\text{.}\) Notice that \(f\in F(a_0,\dots,a_n)[x]\) as well, making \(\a\) algebraic over this extension as well.
As \(a_i\) is algebraic over \(F\) for all \(i\) and \(\a\) is algebraic over \(F(a_0,\dots,a_n)[x]\) we see that each step in this chain has finite degree by the The Degree Formula, \([F(a_0, \dots, a_{n}, \a): F]\) is finite and thus so is \([F(\a):F]\text{.}\) By the Theorem again, \(\a\) is algebraic over \(F\text{.}\) Thus \(\a\) is algebraic over \(F\text{,}\) hence \(\a\in K\text{,}\) making \(K\) algebraically closed.
ActivityC.71.Problem 8.
Let \(n\) be a positive integer and let \(p\) be a prime integer. Consider the polynomial
and define \(K\) to be the splitting field of \(q\) over \(\Z/p\text{.}\)
Let \(E\) denote the set of roots of \(q(x)\text{.}\) Let \(\a\in E\text{.}\) Thus \(\a^{p^n}-\a=0\text{.}\) By the Factor Theorem CITEX there can be at most \(p^n\) roots. Suppose \(\a=\b\) for roots \(\a,\b\text{.}\)
Since \(K\) is splitting field of \(q(x)\) over \(\mathbb{Z}/p[x]\text{,}\)\(q\) factors completely into linear factors over \(K[x]\text{,}\) and thus \(q(x)\) has \(p^n\) roots in \(K\) up to multiplicity. If we can show each root has multiplicity 1, then \(|E|=p^n\text{.}\) We will just show no roots of \(q\) are a root of \(q'\text{.}\)\(q=x^{p^n}-x\) so \(q'=p^nx^{p^n-1}-1=-1\text{,}\) which has no roots. So definitely none of \(a\) in \(E\) are roots of \(q'\text{.}\) Therefore each has multiplicity 1 and thus \(|E|=p^n\text{.}\)
ActivityC.72.Problem 9.
Consider \(f (x) = x^5-4x-2 \in \Q[x]\text{.}\) This polynomial has exactly three real roots, a fact that you may use without proof.
Show that \(f\) is irreducible in \(\Q[x]\text{.}\)
Let \(L\) be a splitting field of \(f\) over \(\Q\text{.}\) Show that \(L/\Q\) is a Galois extension with Galois group \(\Gal(L/\Q)\) isomorphic to the symmetric group \(S_5\text{.}\)
Solution.
Consider \(f (x) = x^5-4x-2 \in \Q[x]\text{.}\)
Using Eisenstein’s Criterion with \(p=2\) we see that \(f\) is indeed irreducible in \(\Q[x]\text{.}\)
Let \(L\) be a splitting field of \(f\) over \(\Q\text{.}\) As \(f\) is monic and irreducible, it is the minimum polynomial for some \(\a\in L\) such that \([\Q(\a):\Q]=5\text{.}\) As \(f\) has complex roots, we know that \(\Q(\a)\neq L\text{,}\) and is thus an intermediate field. By the Fundamental Theorem of Galois Theory there exists a subgroup \(H\) of \(\Gal(L/\Q)\) such that \(H=\Gal(\Q(\a)/\Q)\text{.}\) As \([\Q(\a):\Q]=5\) we know \(|H|=5\text{,}\) making \(H=\langle\s\rangle\) for some \(\s\in H\text{.}\) Thus \(\s\) is an element of order \(5\) in \(\Gal(L/\Q)\text{.}\) As \(f\) is a degree \(5\) polynomial, we know \(\Gal(L/\Q)\) is isomorphic to a subgroup of \(S_5\text{.}\) Thus \(\s\) must be a \(5\)-cycle.
Recall that \(f\) has exactly two complex roots. Thus \(\tau\text{,}\) the complex conjugation automorphism, has order \(2\text{,}\) making it a transposition. From the Gospel of Mark we were told that we did not need to prove that a transposition and an \(n\)-cycle generate all of \(S_n\text{,}\) and thus \(\Gal(L/\Q)\cong S_5\)
