Prove that every group of order \(p^2\) is abelian. (Provide as many details as you can, but you may use main theorems - such as Lagrange’s Theorem or the Class Equation - without proof.)
Prove that there exists a non-abelian group of order \(p^3\) for any prime \(p\text{.}\) (You may use facts about automorphism groups here without proof, although they should be clearly stated.)
Solution.
Show the center is non-trivial, and thus has order \(p\) or \(p^2\text{.}\) If \(p^2\) then \(Z(G)=G\) and we’re done. If \(|Z(G)|=p\text{,}\) then the Class Equation tells us \(p^2=p+\sum[G:C_G(g)]\text{.}\) Since \(|G|=p^2\text{,}\)\(C_G(g)=1,p\) for each \(g\not\in Z(G)\text{.}\) But there are \(p\) elements, so there is only one conjugacy class for these elements. But since \(g\not\in Z(G)\) there exists some \(x\in G\) such that \(gx\neq xg\text{,}\) and thus \(xgx\inv\neq g\text{.}\) Since \(x\) is not in the center and \(x\) is not in the conjugacy class of \(g\text{,}\) we have a contradiction. Alternatively, \(G/Z(G)\) has \(p\) elements, making it cyclic.
Consider \(H=C_p\) and \(K=C_{p^2}\) and the external semidirect product \(H\sdp_\rho K\text{,}\) where \(\rho: K\to\Aut(H)\text{.}\)
ActivityC.56.Problem 2.
Let \(G\) be a group of order \(2m\) for some odd integer \(m > 1\text{.}\) Let \(\Perm(G)\) be the group of permutations on the elements of \(G\) (i.e., the group of bijective functions from \(G\) to \(G\)). You may use without proof that \(\Perm(G)\cong S_{2m}\text{.}\)
Let \(f : G \to \Perm(G)\) be given by \(f (g) = λg\) for all \(g ∈ G\text{,}\) where \(λg : G \to G\) is defined by \(λg(a) = ga\) for all \(a ∈ G\text{.}\) Prove that \(f\) is an injective group homomorphism.
Prove that if \(g ∈ G\) is an element of order \(2\) then \(λg\) is a product of \(m\) transpositions.
Prove \(G\) contains a subgroup of index \(2\) and is consequently not simple. (Tip: Consider the inverse image under \(f\) of the subgroup of \(\Perm(G)\) consisting of the even permutations.)
Solution.
Let \(g,h\in G\text{.}\) Two functions \(\lambda_g,\lambda_h\) are equal iff they agree at all inputs, so let \(a\in G\text{.}\)\((f(g)f(h))(a)=f(g)\circ f(h)(a)=f(g)(\lambda_h(a))=f(g)(ha)=\lambda_g(ha)=gha=\lambda_{gh}(a)=f(gh)(a)\) so it is a homomorhpism.
To show it is injective consider \(ker(f)\text{.}\) Suppose \(g\in ker(f)\text{,}\) then \(f(g)=id_G\) so \(f(g)(a)=a\) for all \(a\in G\text{.}\)\(f(g)(a)=\lambda_g(a)=ga=a\) for all \(a\in A\text{.}\) This implies
so the kernel is trivial \(\iff\)\(f\) is injective.
Let \(g\in G\) such that \(|g|=2\implies g^2=e\text{.}\) Since \(f\) is a hom, we know \(g^2=e\implies f(g^2)=f(g)^2=id_G\) so its order is 2 as well. In a permuation group, elements of order 2 are products of transpositions. As \(g\neq e\) there are no fixed points so all \(2m\) points are moved and thus \(\lambda_g\) is the product of an odd number (\(m\)) of transpositions.
Consider \(f^{-1}(A_{2m})\) (we will use \(A_{2m}\) to represent the alternating group of \(Perm(G)\) since they are isomorphic. By Cauchy’s, CITEX there is an element of order 2 in \(G\) so \(f(G)\) which is a subgroup of \(Perm(G)\) lies exactly half within \(A_{2m}\) and half outside of it. So \(f^{-1}(A_{2m})\) which is a subgroup by properties of homomorhpisms, is half of \(G\) and thus has index 2. Therefore \(G\) is not simple.
ActivityC.57.Problem 3.
Let \(G\) be a finite group and \(p\) a prime dividing the order of \(G\text{.}\) Let \(P\) be a Sylow \(p\)-subgroup.
Suppose \(Q\) is a Sylow \(p\)-subgroup distinct from \(P\text{.}\) Prove that \(PQ\) is not a subgroup of \(G\text{.}\) (Recall that \(PQ = {pq | p ∈ P, q ∈ Q}\text{.}\))
Prove that \(N_G(P ) = N_G(N_G(P))\) where \(N_G(H)\) denotes the normalizer in \(G\) of a subgroup \(H\text{.}\)
Solution.
Let \(G\) be a finite group and \(p\) a prime dividing the order of \(G\text{.}\) Let \(P\) be a Sylow \(p\)-subgroup.
We proceed via the contrapositive. Let \(Q\) be a Sylow \(p\)-subgroup, and suppose \(PQ\leq G\text{.}\) Thus \(|PG|\big||G|\)
Note that \(P\) and \(Q\) both have \(p^k\) elements, where \(p^k\) is the largest power of \(p\) that divides the order of \(G\text{.}\) Recall \(|PQ|=\frac{|P||Q|}{|P\cap Q|}=\frac{p^{2k}}{|P\cap Q|}\text{.}\) As \(p^k\) is the largest power of \(p\) that divides the order of \(G\text{,}\) we see that \(|P\cap Q|\) must be at least \(p^k\text{.}\) Thus \(P=Q\text{.}\)
First, note that \(P\leq N_{G}(P)\leq N_G(N_G(P))\text{.}\) Note that \(N_G(P)\) is the largest subgroup of \(G\) such that \(P\) is normal in \(G\text{,}\) making \(P\) the only Sylow \(p\)-subgroup of \(N_G(P)\text{.}\) Let \(x\in N_G(N_G(P))\text{.}\) Notice that \(xPx\inv\leq xN_G(P)x\inv\text{,}\) but as \(x\in N_G(N_G(P))\) we have \(xN_G(P)x\inv=N_G(P)\text{.}\) As \(xPx\inv\) is a Sylow \(p\)-subgroup that is contained in \(N_G(P)\text{,}\) we see that \(xPx\inv=P\text{,}\) placing \(x\in N_G(P)\text{.}\) Thus \(N_G(P)=N_G(N_G(P))\text{.}\)
SubsectionSection II: Rings, Modules, and Linear Algebra
ActivityC.58.Problem 4.
Let \(R\) be a commutative ring with identity, and assume \(1\neq0\text{.}\) Let \(I\) and \(J\) be ideals such that \(I + J = R\text{.}\)
Prove \(IJ = I\cap J\text{.}\)
Prove the following special case of the Sunzi’s Remainder Theorem: There is an isomorphism of rings of the form \(R/(I \cap J)\cong R/I \times R/J\text{.}\)
Solution.
Let \(R\) be a commutative ring with identity, and assume \(1\neq0\text{.}\) Let \(I\) and \(J\) be ideals such that \(I + J = R\text{.}\)
Let \(x\in IJ\text{.}\) Thus \(x=\sum a_ib_j\text{,}\) where each \(a_i\in I\) and \(b_j\in J\text{.}\) As \(I\) and \(J\) are both ideals, each term in this sum is contained both in \(I\) and \(J\text{.}\) Thus, by absorption, \(x\in I\cap J\text{.}\) Hence \(IJ\subseteq I\cap J\text{.}\)
Let \(x\in I\cap J\text{.}\) Thus \(x\in I\) and \(x\in J\text{.}\) Note that as \(I+J=R\text{,}\) there exists some \(a\in I\) and \(b\in J\) such that \(a+b=1\text{.}\) So \(x=x1=xa+xb\text{.}\) As \(x\in I\cap J\) we see that \(x=ax+xb\text{,}\) with \(a,x\in I\) and \(b,x\in J\text{.}\) Thus \(x\in IJ\text{,}\) yielding \(IJ=I\cap J\text{.}\)
Let \(f:R\to R/I\times R/J\) be defined by \(f(x)=(x+I,x+J)\text{.}\)
Notice that if \(x\in I\cap J\text{,}\) we have \(f(x)=(I,J)\text{,}\) and so \(x\in\ker(f)\text{.}\) Let \(x\in\ker(f)\text{.}\) Thus \(x+I=I\) and \(x+J=J\text{,}\) and so \(x\in I\) and \(x\in J\text{.}\) Hence \(x\in I\cap J\text{,}\) and so \(\ker(f)=I\cap J\text{.}\)
Let \((a+I,b+J)\in R/I\times R/J\text{.}\) As \(R=I+J\text{,}\) we can write \(a\) and \(b\) as \(a_i+a_j\) and \(b_i+b_j\text{.}\) However, as \(a_i\in I\) and \(b_j\in J\text{,}\) we have \((a+I,b+J)=(a_j+I, b_i+J)\text{.}\)
Consider the element \(x=(a_j+b_i)\in R\text{,}\) and observe
Thus \(f\) is a surjective homomorphism. Hence, by the First Isomorphism Theorem, we see \(R/(I \cap J)\cong R/I \times R/J\text{.}\)
ActivityC.59.Problem 5 (*).
Let \(R\) be a non-zero, unital ring, and let \(R^m\) and \(R^n\) be the standard free left \(R\)-modules of finite rank \(m\) and \(n\text{.}\) Assume there is an isomorphism of \(R\)-modules \(R^m \cong R^n.\)
Prove that if \(R\) is commutative then \(m = n\text{.}\) You may assume without justification that this holds in the special case when \(R\) is a field.
Show, by example, than \(m\) need not equal \(n\) if \(R\) is not assumed to be commutative.
Solution.
Let \(I\) be a maximal ideal in \(R\) and consider the residue field \(R/I\text{.}\) As \(R^m\cong R^n\text{,}\) we can tensor both sides to see \(R^m\tensor I\cong R^n\tensor I\text{,}\) and so \(R^m/IR^m\cong R^n/IR^n\text{.}\) As both of these are finitely dimensional vector spaces, they must have the same rank, and thus \(m=n\text{.}\)
In the HW we showed for \(R=\text{End}_{\mathbb{Z}}(\mathbb{Z}^{\infty})\) that \(R\cong R^n\) for all \(n\geq 1\text{.}\)
ActivityC.60.Problem 6.
Find all the ideals of \(\Z[x]\) that contain \((6, x^2 + x + 1)\text{,}\) the ideal of \(\Z[x]\) generated by \(6\) and \(x^2 + x + 1\text{.}\) For each such ideal, give an explicit list of generators, and determine whether the ideal is prime, maximal, or neither.
Solution.
First, notice that \(x^2+x+1\) is irreducible in \(\Z[x]\) as it is a cyclotomic polynomial. Thus the possibilities are the following: - \(I=(3,x^2+x+1)\text{,}\) - \(J=(2,x^2+x+1)\text{,}\) - \(L=(6,x^2+x+1)\text{,}\) - \(\Z[x]\text{.}\) Notice that \(\Z[x]=(1)\text{,}\) which is neither prime nor maximal, as those ideals must be proper. The ideal \(L\) is not maximal as it is contained in both \(I\) and \(J\text{,}\) nor is it prime, given that \(3\cdot2\in L\) but neither \(3\) nor \(2\) are in \(L\text{.}\)
Both \(I\) and \(J\) are generated with irreducible elements, making them irreducible ideals and thus maximal.
SubsectionSection III: Fields and Galois Theory
ActivityC.61.Problem 7.
Suppose \(F\) is any field. Recall that a square matrix \(A\) with entries in \(F\) is nilpotent if \(A^j = 0\) for some positive integer j.
Prove that if \(A\in Mat_{n\times n}(F)\) and \(A\) is nilpotent, then \(A^n = 0\text{.}\)
Find, with justification, the number of similarity classes of \(5 \times 5\) nilpotent matrices with entries in \(F\text{.}\)
Solution.
Let \(F\) be a field.
Assume \(A\in \text{Mat}_{n\times n}(F)\) is nilpotent. Thus there exists a positive integer \(l\) such that \(A^l=0\text{.}\) Thus the set of positive integers \(S=\{z\in \mathbb{Z}|A^z=0\}\) is nonempty, so by the well-ordering-principle we can choose a smallest element of \(S\text{,}\) call it \(m\text{.}\) hus \(A^m=0\) and for each positive integer \(p<m\) we have \(A^p\neq 0\text{.}\)
Thus we have two cases (1) \(m\leq n\) and (2) \(n<m\text{.}\) Case (1) is straightforward. If \(m\leq n\text{,}\) then \(n-m\geq 0\) and we can consider
For case (2): Observe that \(p(x)=x^m\in F[x]\) is in \(I=\{q(x)|q(A)=0\}\) the ideal generated by the minimal polynomial of \(A\text{,}\)\(\text{MinPoly}_A(x)\text{.}\) This tells us that \(\text{MinPoly}_A(x)\vert x^m\implies \text{MinPoly}_A(x)=x^k\) for some \(1\leq k\leq m\) since these are the only monic divisors of \(x^m\text{.}\) Next we know that \(\deg(\text{CharPoly}_A(x))=n\) and also that \(\text{CharPoly}_A(x)\in I\) by Cayley-Hamilton. So \(\text{MinPoly}_A(x)\vert \text{CharPoly}_A(x)\) and therefore degree of the minimal polynomial is lesser than or equal to \(n\text{.}\) Which gives us the following inequality
\begin{equation*}
1\leq k\leq n<m
\end{equation*}
and since \(\text{MinPoly}_A(A)=0\) this is a contradiction since \(m\) is the smallest. So the second case is not possible, and therefore \(m\leq n\) and by case 1 \(A^n=0\text{.}\)
Let \(A\) be a nilpotent \(5\times 5\) matrix. There exists some \(m\in\N\) such that \(A^m=0\text{.}\) Let \(\lambda\) be an eigenvalue of \(A\text{,}\) and so \(\lambda^m\) is an eigenvalue for \(A^m=0\text{.}\) As \(F\) is a field and thus an integral domain, we see that \(\lambda^n=0\) implies that \(\lambda=0\) as well.
As this holds in the algebraic closure of \(F\) as well, we see that when factored into linear terms all the \(\lambda=0\text{.}\) Thus \(\cp_A(x)=\cp_B(x)=x^5\text{.}\)
The possible invariant factors involving \(x^5\) are 1. \(\{x,x,x,x,x\}\text{,}\) 2. \(\{x,x,x,x^2\}\text{,}\) 3. \(\{x,x,x^3\}\text{,}\) 4. \(\{x,x^2,x^2\}\text{,}\) 5. \(\{x,x^4\}\) 6. \(\{x^2,x^3\}\) 7. \(\{x^5\}\) However, in the first case the rank of \(A\) would be \(5\text{,}\) making it invertible, contradicting the fact that 0 is an eigenvalue of \(A\text{.}\) Thus we need only consider the latter cases.
ActivityC.62.Problem 8.
Let \(L\) be the splitting field over \(\Q\) of the polynomial \(f(x) = x^5 + 2 \in \Q[x].\)
Find, with justification, \([L : \Q]\text{.}\)
Note that \(w=-\sqrt[5]{2}\) (where \(\sqrt[5]{2}\) is the unique positive, real fifth root of \(2\)) is one of the roots of \(f (x)\text{.}\) Prove \(\Aut(\Q(w)/\Q)\) is the trivial group.
Solution.
Let \(L\) be the splitting field over \(\Q\) of the polynomial \(f(x) = x^5 + 2 \in \Q[x].\)
Note that \(f\) is irreducible in \(\Q[x]\) by Eisenstein’s Criterion (\(p=2\)). The roots of \(f\) are
As \(f\) is monic and irreducible it is the minimum polynomial of \(\sqrt[5]{2}\in\Q[x]\text{.}\) Let \(F=\Q(\sqrt[5]{2})\text{,}\) and notice \([F:\Q]=5\text{.}\) Note that \(F\subseteq\R\text{.}\)
Let \(\z=e^{\frac{\pi i}{5}}\) be a primitive \(10\th\) root of unity. This is the root a cyclotomic polynomial of degree \(9\) which is irreducible in \(\Q[x]\text{.}\) As this has relatively prime order to \(F\) it is irreducible there as well. Let \(K=F(\z)\text{,}\) and notice \([K:\Q]=40\text{.}\) Notice that \(\a_1\cdot\a_3\inv=\z\text{,}\) and so \(L=K\text{,}\) completing the proof.
\(\mathbb{Q}(-\sqrt[5]{2})\subset \mathbb{R}\) a strict containtment, so \(\zeta_5\not\in \mathbb{Q}(-\sqrt[5]{2})\text{.}\) This means only one root appears in \(\mathbb{Q}(w)\) and since for \(g\in \operatorname{Aut}(\mathbb{Q}(w)/\mathbb{Q})\text{,}\) if \(a\) is a root of a polynomial with rational coefficients(\(p(x)\in \mathbb{Q}[x]\)), then \(p(g(a))\) must also be a root. But \(g(w)\) must be in \(\mathbb{Q}(w)\) so it must be fixed and we find that there’s just the identity automorphism.
ActivityC.63.Problem 9.
Assume \(F\) is field and let \(f(x)\in F [x]\text{.}\) Recall that \(f (x)\) is separable if \(f (x)\) has no repeated roots in an algebraic closure of \(F\text{.}\)
Assume \(\char(F) = 0\text{.}\) Prove that \(f(x)\) is separable if and only if the irreducible factorization of \(f(x)\) in \(F[x]\) has no repeated factors.
Fix a prime integer \(p\text{,}\) let \(\F_p\) be the field with \(p\) elements, and let \(F\) be the field of fractions of the polynomial ring \(\F_{p}[y]\text{.}\) Prove \(x^p-y\) is irreducible in \(F[x]\) but not separable.
Solution.
\((\Rightarrow)\) Suppose that \(f\) is not separable, so \(f\) has a repeated root in \(\overline{F}\text{,}\) which we denote \(\alpha\text{.}\) So \(x-\alpha\) is a factor of \(f(x)\text{.}\) By Corollary 2.96, CITEX \(F(\alpha)\) is separable, so the minimal polynomial of \(\alpha\) in \(F[x]\) has no repeated root in \(\overline{F}\text{.}\) As \(f\) does have a repeated root (by supposition) it cannot be the minimum polynomial of \(\alpha\text{.}\) Thus \(f(x)=\mp_{\alpha,F[x]}(x)g(x)\) for some \(g(x)\in F[x]\) such that \(g(x)\) has \(\alpha\) as a root, otherwise \(f\) would not obtain its repeated root. However, this means that \(\mp_{\alpha,F[x]}(x)\big|g(x)\text{,}\) meaning that \(g(x)\) has \(x-\alpha\) as a factor as well. Thus we see that \(x-\alpha\) is a repeated factor of \(f(x)\text{,}\) one from the minimum polynomial, one from \(g(x)\text{.}\)
\((\Leftarrow)\) Suppose that the prime factorization of \(f(x)\) in \(F[x]\) admits a repeated factor. Thus there exists some prime (and thus irreducible) \(g(x)\) such that \(g(x)g(x)|f(x)\text{.}\) However, \(g\) has a root \(\alpha\) in \(\overline{F}\text{,}\) so in \(F(\alpha)\) we see that \(f(x)\) has \(\alpha\) as a root as well, as \(g(a)=0\text{.}\) But since \(g(x)\) has factor \(x-\alpha\text{,}\) it shows up twice in the factorization of \(f\) because \(g(x)g(x)|f(x)\text{.}\) So \(\alpha\) has multiplicity at least 2, so \(f\) is not separable.
Let \(y,z\) be indeterminants, \(F=\F_{p}[y]\text{,}\) and \(L=\F_{p}[z]\) such that \(z^3=y\) (as seen in Example 2.78). Note then that \(z^3\) is a root of the polynomial \(x^p-y\in F[x]\text{.}\)
Moreover, since \(F\) is the field of fractions of the PID \(R=\F_{p}[y]\) and \(y\) is a prime element of \(R\text{,}\) we may apply Eisenstein’s Criterion (using \(p=y\)) to conclude that \(x^p-y\) is irreducible in \(F[x]\text{.}\) Thus \(x^p-y\) is the minimum polynomial of \(z\) in \(F[x]\text{.}\)
However, as the derivative of this polynomial is \(0\text{,}\) we see that the \(\mp_{z,F[x]}(x)\) is not separable by Proposition 2.72. However, by the Freshman’s Dream, we see that \(x^3-y=x^p-z^p=(x-z)^p\in L\text{.}\) But as \(z\not\in F\text{,}\) we see that the prime factorization of \(x^p-y\) admits no repeated factor.
