Prove that \(C_G(H)\) is a normal subgroup of \(N_G(H)\text{.}\)
Prove that \(N_G(H)/C_G(H)\) is isomorphic to a subgroup of \(Aut(H)\text{,}\) the group of automorphisms of \(H\text{.}\)
Solution.
Let \(\varphi: N_G(H)\to\Aut(H)\) be defined such that \(\varphi(g)=\psi_g\text{,}\) where \(\psi_g\) denotes the inner automorphism (also known as the conjugation automorphism) of \(H\) induced by \(g\text{.}\) Normally this is only well defined when \(g\in H\text{,}\) but as \(g\in x\in N_G(H)\text{,}\) we see \(\psi_x(h)=xhx\inv\in H\) for all \(h\in H\text{.}\)
Let \(x,y\in N_G(H)\text{,}\)\(h\in H\) and observe
Thus \(\varphi\) is operation preserving and a group homomorphism.
Let \(g\in\ker(\varphi)\text{.}\) Thus \(\varphi(g)=\psi_g\) is the identity map on \(h\text{,}\) meaning \(ghg\inv=h\) for all \(h\in H\text{.}\) Mulitplying on the right by \(g\) yields \(gh=hg\text{,}\) placing \(g\) in \(C_G(H)\text{.}\)
Now, let \(g\in C_G(H)\text{.}\) Thus \(\psi_g(h) = ghg\inv = gg\inv h = h\) for every \(h\in H\text{,}\) meaning \(\psi_g=\id_H\) and \(g\in\ker(\varphi)\text{.}\)
Putting this all together, we see that \(C_G(H)\) is precisely the kernel of \(\varphi\text{,}\) making \(C_G(H)\) normal in \(N_G(H)\) by Theorem 3.40.
From the previous part we know that \(\varphi\) is a homomorphism mapping from \(N_G(H)\) to \(\Aut(H)\) with kernel \(C_G(H)\text{.}\) Thus \(N_G(H)/C_G(H)\cong \im(\varphi)\) by the First Isomorphism Theorem, making \(N_G(H)/C_G(H)\) is isomorphic to a subgroup of \(Aut(H)\text{,}\) the group of automorphisms of \(H\text{.}\)
ActivityC.2.Problem 2.
Let \(G\) be a group with center \(Z(G)\text{.}\) Prove that if the quotient group \(G/Z(G)\) is cyclic, then \(G\) is abelian.
Let \(G\) be a non-abelian group of order \(21\text{.}\) Find the number and the sizes of the conjugacy classes of \(G\) with justification.
Solution.
Suppose \(G/Z(G)\) is cyclic. Let \(x\in G\) and suppose by way of contradiction that \(x\not\in Z(G)\text{.}\) 1
I don’t think this actually needs to be a contradiction proof, but I haven’t read it through too carefully and didn’t want to break anything by modifying it unneccessarily.
Thus there exists some \(y\in G\) such that \(xy\neq yx\text{.}\) Note: this means \(y\not\in Z(G)\) as well. As \(G/Z(G)\) is cyclic there exists some \(g\in G\) such that \(xN=g^nN\) and \(yN=g^mN\) for \(n,m\in\N\text{.}\) But then
where \(g_1,g_2,...,g_m\) are a complete non redundant list of conjugacy class representatives. Since \(Z(G)\leq G\text{,}\)\(|Z(G)|\) must divide \(|G|\) so the options for \(|Z(G)|\) are \(1,3,7,21\text{.}\) It can’t be \(21\) because that implies that \(G\) is abelian. Consider if \(|Z(G)|\) equals \(3\) or \(7\text{.}\) Then we would have \([G:Z(G)]\) equals \(7\) or \(3\) respectively. These are both prime, which implies that \(G/Z(G)\) is cyclic and by (a), would mean that \(G\) abelian. So the only option is \(|Z(G)|=1\text{.}\) Then \(21=1+\sum_{i=1}^m[G:C_G(g_i)]\implies \sum_{i=1}^m[G:C_G(g_i)]=20\text{.}\) Each \([G:C_G(g_i)]\) must be greater than \(1\) and divide \(21\text{.}\) The only way to add nontrivial positive divisors of \(21\)
ActivityC.3.Problem 3.
Let \(p\neq q\) be two prime integers. Prove that a group of order \(p^2q\) is not simple.
Solution.
There are 2 cases, \(p<q\) and \(p>q\text{.}\) We will show both. First, when \(p<q\text{,}\)\(n_p=1\) or possibly \(q\) if \(q=pk+1,k\in \mathbb{Z}\text{.}\)\(n_q=1\) or possibly \(p^2\) if \(p^2=ql+1,l\in \mathbb{Z}\text{.}\) It can’t be \(p\) since \(p<q\) and thus \(p\) is not congruent to \(1 mod q\text{.}\) When neither, or ONLY one of the possible conditions are met, it forces at least one of \(n_p\) or \(n_q\) to be \(1\) and \(G\) wouldn’t be simple. So assume both conditions are met. \(n_q=p^2\implies \exists p^2\) Sylow q-subgroups: \(H_1,H_2,...,H_{p^2}\) where \(|H_i|=q\) and \(\forall i\neq j\) we have \(H_i\cap H_j=\{e\}\text{.}\) Then let \(S=H_1\cup H_2\cup \cdots \cup H_{p^2}\subseteq G\) and \(\#S=\underbrace{(q-1)p^2}_\text{elements of order q}+1\) because \(S=\{e\}\cup \{\text{elements of order q}\}\text{.}\) Now \(n_p=q\implies \exists q\) Sylow p-subgroups: \(L_1,L_2,...,L_q\) where \(|L_i|=p^2\) and \(L_i\cap S=\{e\}\text{.}\) Note that nothing of order \(q\) is in any \(L_i\) since \(q\) can’t divide \(p^2\) due to the assumptions \((q=pk+1,p^2=ql+1)\text{.}\)\(|L_i|=p^2\) and \(p^2-1\) of which not the identity so \(G\) has the identity, \(p^2-1\) elements of order \(p^2\) or order \(p\) (which are in \(L_i\)) and \((q-1)p^2\) elements of order \(q\)(which are in \(S\)) which equals \(1+p^2-1+(q-1)p^2=(1-1)+(p^2-p^2)+qp^2=qp^2\text{.}\) This means we have all unique elements of \(G\) which means that \(L_1=L_2=\cdots =L_q\) so \(L_1\trianglelefteq G\text{.}\) Thus \(G\) is not simple.
Now when \(p>q\text{,}\)\(n_q\) is either \(1\) or \(p\) or \(p^2\) if the necessary conditions are met, but \(n_p\) can only be \(1\) since \(p>q\) and \(n_p\) needs to divide \(q\) and \(q\) is not congruent to \(1 mod p\text{.}\) Thus \(n_p=1\implies G\) is not simple.
SubsectionSection II: Rings, Modules, and Linear Algebra
ActivityC.4.Problem 4.
Give an example of a maximal ideal, \(M\text{,}\) in \(\mathbb{Z}[x]\text{,}\) the ring of polynomials with integer coefficients in one variable \(x\text{.}\) Prove that \(M\) is a maximal ideal, including a justification that \(M\neq \mathbb{Z}[x]\)
Solution.
Let \(M=(2,x)\text{.}\) Since \(\mathbb{Z}[x]\) is abelian, \(M=\{2f(x)+xg(x)|f,g\in \mathbb{Z}[x]\}\) which is all polynomials with even constant terms. This is clearly not equal to the original ring.
We will show that that the quotient ring is a field which implies that \(M\) is maximal. Define \(f:\mathbb{Z}[x]\rightarrow \mathbb{Z}_2\) as \(f(p(x))=f(a_0+\cdots +a_nx^n)=[a_0]_2\text{.}\) We will show this is an onto homomorphism with kernel equal to \((2,x)\) and thus the two groups are isomorphic. And since \(\mathbb{Z}_2\) is a field, \(\mathbb{Z}[x]/I\) is as well. \(f(p(x)+q(x))=f(a_0+\cdots +a_nx^n+b_0+\cdots b_mx^m)=f((a_0+b_0)+\cdots +\cdots )=[a_0+b_0]_2=[a_0]_2+[b_0]_2=f(p(x))+f(q(x))\text{.}\) So \(f\) preserves addition. \(f(p(x)q(x))=f((a_0+\cdots +a_nx^n)(b_0+\cdots +b_mx^m))=f(a_0b_0+\cdots )=[a_0b_0]_2=[a_0]_2[b_0]_2=f(p(x))f(q(x))\) so \(f\) preserves multiplication. Then of course \(f(0)=[0]_2\text{.}\) So it maps identity to identity. Thus \(f\) is a homomorphism. This is clearly onto since \(0\rightarrow [0]_2\) and \(1\rightarrow [1]_2\text{.}\)\(ker(f)=\{p(x)|f(p(x))\equiv 0(mod2)\}=\{p(x)|\text{ constant term }\equiv 0(mod2)\}=\{p(x)|\text{ constant term is even}\}\) which by the beginning note is exactly \((2,x)\text{.}\) So the quotient is a field since \(\mathbb{Z}_2\) is a field, and thus \(M\) is maximal.
ActivityC.5.Problem 5.
Prove that a principal ideal domain \(R\) satisfies the ascending chain condition on ideals: every chain of ideals of \(R\text{,}\)\(I_1\subseteq I_2\subseteq \cdots \subseteq I_n\subseteq I_{n+1}\subseteq \cdots\text{,}\) has \(I_n=I_{n+1}\) for sufficiently large integers \(n\text{.}\) Do not invoke any theorems that trivialize the proof.
Solution.
Let \(I_1\subseteq I_2\subseteq \cdots \subseteq I_n\subseteq I_{n+1}\subseteq \cdots\) be a chain of ideals in \(R\text{.}\) Let \(I=\cup_{j\in \mathbb{N}}(I_j)\text{.}\) We will show this is an ideal in \(R\text{.}\) Let \(x,y\in I\text{.}\) Then there exist \(j,k\in \mathbb{N}\) such that \(x\in (I_j)\) and \(y\in (I_k)\text{.}\) Let \(m=max\{j,k\}\) because of the ascending chain \(x,y\in (I_m)\text{.}\)\((I_m)\) is an ideal so \(x-y\in (I_m)\subseteq I\) and for arbitary \(r\in R\text{,}\) we know \(rx,ry\in (I_m)\subseteq I\) and so \(I\) is an ideal. Since \(R\) is a PID, there is a \(z\in R\) such that \(I=(z)\text{.}\) This means that \(z\in I_n\) for some \(n\text{.}\) This implies that \(I\subseteq (I_n)\subseteq I\) so \(I=I_n\) and since \(I\) was the intersection, all ideals further along in the chain are also contained in it so we get for all \(t\geq n, I_t=I_n\text{.}\) So for integers greater than or equal to this \(n\)\(I_n=I_{n+1}\text{.}\)
ActivityC.6.Problem 6.
Let \(A\) be an \(n\times n\) matrix with entries in \(\mathbb{Z}\text{,}\) for some \(n\geq 1\text{,}\) and let \(M\) be the abelian group presented by \(A\text{;}\) that is, let \(M=\mathbb{Z}^n/N\) where \(N\) is the image of the map \(\mathbb{Z}^n\rightarrow \mathbb{Z}^n\) given by \(v\rightarrow Av\text{.}\) Prove \(M\) is an infinite abelian group if and only if \(\det(A)=0\text{.}\)
Solution.
\((\implies)\) Assume \(M\) is infinite. By Thm 1.161 (Mark’s notes) CITEX we can turn \(A\) into its Smith Normal Form
where \(s_{1,1}|s_{2,2}|\cdots |s_{n,n}\text{,}\) using only elementary row and column operations. Then CITEX Corollary 1.163 says that \(M\cong \mathbb{Z}/s_{1,1}\bigoplus \cdots \bigoplus \mathbb{Z}/s_{n,n}\bigoplus \mathbb{Z}^a\) for some \(a\geq 0\text{.}\) Lemma 1.16 CITEX tells us that \(a=n-n=0\) so
unless (at least) one of the \(s_{i,i}=0\text{.}\)\(M\) is infinite so one of them equals 0. CITEX Smith Normal Form is created by elementary row and column operations so the determinant is preserved (up to sign). This is because type 1 operations keep the determinant exactly, type 3 flips it to its negative, and type 2 multiplies it by \(u\) for some unit \(u\) in \(R\text{.}\) CITEX Here \(R\) is \(\mathbb{Z}\) and the only units are \(\pm 1\) so type 2 also either keep the determinant exactly or just flip to the negative. And since \(0=-0\) and \(\text{det}(S)=0\implies \text{det}(A)=0\)\((\impliedby)\) Assume \(\text{det}(A)=0\text{.}\) Then in its Smith Normal Form there is a \(0\) on the diagonal. \(M\) is presented as above, and one of them is 0, so \(M\) is infinite.
SubsectionSection III: Fields and Galois Theory
ActivityC.7.Problem 7.
Prove that every finite field extension is algebraic.
Prove that the converse of (a) is false.
Solution.
Let \(F\subseteq L\) be a finite field extension. To show it is algebraic we need to show that every element \(\alpha \in L\) is algebraic over \(F\) meaning that it is the root of a non-zero polynomial with coefficients in \(F\text{.}\) So take \(\alpha \in L\text{.}\) Since \(F(\alpha)\) is an \(F\)-vector subspace of \(L\) and \([L:F]<\infty\) it must be that \([F(\alpha):F]<\infty\) and we have shown that if that is finite then \(\alpha\) is algebraic over \(F\text{.}\) If \([F(\alpha):F]<\infty\) then the infinite list \(1,\alpha, \alpha^2,...\) of elements of \(F(\alpha)\) cannot by linearly independent over \(F\) so \(a_0+a_1\alpha +\cdots +a_n\alpha^n=0\) for some \(n\) where not all \(a_i\) are 0. So its the root of a nonzero polynomial.
We prove the converse is false by a counter-example. Let \(E_n=\mathbb{Q}(\sqrt[n]{2})\) and set \(F=\cup_nE_n\text{.}\) Then \(F\) is a subfield of \(\mathbb{C}\text{.}\) This is because: \(E_n\subseteq E_m\) provided \(n\vert m\text{,}\) given \(a,b\in F\)\(a\in E_n\) and \(b\in E_m\) for some \(n,m\) and hence \(a,b\in E_{mn}\text{.}\)\(E_{mn}\) is a field, \(a+b,a-b,a\cdot b\) and provided \(a\neq 0\)\(1/a\) are all on \(E_{mn}\) and hence in \(F\text{.}\) So \(F\) is a field extension of \(\mathbb{Q}\text{.}\) It is algebraic over \(\mathbb{Q}\) since each \(E_n\) is. But it is not finite since \([E_n:\mathbb{Q}]=n\) and hence \([F:\mathbb{Q}]\geq n\) for all \(n\text{.}\)
ActivityC.8.Problem 8.
Let \(f\in \mathbb{Q}[x]\) be an irreducible cubic (meaning \(f\) is a degree 3 polynomial) with exactly one real root. Let \(L\) be the splitting field of \(f\) over \(\mathbb{Q}\text{.}\) Show that \(Aut(L/\mathbb{Q})\cong S_3\text{.}\)
Solution.
By the in particular in proposition 2 in worksheet 22 CITEX, since \(L\) is the splitting field over \(\mathbb{Q}\) of \(f\text{,}\)\(\operatorname{Aut}(L/\mathbb{Q})\) is isomorphic to a subgroup of \(S_3\) since \(f\) has degree 3, with only one real root, meaning that it has \(3\) distinct roots since complex roots come in pairs. Furthermore, the 1 real root must be irrational, because otherwise \(f\) would be reducible.
If we order the roots: \(\{\alpha,\zeta_1,\zeta_2\}\) where the two \(\zeta\)’s are complex conjugates, then the conjugate map(restricted to \(L\)) is associated with the permutation \((\zeta_1 \zeta_2)\) or \((23)\text{.}\)
\(L\) is the splitting field of irreducible \(f\text{,}\) so by CITEX corollary 3 from worksheet 22, the action \(g\cdot a\) is transitive, meaning for the root \(\zeta_1\text{,}\) there exists a \(g\in \operatorname{Aut}(L/\mathbb{Q})\) such that \(g\cdot \alpha=\zeta_1\text{,}\) meaning that \(g\cdot \zeta_1=\alpha\) and \(\zeta_2\) is fixed (\((12)\)), or \(g\cdot \zeta_1=\zeta_2\) and \(g\cdot \zeta_2=\alpha\) (\((123)\)). In either case, we have elements that can generate all of \(S_3\) because \(S_3=\langle (123),(23)\rangle=\langle (12),(23)\rangle\text{.}\)
ActivityC.9.Problem 9 (*).
Let \(F\subseteq L\) be a Galois extension of order \(p^2\text{,}\) where \(p\) is a prime integer.
Show that for every intermediate field \(F\subseteq E\subseteq L\text{,}\) the extension \(F\subseteq E\) is Galois.
Show that there must be either \(1\) or \(p+1\) intermediate fields \(F\subset E\subset L\text{.}\)
