Let \(G\) be a finite \(p\)-group for some prime \(p\) and \(N\neq {1}\) a normal subgroup of \(G\text{.}\) Prove that \(N\cap Z(G)\neq{1}\text{,}\) where \(Z(G)\) is the center of \(G\text{.}\)
Solution.
Let \(G\) be a finite \(p\)-group for some prime \(p\) and \(N\neq {1}\) a normal subgroup of \(G\text{.}\)
First, we show that the center of \(G\) is nontrivial. Suppose by way of contradiction that \(Z(G)=\{e\}\text{.}\) We examine the conjugacy classes of \(G\text{.}\) From the Class Equation, we know
Note that \(|G|=p^n\) for some \(n\in\N\text{,}\) meaning that the only divisors of \(|G|\) are powers of \(p\text{.}\) In finite groups, each conjugacy class must divide the order of the group by Theorem 5.41. By Lagrange’s Theorem, \([G : C_G(g_i)]=\frac{|G|}{|C_G(g_i)|}=\frac{p^n}{p^m}=p^k\text{,}\) where \(m,k\in\N\text{.}\) Since \(|Z(G)|=1\text{,}\) we see that \(\sum_i^r |G : C_G(g_i)|\bigg|p^n-1\text{,}\) which is impossible given that \(p^{n-1} \not\mid p^n\text{.}\) Thus \(Z(G)\neq\{e\}\text{.}\)
As \(N\nsg G\text{,}\) by [cross-reference to target(s) "cor-normal-union-of-conjugacy-classes" missing or not unique] it is a union of conjugacy classes of the elements it contains, one of which is \(e\text{.}\) Assume by way of contradiction that \(N\cap Z(G)={e}\text{,}\) meaning that \(e\) is the only element in \(N\) whose conjugacy class is a singleton. This yields
However, by Lagrange’s Theorem\(N\) must also be a \(p-\)group, and thus by an analogous element counting argument as above we see that there exists some \(x\in N\) such that \(C_G(x)=\{x\}\text{,}\) or that \(gxg\inv=x\) for all \(g\in G\text{.}\) Thankfully, this means that \(x\in Z(G)\text{,}\) and thus we have \(Z(G)\cap N\neq 1\text{.}\)
ActivityC.20.Problem 2.
A non-abelian group of order \(27\text{.}\)
Prove there exists a non-abelian group of \(27\text{.}\)
Find, with justification, a presentation of the group you found in part (1).
Hint.
Use a semi-direct product.
Solution.
Let \(C_3 = \langle a \rangle\) be the cyclic group of order \(3\text{,}\) and let \(C_9 = \langle b \rangle\) be the cyclic group of order \(9\text{.}\) Let \(\phi \colon C_9 \to \operatorname{Aut}(C_3)\) be the homomorphism defined by \(\phi(b^i)(a) = a^i\) for \(0 \le i \le 8\text{,}\) where we identify \(\operatorname{Aut}(C_3)\) with \(C_2\) via the isomorphism \(\operatorname{Aut}(C_3) \cong C_2\) given by sending the non-identity automorphism to the generator of \(C_2\text{.}\)
Then the semi-direct product \(G = C_3 \rtimes_{\phi} C_9\) has order \(27\) and is non-abelian. To see that \(G\) is non-abelian, note that \(bab^{-1} \neq a\) for any \(b \in C_9 \setminus {1}\text{,}\) since \(a\) has no non-trivial automorphisms.
To find a presentation of \(G\text{,}\) we need to identify the generators and relations. Since \(C_3\) and \(C_9\) are both cyclic, we can take the generators to be \(a\) and \(b\) respectively. Moreover, since \(C_3\) is normal in \(G\text{,}\) we have \(bab^{-1} = a^i\) for some \(i \in {1, 2}\text{.}\)
To determine the value of \(i\text{,}\) note that \(C_9\) acts on \(C_3\) by automorphisms, and the non-identity automorphism of \(C_3\) is inversion \(a \mapsto a^{-1}\text{.}\) Since \(bab^{-1} \neq a^{-1}\text{,}\) we must have \(i = 2\text{.}\)
Therefore, a presentation of \(G\) is given by \(\langle a, b \mid a^3 = b^9 = 1, bab^{-1} = a^2\rangle\text{.}\)
ActivityC.21.Problem 3.
Let \(G\) be a group of order \(2^5\cdot 7^3\text{.}\) Prove that \(G\) is not simple.
Solution.
By Sylow’s Theorems we know that \(\Syl_7(G)\equiv 1\mod{7}\) and \(\Syl_7(G)|2^5=32\text{.}\) Thus our options are \(1\) and \(8\text{.}\) Suppose that \(\Syl_7(G)=8\text{.}\)
Let \(G\) act on \(\Syl_7(G)\) by conjugation, yielding the homomorphism \(\rho: G \to S_8\) via the Permutation Representation. This map is non-trivial from part (2) of Sylow’s Theorems, but \(2^5\cdot 7^3\) does not divide \(|S_8|=8!\text{,}\) and thus \(\rho\) cannot be injective. Then the kernel of this homomorphism is non-trivial, normal subgroup of \(G\) by Theorem 3.40. Thus \(G\) is not simple.
SubsectionSection II: Rings, Modules, and Linear Algebra
ActivityC.22.Problem 4.
Let \(R\) be a commutative ring with identity, and assume \(1\neq 0\text{.}\) Let \(I\) and \(J\) be ideals such that \(I + J = R\text{.}\)
Prove \(IJ = I\cap J\text{.}\)
Prove the following special case of 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 for Rings, we see \(R/(I \cap J)\cong R/I \times R/J\text{.}\)
ActivityC.23.Problem 5.
Prove the Rank-Nullity Theorem: If \(F\) is a field, \(V\) and \(W\) are a finite dimensional \(F\)-vector spaces, and \(g : V \to W\) is an \(F\)-linear transformation, then
Let \(F\) be a field and \(n\) a positive integer. We say an \(n\times n\) matrix \(A\) with entries in \(F\) is unipotent if \(A-In\) is nilpotent (i.e., \((A-In)^k = 0\) for some \(k \geq 1\)). For the field \(F = \Q\text{,}\) find (with complete justification) the number of similarity classes of \(4\times 4\) unipotent matrices and give an explicit representative for each class.
Solution.
Let \(F=\Q\text{,}\)\(n\) a positive integer, and \(A\) a unipotent \(4\times 4\) matrix with entries in \(F\text{.}\) Thus \(A-I\) is nilpotent. Let \(\l\) be an eigenvalue of \(A-I\text{.}\) Then \((A-I)v=\l v\text{,}\) so \(Av-Iv=\l v\) and \(Av=\l v+Iv\text{.}\) As \(Iv=v\text{,}\) we have \(Av=\l v+v\) and \(Av=(\l+1)v\text{.}\)
Notice that as \(\l\) is an eigenvalue of \(I\text{,}\) we have \(\l +1\) as an eigenvalue of \(A\text{.}\)
Assume inductively that \(\l^{n-1}\) is an eigenvalue of \((A-I)^{n-1}\text{.}\) Notice
making \(\l\) an eigenvalue of \((A+I)^n\text{.}\) Thus if \(\l\) is an eigenvalue of \(A-I\text{,}\) it is an eigenvalue of \((A-I)^n\) as well. As \((A-I)\) is nilpotent, there exists some \(k\) such that \((A-I)^k=0\text{.}\) This means that \(\l^kv=0\text{.}\) As \(v\neq 0\) and \(\l^k\) is a scalar in a field (and hence integral domain) we have \(\l=0\text{.}\) Thus the only eigenvalue of \((A-I)\) is \(0\text{,}\) meaning that the only eigenvalue of \(A\) is \(1\text{.}\)
Eigenvalues of \(A\) correspond to the roots of \(\cp_A(x)\text{,}\) which is a monic quartic polynomial, as \(A\) is a \(4\times 4\) matrix. Thus \(\cp_A(x)=(x-1)^4\text{,}\) as all roots must be \(1\text{.}\)
Two matrices are similar if and only if they share the same invariant factors. Given that invariant factors divide \(\cp_A(x)\) and each invariant factor must divide the following one, the possible sets of invariant factors for \(A\) are the following:
\(\{(x-1)^4\}\text{,}\)
\(\{(x-1),(x-1)^3\}\text{,}\)
\(\{(x-1), (x-1), (x-1)^2\}\text{,}\)
\(\{(x-1)^2, (x-1)^2\}\text{,}\) and
\(\{(x-1), (x-1), (x-1), (x-1)\}\text{.}\)
We identify the companion matrices for each possible invariant factor:
As \(f\) is monic and irreducible, it is the minimum polynomial of \(\a_5\text{.}\) Let \(E=\Q(\a_5)\) and notice \([E:\Q]=5\text{.}\)
Notice that \(\z\text{,}\) a primitive \(5\)-th root of unity, is the root of \(g(x)=x^4+x^3+x^2+x+1\text{,}\) the fourth cyclotomic polynomial, which is monic and irreducible in \(\Q[x]\) by [provisional cross-reference: prop-cylcotomic-irreducible]. Thus \(g\) is the minimum polynomial of \(\z\) and \([F:\Q]=4\text{.}\)
We know that \(L=\Q(\sqrt[5]{11},\z)\text{,}\) as every generator of \(L\) can be written using \(\z\) and \(\sqrt[5]{11}.\) Let \([L:\Q]=n\text{,}\)\([L:F]=p\text{,}\) and \([L:E]=q\text{.}\) By the The Degree Formula, we have
\([L:\Q]=[L:F][F:\Q]\text{,}\) and thus \(n=4p\)
\([L:\Q]=[L:E][E:\Q]\text{,}\) and thus \(n=5q\)
As \(E\neq L\) and \(F\neq L\) we know that \(p,q>1\text{.}\)
Thus we have \(n=20\text{.}\)
Let \(F = \Q(\zeta)\) where \(\zeta = e^{2\pi i/5}\text{.}\) From Part (a) we know \([L:F]=5\text{.}\) As \(L=F(\a)\text{,}\) we know that \(\mp_\a(x)\) has degree \(5\text{.}\) As \(f\) is a monic polynomial of degree \(5\text{,}\) we see that it is the minimum polynomial and thus irreducible.
ActivityC.26.Problem 8.
Let \(L\) be the splitting field of \(x^4-2022\) over \(\Q\text{.}\) Prove there exists a unique intermediate field \(\Q\subseteq K\subseteq L\) such that \([K : \Q] = 4\) and \(\Q\subseteq K\) is a Galois extension.
Solution.
Let \(L\) be the splitting field of \(f(x)=x^4-2022\) over \(\Q\text{.}\) Let \(\z\) be a primitive fourth root of unity. Thus \(\z=e^{2\pi i/4}=e^{\pi/2}=\cos(\pi/2)+i\sin(\pi/2)=i\text{.}\)
Notice that the roots of \(f\) are the following:
\(\sqrt[4]{2022} i\text{,}\)
\(\sqrt[4]{2022}i^2=-\sqrt[4]{2022}\text{,}\)
\(\sqrt[4]{2022}i^3=-\sqrt[4]{2022}i\text{,}\) and
\(\sqrt[4]{2022}i^4=\sqrt[4]{2022}\text{.}\)
Thus \(L=\Q(\sqrt[4]{2022},i)\text{.}\)
Using Eisenstein’s Criterion with \(p=2\) we see that \(f\) is irreducible in \(\Q[x]\text{.}\) Let \(\a=\sqrt[4]{2022}\text{,}\) and notice that \(f\) is the minimum polynomial of \(\a\text{.}\) Let \(K=\Q(\a)\) and observe \([K:\Q]=4\text{.}\)
As \(i\) is the root of the monic irreducible polynomial \(x^2+1\) we have \([L:F]=2\) and \([L:\Q]=[L:F][F:\Q]=2\cdot4=8\text{.}\) Thus \(G=\Gal(L/\Q)\) is isomorphic to a subgroup of \(S_4\) of order \(8\text{,}\) making it \(D_8\text{.}\)
Notice that \(K\) is an extension of degree \(24\text{,}\) and thus by the Fundamental Theorem of Galois Theory we have a subgroup \(H\leq G\) such that \(H=\Gal(L/K)\text{.}\) As \([L:\Q]=[L:K][K:\Q]\) we have \(8=[L:K]\cdot 4\text{,}\) and thus \(|H|=2\text{.}\) This makes \(H\) the cyclic subgroup of \(D_8\) generated by a reflection, the only element of order \(2\) in \(D_8\text{,}\) making it unique. This is also a normal subgroup of \(D_8\text{,}\) making \(K\) Galois over \(\Q\text{.}\)
ActivityC.27.Problem 9.
Assume \(F\subseteq L\) is an algebraic field extension such that every non-constant polynomial in \(F[x]\) splits completely into linear factors in \(L[x]\text{.}\) Prove \(L\) is an algebraic closure of \(F\text{.}\)
Solution.
Let \(F\subseteq L\) be an algebraic field extension such that every non-constant polynomial in \(F[x]\) splits completely into linear factors in \(L[x]\text{.}\)
Let \(f\) be a polynomial in \(L[x]\text{,}\) so \(f=a_nx^n+\dots+a_1x+a_0\text{,}\) with \(a_i\in L\text{,}\) and let \(\a\) be a root of \(f\text{.}\) There exists some field extension \(L\subseteq K\) such that \(K\) is algebraically closed, meaning \(\a\in K\text{.}\) Notice that \(\a\) is algebraic over \(L(\a)\text{.}\)
making \(\a\) algebraic over \(F\) as well. As every non-constant polynomial in \(F[x]\) splits completely into linear factors in \(L[x]\text{,}\) this yields \(\a\in L\text{.}\)
