My name is Hanzhang Yin, and I have developed this website as a resource to facilitate the review of key concepts in abstract algebra, and it is not fully completed. Feel free to email me at hanyin@ku.edu if there are any errors or suggestions for improvement.
1. Let \( G := \{ I := C_1, C_2, \ldots, C_r \} \) be a finite set of invertible \( n \times n \) complex matrices, such that \( G \) is a finite group under matrix multiplication. Here \( I \) denotes the \( n \times n \) identity matrix.
(i) Prove that each \( C_i \) is diagonalizable.
(ii) Show that \( \text{trace}(C_i^{-1}) = \overline{\text{trace}(C_i)} \), for all \( i \), where \( \overline{z} \) denotes the complex conjugate of \( z \in \mathbb{C} \).
(iii) Let \( \langle -, - \rangle \) denote the standard inner product on \( \mathbb{C}^n \). Define \( [v, w] := \frac{1}{|G|} \sum_{i=1}^r \langle C_i v, C_i w \rangle \), for all \( v, w \in \mathbb{C}^n \). Prove that \( [ -, - ] \) is an inner product on \( \mathbb{C}^n \) and that \( [v, w] = [C_i v, C_i w] \), for all \( v, w \in \mathbb{C}^n \) and \( C_i \in G \).
Proof (i). Given that \(G\) is a finite group with order \(r\). Hence, we can see that for any \(C_i\), we have \(C_i^r = I\). Then, we have \(C^r - I = 0\). Let \(f(x) = x^r - 1\) and we have \(f(C_i) = 0\) for any \(i\). Denote the minimal polynomial of \(C_i\) as \(\mu_{C_i}(x)\). We can know that \(\mu_{C_i}(x)\mid f(x)\). Since \(f'(x) = rx^{r-1}\) and \[ x^r - 1 = \frac1r x\cdot rx^{r-1} - 1, \] we can know that \(\gcd(f(x), f'(x)) = 1\), which implies that \(x^r - 1\) is separable in \(\mathbb{C}\). Hence, we can know that \(f(x)\) has distinct roots in \(\mathbb{C}\). Given that \(\mu_{C_i}(x)\mid f(x)\), we can know that \(\mu_{C_i}(x)\) has distinct roots in \(\mathbb{C}\). Since the minimal polynomial contains all the eigenvalues of \(C_i\) as roots. Let \(\mu_{C_i}(x) = (x - \lambda_1)(x - \lambda_2)\cdots(x - \lambda_{n_i})\), where \(\lambda_j\) are the eigenvalues of \(C_i\). Since each \(\lambda_j\) is distinct, we have \[ V = \ker(C_i - \lambda_1I) \oplus \ker(C_i - \lambda_2I) \oplus \cdots \oplus \ker(C_i - \lambda_{n_i}I). \] Since the algebraic multiplicity of each eigenvalue is equal to the geometric multiplicity, we have \(C_i\) is diagonalizable. \(\blacksquare\)
Proof (ii). Since we already showed each \(C_i\) is diagonalizable, we write \(C_i = P_iD_iP_i^{-1}\), where \(D_i\) is a diagonal matrix and \(P_i\) is an invertible matrix. Since \(C_i\) is invertible, we have \(D_i\) is invertible. Then, we can have \(C_i^{-1} = P_iD_i^{-1}P_i^{-1}\), where \(D_i^{-1}\) is a diagonal matrix and \(P_i\) is an invertible matrix. Moreover, the diagonal entries of \(D_i^{-1}\) is the reciprocal of the diagonal entries of \(D_i\). Since the diagonal entries of \(D_i\) are eigenvalues of \(C_i\), we have \[ \begin{align} \text{trace}(C_i) &= \text{trace}(P_iD_iP_i^{-1}) = \text{trace}(D_iP_i^{-1}P_i) = \text{trace}(D_i) = \sum_{j=1}^n \lambda_j, \\ \text{trace}(C_i^{-1}) &= \text{trace}((P_iD_iP_i^{-1})^{-1}) = \text{trace}(P_iD_i^{-1}P_i^{-1}) = \text{trace}(D_i^{-1}P_i^{-1}P_i) = \text{trace}(D_i^{-1}) = \sum_{j=1}^n \frac1{\lambda_j}. \end{align} \] Again, we can know that \((\lambda_i)^r - 1 = 0\), since the eigenvalue is the root of \(x^r - 1\) from the previous part. Then, \[ \overline{\lambda_i^r - 1} = \overline{\lambda_i^r} - \overline{1} = \overline{\lambda_i}^r - 1 = 0. \] Thus, we can know that \(\lambda_i^r\) = 1 and \(\overline{\lambda_i}^r = 1\). Hence, we have \[ \lambda_i^r\cdot \overline{\lambda_i}^r = (\lambda_i\overline{\lambda_i})^r = 1. \] Since \(\lambda_i\cdot \overline{\lambda_i} = |\lambda_i|^2\geq 0\), we have \((\lambda_i\cdot \overline{\lambda_i} =1\), for every eigenvalue \(\lambda_i\) of \(C_i\). Thus, we have \[ \text{trace}(C_i^{-1}) = \sum_{j=1}^n \frac1{\lambda_j} = = \sum_{j=1}^n \frac{\overline{\lambda_j}}{\lambda_j\overline{\lambda_j}} = \sum_{j=1}^n \overline{\lambda_j} = \overline{\sum_{j=1}^n \lambda_j} = \overline{\text{trace}(C_i)}. \blacksquare \]
Solution (c).Firstly, we want to show that \([v, v]\geq 0\) for any \(v\). Given that \[ [v, v] = \frac{1}{|G|}\sum_{i = 1}^r \langle C_i v, C_i v \rangle = \frac{1}{|G|}\sum_{i = 1}^r \|C_iv\|^2 \geq 0. \] Then, we want to show that \([v, v] = 0\) if and only if \(v = 0\). Suppose that \([v, v] = 0\), then we have \(\|C_iv\|^2 = 0\) for all \(i\), which implies that \(C_i v = 0\) for all \(i\).
2. Let \( T \) be a linear operator on a finite dimensional complex inner product space \( V \). Recall that \( T \) is a normal operator if \( TT^* = T^*T \), where \( T^* \) denotes the adjoint of \( T \). Prove that \( T \) is a normal operator if and only if \( \|T(v)\| = \|T^*(v)\| \), for all \( v \in V \). (Hint: use the fact that \( T^*T - TT^* \) is self-adjoint.) Give an example of a normal operator on a three-dimensional complex space that is not self-adjoint.
Proof. Suppose that \(T\) is a normal operator. Then, we have \(TT^* = T^*T\). Let \(v\in V\), then we have \[ \|T(v)\|^2 = \langle T(v), T(v) \rangle = \langle T^*T(v), v \rangle = \langle TT^*(v), v \rangle = \langle T^*(v), T^*(v) \rangle = \|T^*(v)\|^2. \] Hence, we have \(\|T(v)\| = \|T^*(v)\|\) for all \(v\in V\). For the other direction, let \(v\) be any vector in \(V\), then we have \(\|T(v)\| = \|T^*(v)\|\), which implies that \(\|T(v)\|^2 = \|T^*(v)\|^2\). Then, we have \[ \begin{align} \|T(v)\|^2 - \|T^*(v)\|^2 &= 0\\ \langle T(v), T(v) \rangle - \langle T^*(v), T^*(v) \rangle &= 0\\ \langle T^*T(v), v \rangle + \langle -TT^*(v), v \rangle &= 0\\ \langle T^*T(v) - TT^*(v), v \rangle &= 0 \\ \langle (T^*T - TT^*)(v), v \rangle &= 0. \end{align} \] Since \(v\) is any vector, it forces \((T^*T - TT^*)(v) = 0\). Again, because of \(v\) is any vector, we can have \(T^*T - TT^* = 0\), which implies that \(T\) is a normal operator. \(\blacksquare\)
4. Let \( G \) be a group of order \( 3n \) where \( n \) is an odd number.
(a) Prove that any subgroup of order \( n \) in \( G \) is normal.
(b) Give an example, with full justification, to show that the statement in part (a) is no longer true if \( n \) is even.
Proof (a). Suppose that \(n\) is an odd number and there exists a subgroup of \(G\), \(H\), such that the order is \(n\). Then, we can know that \([G:H] = 3\). Since \(n\) is an odd number, we can know that \(2\not\mid n\), which implies that \(3\) is the smallest prime dividing \(3n\). Hence, we can conclude that \(H\) is normal in \(G\). \(\blacksquare\)
5. Let \( R \) be a PID. Let \( I \) be a nonzero proper ideal in \( R \) and \( S = R/I \). Prove that the following are equivalent:
(a) \( S \) is indecomposable. (Namely, \( S \) cannot be written as direct sum of two non-zero ideals \( S = I \oplus J \).)
(b) \( I = (f^n) \) where \( f \) is a prime element and \( n \) is some positive integer.
Proof. Firstly, we want to show one direction, namely, (a) implies (b), by contradiction. Suppose that \(I = (f^n)\) where \(f\) is a prime element and \(n\) is some positive integer. According to the \(\textbf{Correspondence Theorem of Rings}\), we know that the maximal ideal of \(R\) containing \(I\) can be mapped to maximal ideal of \(S\). Suppose that \(\mathcal{m}\) is a maximal ideal of \(R\) containing \(I\), then we have \(f^n\in \mathcal{m}\). We also know that \(\mathcal{m}\) is a maximal ideal, which implies that \(\mathcal{m}\) is a prime ideal. Thus, we can know that \(f\in \mathcal{m}\). In that case, we can know that \(\mathcal{m}\) contains \(f\). Hence, we can know that \((f)\subset \mathcal{m}\). Since \(f\) is prime, we have \((f)\) is a prime ideal. Given that \(R\) is a PID, we can know that \((f)\) is a maximal ideal. Thus, we can know that the only maximal ideal containing \(I\) is \((f)\). In that case, we can know that \((f)/(f^n)\) is the only maximal ideal of \(S\). Now suppose that \(S = J\oplus K\) where \(J\) and \(K\) are non-zero ideals of \(S\). Hence, we want to show that \(J\) and \(K\) are proper ideals. If \(J\) is a whole ring, we can know and \(J\cap K = K = \{0\}\), which is a contradiction. Now, we know that \(J\) and \(K\) are both proper ideal of \(S\) and there is only one maximal ideal of \(S\), which is \((f)/(f^n)\). Thus, \(J\subset (f)/(f^n)\) and \(K\subset (f)/(f^n)\). Then, we can know that \(J + K\subset (f)/(f^n)\), since \((f)/(f^n)\) is a ring. Moreover, we can know that \(J + K\) is a proper ideal of \(S\), which is a contradiction. Hence, we can know that \(S\) is indecomposable.
Question 6. Construct a field \( F \) with nine elements, with full justification. Then prove that the map \( f : F \to F \) defined by \( f(\alpha) = \alpha^3 \), for all \( \alpha \in F \), is an automorphism (i.e., an isomorphism from \( F \) to itself).
Solution. We can know that \(F_3 = \mathbb{Z}/3\mathbb{Z}\) is a field with three elements. Moreover, \(p(x) = x^2 + 1\) is an irreducible polynomial over \(\mathbb{Z}/3\mathbb{Z}\). We can show that \(p(x)\) is irreducible by contradiction. Suppose that \(p(x) = (x - a)(x -b)\), which implies that there exists roots in \(\mathbb{Z}/3\mathbb{Z}\). However, we have \(p(0) = 1, p(1) = 2, p(2) = 2\), which implies that \(p(x)\) has no roots in \(\mathbb{Z}/3\mathbb{Z}\). Thus, we can know that \(p(x)\) is irreducible over \(\mathbb{Z}/3\mathbb{Z}\). Hence, the elements of \(F_3[x]/(p(x))\) are of the form \(a + bx\), where \(a, b\in \mathbb{Z}/3\mathbb{Z}\). Hence, there are \(3\cdot 3 = 9\) elements in \(F_3[x]/(p(x))\). Since \(i^2 + 1 = 0\), we can know that \[ F = \mathbb{Z}/3\mathbb{Z}[x]/(x^2 + 1)\cong \mathbb{Z}/3\mathbb{Z}[i]. \]
Proof. Now, we want to show that \(f: F\to F\) defined by \(f(\alpha) = \alpha^3\) is an automorphism. Firstly, we can see that \(f\) is well-defined since for any \(a = b\in F\) we have \(f(a) = a^3 = b^3 = f(b)\). Then, we want to show that \(f\) is a homomorphism. Let \(a, b\in F\), then we have \[ f(a)f(b) = a^3b^3 = (ab)^3 = f(ab). \] Now, we want to show that \(f\) is injective. Suppose that \(f(a) = f(b)\) for some \(a, b\in F\). We will have \[ \begin{align} a^3 &= b^3 \\ a^3\cdot b^{-3} &= 1 \\ (ab^{-1})^3 &= 1. \end{align} \] Since we know that \(F\setminus\{0\}\) is a group under multiplication with order of \(8\). Given that \((ab^{-1})^3 = 1\), we can know that \(ab^{-1}\) has order either \(1\) or \(3\). If \(ab^{-1}\) has order of \(3\), then we can know that \(3\mid 8\), which is a contradiction. Hence, we know that \(ab^{-1} = 1\), which implies that \(a = b\). And we have \(f\) is injective. Since \(f\) is a map from \(F\) to \(F\) and \(F\) is finite, we can know that \(f\) is surjective by \(\textbf{the pigeonhole principle}\). Therefore, we have \(f\) is an automorphism. \(\blacksquare\)