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 \( G \) be a finite group with \( |G| = p^n \), \( p \) prime.
(i) Prove that \( Z(G) \), the center of \( G \), is non-trivial.
(ii) Prove that if \( N \subseteq G \) is a normal subgroup of order \( p \), then \( N \subseteq Z(G) \). Hint: Let \( G \) act on \( N \).
(iii) Give an example of a non-abelian group of order \( p^n \) whose center contains more than one normal subgroup of order \( p \).
Proof (i).
Proof (ii). Firstly, we define a map \(\varphi: G\to \text{Aut}(N)\) such that \(\varphi(g) = f_g = gng^{-1}\) for any \(g\in G\) and \(n\in N\). The reason that \(f_g\) is in \(\text{Aut}(G)\) is \(N\) is a normal subgroup such that \(gNg^{-1} = N\). We can see that \(\varphi\) is well-defined since if \(g = h\), we will have \(f_g = f_h\). Suppose that \(g, h\in G\), we will have \[ \varphi(g)\varphi(h) = \varphi(g)\circ\varphi(h) = f_g\circ f_h = f_g(hnh^{-1}) = g h n h^{-1}g^{-1} = (gh)n(gh)^{-1} = f_{gh}(n) = \varphi(gh), \] which implies that \(\varphi\) is a group homomorphism. Hence, we can see that \(\varphi(G)\) is a subgroup of \(\text{Aut}(N)\). Given that \(N\) is a subgroup of order \(p\), we have \(N\cong \mathbb{Z}_p\). Hence, we have \(N\) is cyclic. For any \(f\in \text{Aut}(N)\), any nontrivial \(g\in N\) is a generator of \(N\), and once determine \(f(g)\), we can determine \(f\) completely. Hence, there will be \(p - 1\) automorphisms of \(N\) since each nontrivial element has \(p-1\) choices to be mapped. Thus, we can know that \(|\text{Aut}(N)| = p - 1\). Since \(\varphi(G)\) is a subgroup of \(\text{Aut}(G)\), we have \(|\varphi(G)|\) divides \(|\text{Aut}(N)|\), which implies that \(|\varphi(G)|\) divides \(p - 1\). By the \(\textbf{First Isomorphism Theorem}\), we have \(G/\ker(\varphi) \cong \varphi(G)\). Since \(|G/\ker(\varphi)| = [G:\ker(\varphi)]\), we have \(|G/\ker(\varphi)|\) divides \(p^n\). Hence, we have \(|G/\ker(\varphi)| = p^k\) for some \(0\leq k\leq n\). Since \(|\varphi(G)|\) divides \(p - 1\) and \(|G/\ker(\varphi)| = |\varphi(G)| = p^k\), we have \(p^k\mid p-1\). Thus, \(p^k\leq p-1\), which implies that \(k = 0\). Thus, we can know that \(|\varphi(G)| = 1\), which implies that \(\varphi(G) \subset \text{Aut}(N)\) is trivial. Hence, we can know that \(gng^{-1} = n\) for any \(g\in G\) and \(n\in N\), which implies that \(N\subset Z(G)\). \(\blacksquare\)
2. Let \( R \) be an integral domain. Suppose there exists a nonzero, non-unit \( a \in R \) such that for every \( r \in R \), there exists a unit \( u \) and \( n \geq 0 \) such that \( r = ua^n \). Such a ring is called a discrete valuation ring. Set \( P := aR \). Let \( R[\frac{1}{a}] \) denote the ring of polynomial expressions in \(\frac{1}{a}\) with coefficients in \( R \). Note, \(\frac{1}{a} \notin R \).
(i) Prove that \( P \) is a maximal ideal, and in fact, the only maximal ideal.
(ii) Prove that \(\bigcap_{n=1}^{\infty} P^n = (0)\).
(iii) Prove that \( R[\frac{1}{a}] \) is a field.
(iv) Let \( R \) be the ring of formal power series over \(\mathbb{Q}\). Thus, a typical element in \( R \) is of the form \( \sum_{n=0}^{\infty} a_n x^n \) where \( a_n \in \mathbb{Q} \). Let \( a = x \). Prove that \( R \) is a discrete valuation ring.
Proof (i). Given that \(a\) is not a unit, we can know that \(a^n\) will never be a unit for any \(n\geq 1\). In that case, we have \(ar\) is not a unit for any \(r\in R\). It is because if \(ar\) is a unit, then there exists \(s\in S\) such that \((ar)s = a(rs) = 1\), which implies \(a\) is a unit. Thus, we can know that \(aR\) does not contain a unit; hence, \(aR\) is a proper ideal. Now, we want to show that \(aR\) is a maximal ideal by contradiction. Suppose that \(aR\) is not a maximal ideal. Since every proper ideal is contained in a maximal ideal. Thus, we can know that \(aR\) is contained in a maximal ideal \(M\) such that \(aR\neq M\). Hence, there exists \(m\in M\) such that \(m\) is not a unit and \(m\notin aR\). However, we know that \(m = ua^n\) for some unit \(u\) and \(n\geq 0\) by assumption, which means that \(m\in aR\). It is a contradiction. Hence, we can know that \(aR\) is a maximal ideal. Actually, for any non-unit \(r\in R\), we have \(r = ua^n\) for some unit \(u\) and \(n\geq 0\), which implies that \(r\in aR\). In short, \(aR\) contains all the non-units in \(R\). And we know that every maximal ideal is a proper ideal. In other words, every element in each maximal ideal is a non-unit. Hence, if \(M\) is a maximal ideal of \(R\), we have \(M\subset aR\), which implies that \(M = aR\). Therefore, \(aR\) is the only maximal ideal. \(\blacksquare\)
3. Let \(\mathbb{Z}^n\) denote the free abelian group of rank \( n \), its elements being row vectors of length \( n \). Let \( A \) be an \( r \times n \) matrix over \(\mathbb{Z}\) and write \( K_A \) for the subgroup of \(\mathbb{Z}^n\) generated by the rows of \( A \).
(i) Suppose \( B := PAQ \), where \( P \) is an \( r \times r \) invertible matrix over \(\mathbb{Z}\) and \( Q \) is an invertible \( n \times n \) matrix over \(\mathbb{Z}\). Prove that \(\mathbb{Z}^n / K_A\) and \(\mathbb{Z}^n / K_B\) are isomorphic as abelian groups.
(ii) Suppose \( A = \begin{pmatrix} 4 & -2 & 4 \\ 2 & 4 & 4 \end{pmatrix} \). Write \(\mathbb{Z}^3 / K_A\) as a direct sum of cyclic groups.
Proof (i). Given that \(K_A\) is the subgroup of \(\mathbb{Z}^n\) generated by the rows of \(A\), we have \(K_A = \{x\in \mathbb{Z}^n\,\mid\, x = \sum_{i=1}^r a_iA_i\}\), where \(A_i\) is the \(i\)-th row of \(A\). And we can know that \(K_A\) is isomorphic to the subgroup of \(\mathbb{Z}^r\) generated by the columns of \(A^T\), where \(A^T\) is the transpose of \(A\), which is a \(n\times r\) matrix. Now, we have \(B^T = (PAQ)^T = Q^T A^T P^T\). We know that, if \(P\) is invertible, then \(p^T\) is invertible and \((P^T)^{-1} = (P^{-1})^T\). If \(y\in K_A\), we have \(y = A^Tx\) for some \(x\in \mathbb{Z}^r\). Then, we can have \(y = A^T(P^T(P^T)^{-1})x = (A^TP^T)((P^T)^{-1}x)\), where \((P^T)^{-1}x\in \mathbb{Z}^r\). Hence, we have \(K_A \subset K_{PA}\). For the other direction, if \(y\in K_{PA}\), we can know that \(y = A^TP^Tx\) for some \(x\in \mathbb{Z}^r\). Since \(y = A^T(P^Tx)\) where \(P^Tx\in \mathbb{Z}^r\), we have \(K_{PA} \subset K_A\). We can conclude that \(K_A = K_{PA}\). Then, we want to show that for any \(r\times n\) matrix \(M\) and invertible matrix \(Q\), we have \(K_{MQ} \cong K_M\). Let \(\varphi: K_{MQ}\to K_{M}\) be a map such that \(\varphi(x) = Q^Tx\) for any \(x\in K_{MQ}\). Since \(Q\) is invertible, we have \(Q^T\) is invertible, which implies that \(\varphi\) is a bijection. Hence, we can know that \(K_{M}\cong K_{MQ}\). Now, replace \(M\) with \(PA\), we have \(K_A = K_{PA} \cong K_{PAQ} = K_B\). Therefore, we can conclude that \(\mathbb{Z}^n / K_A\) and \(\mathbb{Z}^n / K_B\) are isomorphic as abelian groups. \(\blacksquare\)
Solution (ii). Firstly, we define \(Q_1\) as the following \[ Q = \begin{pmatrix} 1 & 1 & 0 \\ -1 & 0 & 2 \\ -1 & -1 & 1 \end{pmatrix}. \] Hence, we have \[ (AQ_1)^T = Q^TA^T = \begin{bmatrix} 1 & - 1 & - 1 \\ 1 & 0 & -1 \\ 0 & 2 & 1 \end{bmatrix}\cdot \begin{bmatrix} 4 & 2 \\ -2 & 4 \\ 4 & 4 \end{bmatrix} = \begin{bmatrix} 2 & -6 \\ 0 & -2 \\ 0 & 12 \\ \end{bmatrix}. \] Now, define \(P\) as \[ P = \begin{bmatrix} 1 & 0 \\ 3 & 1 \end{bmatrix}. \] Then, we have \[ (PAQ_1)^T = Q^TA^TP^T = \begin{bmatrix} 2 & -6 \\ 0 & -2 \\ 0 & 12 \\ \end{bmatrix}\cdot \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & -2 \\ 0 & 12 \end{bmatrix}. \] Now, we define \(Q_2\) as \[ Q_2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 5 & 6 \\ 0 & 1 & 1 \end{bmatrix}. \] Then, we have \[ (PAQ_1Q_2)^T = Q_2^TQ_1^TA^TP^T = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 5 & 1 \\ 0 & 6 & 1 \end{bmatrix}\cdot \begin{bmatrix} 2 & 0 \\ 0 & -2 \\ 0 & 12 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \\ 0 & 0 \\ \end{bmatrix}. \] Since \(\det(Q_1) = \det(P) = 1, \det(Q_2\) = -1\), thus we know that \(Q_1Q_2\) and \(P\) are invertible. Thus, we can know that \(\mathbb{Z}^3 / K_A \cong \mathbb{Z}^3 / K_{PAQ_1Q_2}\). Given that \(PAQ_1Q_2 = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \end{bmatrix}\). Hence, we have \(K_{PAQ_1Q_2} = 2\mathbb{Z}\oplus 2\mathbb{Z}\). Therefore, \[ \mathbb{Z}^3 / K_A \cong \mathbb{Z}^3 / K_{PAQ_1Q_2} \cong (\mathbb{Z}\oplus\mathbb{Z}\oplus \mathbb{Z})/(2\mathbb{Z}\oplus 2\mathbb{Z}) \cong \mathbb{Z}_2\oplus \mathbb{Z}_2\oplus \mathbb{Z}\cong \mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}. \]
4. Let \( f(x) := x^3 - 9x + 3 \in \mathbb{Q}[x] \) and \( \alpha \) a root of \( f(x) \).
(i) Prove that \( f(x) \) is irreducible over \( \mathbb{Q} \).
(ii) In the field \( \mathbb{Q}(\alpha) \), write \( (3\alpha^2 + 2\alpha + 1)^{-1} \) in terms of the basis \( 1, \alpha, \alpha^2 \).
Proof (i).
Solution (ii). Since \(\alpha^3 - 9\alpha + 3 = 0\), we have \(\alpha^3 = 9\alpha - 3\) and \(\alpha^4 = 9\alpha^2 - 3\alpha\). Suppose that \((3\alpha^2 + 2\alpha + 1)^{-1} = x\alpha^2 + y\alpha + z\) for some \(x, y, z\in \mathbb{Q}\). Then we have \[ \begin{align} (3\alpha^2 + 2\alpha + 1)(x\alpha^2 + y\alpha + z) &= 1 \\ 3x\alpha^4 + (3y + 2x)\alpha^3 + (3z + 2y + z)\alpha^2 + (2z + y)\alpha + z &= 1\\ 3x(9\alpha^2 - 3\alpha) + (3y + 2x)(9\alpha - 3) + (3z + 2y + z)\alpha^2 + (2z + y)\alpha + z &= 1\\ (28x + 3z + 2y)\alpha^2 + (28y + 9x + 2z)\alpha -6x - 9y + z &= 1 \\ (28x + 3z + 2y)\alpha^2 + (28y + 9x + 2z)\alpha -6x - 9y + z - 1&= 0. \end{align} \] Since \(\alpha^2, \alpha, 1\) form a basis of \( \mathbb{Q}(\alpha) \), we have \[ \begin{align} 28x + 3z + 2y &= 0 \tag*{(1)}\\ 28y + 9x + 2z &= 0 \tag*{(2)}\\ -6x - 9y + z - 1 &= 0 \tag*{(3)} \end{align} \] From \((3)\), we have \(z = 6x + 9y + 1\). Plug it into \((1)\) and \((2)\), we have \[ \begin{align} 46x + 27y + 3 &= 0, \\ 21x + 46y + 2 &= 0. \end{align} \] Hence, we have \[ \begin{align} x &= -\frac{84}{1549}, \\ y &= -\frac{29}{1549}, \\ z &= \frac{784}{1549}. \end{align} \] Therefore, we have \((3\alpha^2 + 2\alpha + 1)^{-1} = -\frac{84}{1549}\alpha^2 - \frac{29}{1549}\alpha + \frac{784}{1549}\).
5. Let \( A \) be an \( n \times n \) matrix over the field \( F \) and \( F^n \) denote the vector space of column vectors of length \( n \). Suppose \( A \) is idempotent, i.e., \( A^2 = A \).
(i) Prove that \( F^n = U \oplus W \), where \( A \cdot u = 0 \), for all \( u \in U \) and \( A \cdot w = w \), for all \( w \in W \).
(ii) If \( F = \mathbb{Z}_p \), how many idempotent \( 3 \times 3 \) matrices are there?
Proof (i). Given that \(A^2 = A\), we have \(A^2 - A = A(A - I) = 0\). Hence, we define \(p(x) = x(x - 1)\). If \(p(x)\) is the minimal polynomial of \(A\), then we can know that \[ F^n = \ker(A) \oplus \ker(A - I). \] Let \(U = \ker(A)\) and \(W = \ker(A - I)\). Hence, for any \(u\in U\), we have \(A\cdot u = 0\) by the definition of the kernel. Let \(w\in W\), then we have \((A - I)\cdot w = 0\). Thus, we have \(A\cdot w = w\). Suppose that \(p(x)\) is not the minimal polynomial of \(A\). Then, we can know that the minimal polynomial of \(A\) divides \(p(x)\). Hence, we have two options: either it is \(x\) or \(x - 1\). (To be continued...)
Proof (ii). Firstly, we know that \(V = \ker(A) \oplus \ker(A - I)\). Hence, we can know that \(A\) is diagonalizable. And we can know that \(A\) has only two possible eigenvalues: \(0\) and \(1\). Let \(G\) be the set of all nonsingular matrices over \(\mathbb{Z}_p\). Then we have \(|G| = (p^3 - 1)(p^3 - p)(p^3 - p^2)\). And we denote \(X\) be the set such as \[ X = \left\{x_0 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}, x_1 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}, x_2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix},x_3 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix}\right\}. \] Since \(A\) is diagonalizable, we have \(A = gxg^{-1}\) for some \(g\in G\) and \(x\in X\). Now, we try to find the centralizer for each element in \(X\). For identity matrix and zero matrix, we know they are in the center of \(G\). Hence, we have \(|C_G(x_0)| = |C_G(x_3)| = |G| = (p^3 - 1)(p^3 - p)(p^3 - p^2)\). For the matrix \(x_1 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}\), suppose that \(g = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix}\), such that \[ gxg^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}. \] Hence, \[ \begin{align} \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} &= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix},\\ \begin{bmatrix} a & 0 & 0 \\ d & 0 & 0 \\ g & 0 & 0 \\ \end{bmatrix} &= \begin{bmatrix} a & b & c \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix},\\ g &= \begin{bmatrix} a & 0 & 0 \\ 0 & e & f \\ 0 & h & i \\ \end{bmatrix}. \end{align} \] Then, we can see that \(a\neq 0\), which implies that there are \(p - 1\) choices for \(a\). And the principal submatrix \(\begin{bmatrix}e & f \\ h & i\end{bmatrix}\) has to be nonsingular, which implies that there are \(p^2 - 1\) choices for \(e\) and \(h\) and \(p^2 - p\) choices for \(f\) and \(i\). In total, there are \((p - 1)(p^2 - 1)(p^2 - p)\) choices for \(g\). Now, we consider the matrix \(\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}\). Suppose that \(g = \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix}\), such that \[ gxg^{-1} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}. \] Hence, \[ \begin{align} \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix}\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} &= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{bmatrix},\\ \begin{bmatrix} a & b & 0 \\ d & e & 0 \\ g & h & 0 \\ \end{bmatrix} &= \begin{bmatrix} a & b & c \\ d & e & f \\ 0 & 0 & 0 \\ \end{bmatrix} \end{align}. \] Then, we have \[ g = \begin{bmatrix} a & b & 0 \\ d & e & 0 \\ 0 & 0 & i \\ \end{bmatrix}. \] Similarly, we have \(i\neq 0\), which implies that there are \(p - 1\) choices for \(i\). And the principal submatrix \(\begin{bmatrix}a & b \\ d & e\end{bmatrix}\) has to be nonsingular, which implies that there are \(p^2 - 1\) choices for \(a\) and \(d\) and \(p^2 - p\) choices for \(b\) and \(e\). In total, there are \((p - 1)(p^2 - 1)(p^2 - p)\) choices for \(g\). Hence, we have \(|C_G(x_1)| = |C_G(x_2)| = (p - 1)(p^2 - 1)(p^2 - p)\). And we can get the \[ \begin{align} \text{orb}(x_0) &= 1,\\ \text{orb}(x_1) &= \dfrac{|G|}{|C_G(x_1)|} = \frac{(p^3 - 1)(p^3 - p)(p^3 - p^2)}{(p - 1)(p^2 - 1)(p^2 - p)} = p^2(p^2 +p + 1) = p^4 + p^3 + p^2,\\ \text{orb}(x_2) &= \dfrac{|G|}{|C_G(x_2)|} = \frac{(p^3 - 1)(p^3 - p)(p^3 - p^2)}{(p - 1)(p^2 - 1)(p^2 - p)} = p^2(p^2 +p + 1) = p^4 + p^3 + p^2,\\ \text{orb}(x_3) &= 1. \end{align} \] Then, we have the number of idempotent \(3 \times 3\) matrices is \(2p^4 + 2p^3 + 2p^2 + 2\).