7 problems found
Let \(a\) be a non-zero real number and define a binary operation on the set of real numbers by $$ x*y = x+y+axy \,. $$ Show that the operation \(*\) is associative. Show that \((G,*)\) is a group, where \(G\) is the set of all real numbers except for one number which you should identify. Find a subgroup of \((G,*)\) which has exactly 2 elements.
Solution: Claim: \(*\) is associative. Proof: Then \(x*(y*z) = x*(y+z+ayz) = x + (y+z+ayz) + ax(y+z+ayz) = x + y + z + a(yz + xy + zx) + a^2xyz\) and \((x*y)*z = (x+y+axy)*z = (x+y+axy) + z+ a(x+y+axy)z = x + y + z + a(yz + xy + zx) + a^2xyz\) so \(x*(y*z) = (x*y)*z\) and we are done. Let \(G = \mathbb{R} \setminus \{-\frac1{a} \}\) In order to show that \((G, *)\) is a group we need to check:
Solution:
Consider the following sets with the usual definition of multiplication appropriate to each. In each case you may assume that the multiplication is associative. In each case state, giving adequate reasons, whether or not the set is a group.
Solution:
| 1 | 3 | 5 | 7 | |
| 1 | 1 | 3 | 5 | 7 |
| 3 | 3 | 1 | 7 | 5 |
| 5 | 5 | 7 | 1 | 3 |
| 7 | 7 | 5 | 3 | 1 |
| 1 | 2 | 3 | 4 | |
| 1 | 1 | 2 | 3 | 4 |
| 2 | 2 | 4 | 1 | 3 |
| 3 | 3 | 1 | 4 | 2 |
| 4 | 4 | 3 | 2 | 1 |
| (i) | (iii) | (iv) | (vi) | |
| (i) | \(\checkmark\) | \(\checkmark\) consider \(z \mapsto \begin{pmatrix} \cos \arg (z) | - \sin \arg(z) | |
| \sin \arg(z) | \cos \arg(z) \end{pmatrix}\) | not finite | not finite | |
| (iii) | \(\checkmark\) | not finite | not finite | |
| (iv) | \(\checkmark\) | no element order \(4\) | ||
| (vi) | \(\checkmark\) |
Explain what is meant by the order of an element \(g\) of a group \(G\). The set \(S\) consists of all \(2\times2\) matrices whose determinant is \(1\). Find the inverse of the element \(\mathbf{A}\) of \(S\), where \[ \mathbf{A}=\begin{pmatrix}w & x\\ y & z \end{pmatrix}. \] Show that \(S\) is a group under matrix multiplication (you may assume that matrix multiplication is associative). For which elements \(\mathbf{A}\) is \(\mathbf{A}^{-1}=\mathbf{A}\)? Which element or elements have order 2? Show that the element \(\mathbf{A}\) of \(S\) has order 3 if, and only if, \(w+z+1=0.\) Write down one such element.
Solution: The order of an element \(g\) is the smallest positive number \(k\) such that \(g^k = e\). $\mathbf{A}^{-1} = \begin{pmatrix}z & -x\\ -y & w \end{pmatrix}$. Claim, \(S\) is a group. \begin{enumerate} \item (Closure) The product of two \(2\times2\) matrices is always a \(2\times 2\) matrix so we only need to check the determinant. Suppose \(\det(\mathbf{A}) = \det (\mathbf{B}) = 1\), then \(\det(AB) = \det(A)\det(B) = 1\), so our operation is closed \item (Associativity) Inherited from matrix multiplication \item (Identity) $\mathbf{I} =\begin{pmatrix}1 & 0\\ 1 & 1 \end{pmatrix}\( has determinant \)1$. \item (Inverses) The inverse is always fine since the matrix of cofactors always contains integers and the determinant is one, so we never end up with anything which isn't an integer. \end{itemize} If \(\mathbf{A}^-1 = \mathbf{A}\) then assuming $\mathbf{A} = \begin{pmatrix}a & b\\ c & d \end{pmatrix}\( then \)\mathbf{A}^{-1} = \begin{pmatrix}d & - b\\ -c & a \end{pmatrix}\( so we must have \)a=d, -b=b, -c=c\(, so \)b = c = 0\( and \)a = d\(. For the determinant to be \)1\( we must have \)ad = a^2 = 1\(, ie \)a = \pm 1\(. Therefore we must have \)\mathbf{A} = \begin{pmatrix}1 & 0\\ 0 & 1 \end{pmatrix}\( or \)\mathbf{A} = \begin{pmatrix}-1 & 0\\ 0 & -1 \end{pmatrix}$. For an element to have order \(2\) then \(\mathbf{A}^2 = \mathbf{I}\) ie, \(\mathbf{A} = \mathbf{A}^{-1}\) and \(\mathbf{A} \neq \mathbf{I}\) therefore the only element of order \(2\) is $\begin{pmatrix}-1 & 0\\ 0 & -1 \end{pmatrix}$. For an element to have order \(3\) we must have \(\mathbf{A}^2 = \mathbf{A}^{-1}\), ie $\begin{pmatrix}w^2 + xy & x(w+z)\\ y(w+z) & z^2 + xy \end{pmatrix} = \begin{pmatrix}z & -x\\ -y & w \end{pmatrix}$. Therefore \(w^2 + xy = z, x(w+z) = -x, y(w+z) = -y, z^2+xy = w\). The second and third equations are satisfied iff \(w+z+1 = 0\) or \(x = 0\) and \(y = 0\), but if \(x = 0\) and \(y = 0\) then we aren't order \(3\), so we just need to check this is sufficient for the first and last equations. Since \(\det(\mathbf{A}) = 1\) we have \(wz =xy +1\), so the first and last equations are equivalent to \(w^2 + wz - 1 = z\) and \(x^2 + wz-1 = w\) which are equivalent to \(w(w+z) = z+1\) or \(w + z+ 1 = 0\) as required
Let \(G\) be the set of all matrices of the form \[ \begin{pmatrix}a & b\\ 0 & c \end{pmatrix}, \] where \(a,b\) and \(c\) are integers modulo 5, and \(a\neq0\neq c\). Show that \(G\) forms a group under matrix multiplication (which may be assumed to be associative). What is the order of \(G\)? Determine whether or not \(G\) is commutative. Determine whether or not the set consisting of all elements in \(G\) of order \(1\) or \(2\) is a subgroup of \(G\).
Solution: Claim \(G\) is a group under matrix multiplication
The set \(S\) consists of \(N(>2)\) elements \(a_{1},a_{2},\ldots,a_{N}.\) \(S\) is acted upon by a binary operation \(\circ,\) defined by \[ a_{j}\circ a_{k}=a_{m}, \] where \(m\) is equal to the greater of \(j\) and \(k\). Determine, giving reasons, which of the four group axioms hold for \(S\) under \(\circ,\) and which do not. Determine also, giving reasons, which of the group axioms hold for \(S\) under \(*\), where \(*\) is defined by \[ a_{j}*a_{k}=a_{n}, \] where \(n=\left|j-k\right|+1\).
Solution:
Let \((G,*)\) and \((H,\circ)\) be two groups and \(G\times H\) be the set of ordered pairs \((g,h)\) with \(g\in G\) and \(h\in H.\) A multiplication on \(G\times H\) is defined by \[ (g_{1},h_{1})(g_{2},h_{2})=(g_{1}*g_{2},h_{1}\circ h_{2}) \] for all \(g_{1},g_{2}\in G\) and \(h_{1},h_{2}\in H\). Show that, with this multiplication, \(G\times H\) is a group. State whether the following are true or false and prove your answers.
Solution: Claim: \(G \times H\) is a group. (Called the product group). Proof: Checking the group axioms: