1 Group Homomorphism
Suppose \(\phi: G \to H\) is a group homomorphism.
1.1 Show that \(\phi(e_G) = e_H\).
\[\begin{align*} e_H \phi(e_G) &= \phi(e_G) \\ &= \phi(e_G e_G) \\ &= \phi(e_G) \phi(e_G) \\ e_H &= \phi(e_G) \tag*{$\square$} \end{align*}\]
1.2 Show that \(\phi(g^{-1}) = \phi(g)^{-1}\).
By earlier tutorial/homework it suffices to show that for any \(g\in G\), \[ \phi(g^{-1}) \phi(g) = e_H \] which is true because \[\begin{align*} \phi(g^{-1}) \phi(g) &= \phi(g^{-1} g) \\ &= \phi(e_G) \\ &= e_H \tag*{$\square$} \end{align*}\]
2 Group isomorphism.
\(I\circ I = I\)
Homework 1.
Homework 1.
3 \(\mathbb{Z}/n\mathbb{Z}\cong \mu_n\)
Verify that \(\phi^{-1}\) is given by \(z \mapsto \frac{n}{2\pi i} \log z\).
Let \(k, k' \in \mathbb{Z}/nZ\), verify that
- Case \(k +_\mathbb{Z}k' < n\), \(\phi(k+k') = e^{2\pi ik/n}e^{2\pi ik'/n}\)
- Case \(k +_\mathbb{Z}k' \geq n\), then \(\phi(k+k') = \phi(k+_\mathbb{Z}k' -_\mathbb{Z}n) = e^{2\pi i(k+k'-n)/n} = e^{2\pi ik/n}e^{2\pi ik'/n}\)
4
Let \(\phi: \mathbb{Z}/12\mathbb{Z}\to \mathbb{Z}/12\mathbb{Z}\) be a group isomorphism.
4.1 \(\phi(0) = 0\).
By question 1a.
4.2 \(\phi(6) = 6\).
We have \(\phi(6 + 6) = \phi(0) = 0 = \phi(6) + \phi(6)\). Verify that in \(\mathbb{Z}/12\mathbb{Z}\), only \(0\) and \(6\) are solutions to the equation \(t + t = 0\), but as \(\phi\) is an isomorphism \(\phi(6) \neq \phi(0)\), so we have \(\phi(6) = 6\).
4.3 Show that \(\phi(1) \notin \left\{\, 2,3,4,6,8,9,10 \,\right\}\)
Let \(x\in \left\{\, 2,3,4,6,8,9,10 \,\right\}\), then \(d = \gcd(x,12) > 1\). Define \(a := \frac{12}{d}\), as \(d > 1\), \(a < 12\). Since \(ax = \operatorname{lcm}(12,x)\), we have \[\begin{align*} a x &\equiv 0 \pmod{12} \\ \underbrace{x + x + \dots + x}_{a\text{ times}} &= 0 \end{align*}\] Suppose for a contradiction \(\phi(1) = x\), then \[\begin{align*} \phi(\underbrace{1 + 1 + \dots + 1}_{a\text{ times}}) &= \underbrace{x + x + \dots + x}_{a\text{ times}} \\ \phi(a) &= 0 \end{align*}\] but we showed earlier that \(a \ne 0 \in \mathbb{Z}/12\mathbb{Z}\), so this contradicts fact that \(\phi\) is an isomorphism. \(\square\)
4.4 Show that there is a group isomorphism \(\phi\) such that \(\phi(1) = 5\).
As \(\gcd(5,12) = 1\), \(\operatorname{lcm}(5,12) = 60 = 5\times 12\). So for all \(x\in \left\{\, 0,\dots,11 \,\right\}\), \(5x \not\equiv 0 \pmod{12}\).
DO THIS
5 \(\operatorname{GL}_n\)
5.1 \(\operatorname{GL}(n,\mathbb{R})\) is a group
- Associative
- MA1101R
- Identity
- Verify that identity matrix \(I_n\) does what we want and \(\det(I_n) = 1 \ne 0\).
- Inverse
- Linear algebra.
5.2 \(\operatorname{SL}_n\) is a subgroup.
By multiplicativity of determinant, let \(A, B\in \operatorname{SL}_n\), and \(\det(AB^{-1}) = \det(A)\det(B^{-1}) = 1\), win.
6 Let \(G = \mathbb{Z}/100\mathbb{Z}\).
6.1 Let \(d \mid 100\), show that \(T = \left\{\, 0,d, 2d,\dots, 100-d \,\right\}\) is a subgroup of \(G\).
Notice that \(100-d = \frac{100}{d} - 1\). Define \(k = \frac{100}{d}\). Take \(ad, bd\in T\) where \(a,b\in\left\{\, 0,1,\dots,k-1 \,\right\}\), then \[\begin{align*} ad + bd &= \begin{cases} ad + bd &\text{if } ad + bd < 100 = kd \\ ad + bd - 100 &\text{otherwise} \end{cases} \\ &= \begin{cases} (a + b)d &\text{if } a + b < k \\ (a + b - k)d &\text{otherwise} \end{cases} \in T \end{align*}\] similar argument for inverse.
6.2 \(H < G\), \(H \ne \left\{\, 0 \,\right\}\), show \(H = T\) and \(d \mid 100\) where \(d = \min(H\setminus\left\{\, 0 \,\right\})\).
To show \(d\mid 100\), consider \(\langle d \rangle\), we have \(\left\lvert \langle d \rangle \right\rvert \mid 100\) so there exists an integer which multiplied by \(d\) gives \(100\), so \(d\mid 100\).
\(T = \langle d \rangle \subseteq H\) is clear. Suppose for a contradiction \(T \subsetneq H\), then take \(h\in H\setminus T\). By choice of \(d\), \(d < h\). Consider the largest \(x \in \mathbb{Z}\) such that \[\underbrace{d + d + \dots + d}_{x \text{ times}} < h. \] LHS and RHS both are in \(H\), so take \(h - xd\) which is still in \(H\) but is a number less than \(d\), this contradicts our choice of \(d\). \(\square\)