Cosets and Lagrange's Theorem
Table of Contents
\(\newcommand{\qed}{\tag*{\(\blacksquare\)}}\)
1. Exercises
Describe the left cosets of \(SL_2(\mathbb{R})\) in \(GL_2(\mathbb{R})\). What is the index of \(SL_2(\mathbb{R})\) in \(GL_2(\mathbb{R})\)?
\(Solution.\) A left coset of \(SL_2(\mathbb{R})\) in \(GL_2(\mathbb{R})\) has the form \[ASL_2(\mathbb{R}) = \{AB \big | B \in SL_2(\mathbb{R})\}\] where \(A \in GL_2(\mathbb{R})\) and \(\text{det}(A) \not = 0\). If \(B \in SL_2(\mathbb{R})\), then \(\text{det}(B) = 1\). \(\text{det}(AB) = \text{det}(A)\text{det}(B) = \text{det}(A)\), thus all the elements in \(ASL_2{\mathbb{R}}\) have their determinant equal to \(d\).
Conversely, suppose that \(C \in GL_2{\mathbb{R}}\) and \(\text{det}(C) = \text{det}(A)\). \(\text{det}(A^{-1}C) = 1\), so \[C = AA^{-1}C \in ASL_2(\mathbb{R})\] Hence, a left coset of \(SL_2(\mathbb{R})\) is a collection of matrices with fixed determinant.
Obviously, the index of \(SL_2(\mathbb{R})\) in \(GL_2(\mathbb{R})\) is infinite.
\(\blacksquare\) Suppose that \([G:H] = 2\). If \(a\) and \(b\) are not in \(H\), show that \(ab \in H\).
\(Proof.\) If \(a\) and \(b \not \in H\), then \(a = gh_1\), \(b = gh_2\) where \(g \not \in H\). If \(ab = gh_1gh_2 \not \in H\), then \(ab\) must be in \(gH\), thus \(\exists h \in H\), \(ab = gh_1gh_2 = gh\). Finally, we can get \(g \in H\) which contradict with that \(g \not \in H\).
Hence, \(ab \in H\).
\(\blacksquare\) Let \(n = p_1^{e_1}p_2^{e_2} \cdots p_k^{e_k}\), where \(p_1, p_2, \cdots , p_k\) are distinct primes. Prove that \[\phi(n) = n\left(1 - \frac{1}{p_1} \right)\left(1 - \frac{1}{p_2} \right) \cdots \left(1 - \frac{1}{p_k} \right)\].
\(Proof.\) For any prime \(p\), \(\phi(p) = p - 1\). \(\phi(p^e) = p^e - p^{e - 1} = p^e(1 - \dfrac{1}{p})\).
Let \(G = \langle g \rangle \) is a cyclic group of order \(n\) and \(H_i = \langle g^{p_i} \rangle \), thus \(|H_i| = \dfrac{n}{p_i}\).
According to inclusion-exclusion principle, we have
\begin{aligned} \phi(n) &= |G| - |\bigcup_{i = 1}^{k} H_i| \\ &= |G| - (\sum_{i = 1}^{k}|H_i| - \sum_{1 \le i,j \le k} |H_i \cap H_j| + \cdots + (-1)^{k + 1}|\bigcap_{i = 1}^{k} H_i|). \end{aligned}Because
\[|\bigcap_{a \le i \le b} H_i| = \frac{n}{p_a \cdots p_b}\] we can finally get
\begin{align*} \phi(n) &= n(1 - \sum_{i = 1}{k}\frac{1}{p_i} + \cdots + (-1)^k\frac{n}{p_1 \cdots p_k}) \\ &= n\left(1 - \frac{1}{p_1} \right)\left(1 - \frac{1}{p_2} \right) \cdots \left(1 - \frac{1}{p_k} \right) \qed \end{align*}Show that \[n = \sum_{d|n}\phi(d)\] for all positive integers \(n\).
\(Proof.\) If \(|a^k| = d\), then
\[\frac{n}{\text{gcd}(n, k)} = d \text{, } (n, k) = \frac{n}{d} \text{.}\]
thus \(k = q \dfrac{n}{d}\) where \(q\) is a positive integer. Hence \((n, k) = (n, q\dfrac{n}{d}) = \dfrac{n}{d}(d, q)\), so \((d, q) = 1\). \(|\{1 \le q \le d | (d, q) = 1\}| = \phi(d)\). The order of elements in a cyclic group must be a factor of the group,so
\[ n = \sum_{d|n}\phi(d) \qed \]