*Algebra*

## Section 2.12. Quotient Groups

Exercise 2.12.1 Suppose \(H\) is a subgroup of \(G\) and that the product of two left cosets is always a left coset. Suppose \(g \in G, h \in H\) and consider the cosets \(gH\) and \(g^{-1}H\). Their product contains the identity, so it must be \(H\) itself; it also contains \(ghg^{-1}h^{-1}\), which must therefore be in \(H\), implying that \(ghg^{-1} \in H\) as well. Since this holds for all choices of \(g\) and \(h\), we conclude that \(H\) is normal. This is the contrapositive to what we wanted to prove.

Exercise 2.12.2 One can easily verify that \(I \in H\), that the product of two upper triangular matrices is upper triangular, and that \begin{equation} \begin{bmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & -a & ac-b \\ 0 & 1 & -c \\ 0 & 0 & 1 \end{bmatrix} \label{eqn:e1} \end{equation} establishing that \(H\) is a subgroup of \(GL_3(\mathbb{R})\). Obviously \(K\) is a subset of \(H\) containing the identity, and elements of \(K\) can be multiplied by adding the upper-right elements together, implying that \(K\) satisfies the conditions to be a subgroup of \(H\). Note that multiplication between elements of \(K\) and elements of \(H\) is given by \begin{align*} \begin{bmatrix} 1 & 0 & x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix} &= \begin{bmatrix} 1 & a & b + x \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix} \\ \begin{bmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} &= \begin{bmatrix} 1 & a & x + b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix} \end{align*} therefore \(K\) is contained in the centre of \(H\), implying that \(K\) is a normal subgroup of \(H\). Moreover, let \(h \in H\); then the coset \(hK\) consists of precisely those elements of \(H\) that are equal to \(h\) at the 12 and 23 entries. Call this coset \(C_{ac}\) where \(a = h_{12}\) and \(c = h_{23}\). To see how these cosets behave under multiplication, we compute the product of two arbitrary elements: \begin{equation} \begin{bmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & d & e \\ 0 & 1 & f \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & d+a & e+af+b \\ 0 & 1 & f + c \\ 0 & 0 & 1 \end{bmatrix} \label{eqn:e2} \end{equation} so the product of the cosets \(C_{ac}\) and \(C_{df}\) is \(C_{d+a,f+c}\); so the quotient group \(H/K\) is isomorphic to \(\mathbb{R}^+ \times \mathbb{R}^+\). Using \((\ref{eqn:e2})\) we can also see that multiplication in the opposite order yields the same matrix but with \(ce\) in place of \(af\), therefore the two matrices commute iff \(af = cd\). If either \(a\) or \(c\) is nonzero, it is always possible to choose \(f\) and \(d\) such that \(af \neq cd\), therefore the centre of \(H\) is precisely \(K\).

Exercise 2.12.3 For each \(g \in G\), write \(P_g\) for the equivalence class of \(g\) (that is, the element of the partition \(P\) that contains \(g\)). We are given that \(N = P_1\). Another useful fact is that for any \(a, b \in G\), the product \(P_a P_b\) is contained within \(P_{ab}\). (Since \(ab \in P_a P_b\) and the RHS is contained within some set of \(P\), the latter must be the unique element of \(P\) that contains \(ab\), namely \(P_{ab}\).)

We are given that \(1 \in N\). For all \(n \in N\), we have \(P_{n^{-1}}P_n \subseteq P_1\). But \(P_{n^{-1}}P_n\) also contains \(n^{-1}1\) since \(1 \in P_n\). Therefore \(n^{-1} \in P_1 = N\). So \(N\) is closed under inverses.

Also, \(P_1 P_1 \subseteq P_1\), but \(N = P_1\), so \(NN \subseteq N\), that is, \(N\) is closed under the operation of \(G\). So \(N\) is a subgroup of \(G\).

Moreover, for all \(g \in G\), we have \(gNg^{-1} \subseteq P_g P_1 P_{g^{-1}} \subseteq P_1 = N\). So \(N\) is a normal subgroup of \(G\).

Take \(a \in G\); then \(aN \subseteq P_a P_1 \subseteq P_a\). Also, \(a^{-1} P_a \subseteq P_{a^{-1}} P_a \subseteq P_1 = N\), therefore \(P_a \subseteq aN\). Taken together, these results imply that \(aN = P_a\). Thus, each coset of \(N\) exactly coincides with one of the elements of \(P\), and since the cosets cover \(G\), there are no elements of \(P\) that are not cosets of \(N\).

Exercise 2.12.4 The coset \(zH\) consists of the four complex numbers \(\{z, iz, -z, -iz\}\). On the complex plane, these are four points with the same distance from the origin, spaced out evenly on the circumference of the circle. We claim that an isomorphism between \(G\) and \(G/H\) is given by \(\varphi(re^{i\theta}) = [re^{i\theta/4}]\).

First, we need to check that this is well-defined. If \(z = re^{i\theta} = r'e^{i\theta'}\), then \(r' = r\) and \(\theta' - \theta = 2\pi i n\) for some \(n \in \mathbb{Z}\). In that case, \(re^{i\theta'/4} = e^{2\pi i n/4} re^{i\theta/4}\), but \(e^{2\pi i n/4} \in H\), so \([re^{i\theta'/4}] = [re^{i\theta/4}]\). So \(\varphi\) is indeed well-defined.

Now let \(z = re^{i\theta}, z' = r'e^{i\theta'}\). Then \(\varphi(zz') = \varphi(rr' e^{i(\theta + \theta')}) = [rr' e^{i(\theta + \theta')/4}]\), and \(\varphi(z)\varphi(z') = [re^{i\theta/4}][r'e^{i\theta'/4}]\); this latter coset is easily seen to contain the element \(rr'e^{i(\theta + \theta')/4}\), so it is the same coset as \(\varphi(zz')\). Therefore \(\varphi\) is a homomorphism.

Finally, suppose \(\varphi(re^{i\theta}) = [1]\). Then \([re^{i\theta/4}] = [1]\), implying \(re^{i\theta/4} = e^{2\pi i n/4}\) for some \(n \in \mathbb{Z}\), implying \(r = 1\) and \(\theta/4 - 2\pi n/4 = 2\pi m\) for some \(m \in \mathbb{Z}\), so \(\theta\) is an integer multiple of \(2\pi\) and \(re^{i\theta} = 1\). That is, the kernel of \(\varphi\) is trivial; \(\varphi\) is an isomorphism as we claimed.