Brian Bi

## 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 $$\begin{bmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix}^{-1} = \begin{bmatrix} 1 & -a & ac-b \\ 0 & 1 & -c \\ 0 & 0 & 1 \end{bmatrix} \label{eqn:e1}$$ 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: $$\begin{bmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & d & e \\ 0 & 1 & f \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & d+a & e+af+b \\ 0 & 1 & f + c \\ 0 & 0 & 1 \end{bmatrix} \label{eqn:e2}$$ 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.