Brian Bi

## Section 7.10. Generators and Relations

Exercise 7.10.1 Let $$\mathcal{F}$$ be the free group on a set of generators $$S$$, and let $$f : S \to G$$ be a function from $$S$$ to a group $$G$$. It is clear that there is at most one way to extend $$f$$ to a group homomorphism $$\varphi : \mathcal{F} \to G$$, since $$S$$ generates $$\mathcal{F}$$, namely, by writing each element of $$\mathcal{F}$$ as a reduced word, replacing each instance of a generator $$s \in S$$ by $$f(s)$$, and replacing each instance of $$s^{-1}$$ by $$f(s)^{-1}$$, where $$s \in S$$. We need to show that this is indeed a valid homomorphism. Let $$g_1, g_2$$ be arbitrary elements of $$\mathcal{F}$$; write them as reduced words: \begin{align*} g_1 &= s_1^{d_1} \ldots s_m^{d_m} \\ g_2 &= t_1^{e_1} \ldots t_n^{e_n} \end{align*} so that \begin{align} \varphi(g_1) &= f(s_1)^{d_1} \ldots f(s_m)^{d_m} \nonumber \\ \varphi(g_2) &= f(t_1)^{e_1} \ldots f(t_n)^{e_n} \nonumber \\ \varphi(g_1) \varphi(g_2) &= f(s_1)^{d_1} \ldots f(s_m)^{d_m} f(t_1)^{e_1} \ldots f(t_n)^{e_n} \label{eqn:phig1g2} \end{align} Suppose that in forming the product $$g_1 g_2$$ and reducing, we need to cancel $$k$$ factors from the right of $$g_1$$ and $$k$$ factors from the left of $$g_2$$. Then the reduced word form of $$g_1 g_2$$ is $g_1 g_2 = s_1^{d_1} \ldots s_{m-k}^{d_{m-k}} t_{k+1}^{e_{k+1}} \ldots t_n^{e_n}$ and $\varphi(g_1 g_2) = f(s_1)^{d_1} \ldots f(s_{m-k})^{d_{m-k}} f(t_{k+1})^{e_{k+1}} f(t_n)^{e_n}$ The cancellation implies that for each $$j$$ from 1 to $$n$$, we had $$s_{m-j+1} = t_j$$ and $$d_{m-j+1} = -e_j$$. If that's the case, then it's also true that $$f(s_{m-j+1})^{d_{m-j+1}} f(t_j)^{e_j} = f(t_j)^{-e_j} f(t_j)^{e_j} = 1$$. So we may cancel $$k$$ pairs of factors in $$(\ref{eqn:phig1g2})$$ to obtain $\varphi(g_1)\varphi(g_2) = f(s_1)^{d_1} \ldots f(s_{m-k})^{d_{m-k}} f(t_{k+1})^{e_{k+1}} \ldots f(t_n)^{e_n} = \varphi(g_1 g_2)$ as required. So $$\varphi$$ is indeed a homomorphism.

We now turn our attention to the mapping property for free groups. Using the notation in Proposition 7.10.13, we will prove that there is a unique homomorphism $$\overline{\varphi} : \overline{G}' \to G$$ such that $$\varphi = \overline{\varphi} \circ \pi$$. The group $$\overline{G}'$$ is the group of cosets of $$N$$ in $$G'$$. For each $$C \in \overline{G}'$$, there is some $$a \in G'$$ such that $$C = [aN]$$, so that $$C = \pi(a)$$. Since we have $$\varphi(a) = \overline{\varphi}(\pi(a))$$, this implies $$\overline{\varphi}(C) = \varphi(a)$$. So knowing $$\varphi$$ allows us to deduce what $$\overline{\varphi}$$ must be. This is well-defined, because if $$a, b \in G'$$ belong to the same coset, then $$ab^{-1} \in N \subseteq K$$, so $$\varphi(a) = \varphi(b)$$. By construction, it is therefore the case that $$\varphi = \overline{\varphi} \circ \pi$$ as required. It remains to show that $$\overline{\varphi}$$ is a homomorphism. Let $$C_1, C_2 \in \overline{G}'$$ and write $$C_1 = [a_1 N], C_2 = [a_2 N]$$. Then $$C_1 C_2 = [a_1 a_2 N]$$, so $$\overline{\varphi}(C_1 C_2) = \varphi(a_1 a_2) = \varphi(a_1) \varphi(a_2) = \overline{\varphi}(C_1) \overline{\varphi}(C_2)$$, as required. We have verified that $$\overline{\varphi}$$ is a homomorphism.

Exercise 7.10.2 Let $$g \in G$$. Then $$\varphi(g)$$ is a product of the generators $$\varphi(S)$$, say, $$\varphi(g) = \varphi(s_1)^{d_1} \ldots \varphi(s_m)^{d_m}$$ where each $$s_i$$ is some element of $$S$$ and each $$d_i$$ is $$\pm 1$$. This implies $$\varphi(g) = \varphi(s_1^{d_1} \ldots s_m^{d_m})$$. Since the elements $$g$$ and $$s_1^{d_1} \ldots s_m^{d_m}$$ have the same image, it must be that $$g = k s_1^{d_1} \ldots s_m^{d_m}$$ for some $$k \in \ker \varphi$$. But then $$k = t_1^{e_1} \ldots t_n^{e_n}$$ where each $$t_i \in T$$ and each $$e_i$$ is $$\pm 1$$. So $$g = t_1^{e_1} \ldots t_n^{e_n} s_1^{d_1} \ldots s_m^{d_m}$$. Since every $$g \in G$$ can be written in this form, $$S \cup T$$ generates $$G$$.

Exercise 7.10.3 Yes. If $$G = \{g_1, \ldots, g_n\}$$ and the composition law is $$g_i g_j = g_{a_{ij}}$$ for some matrix $$a$$, then a presentation of $$G$$ is given by $$\langle g_1, \ldots, g_n \mid g_i g_j g_{a_{ij}}^{-1} \rangle$$ where $$i, j$$ range over all pairs of indices from 1 to $$n$$. The assumption that $$G$$ exists and satisfies these relations establishes that this presentation does indeed describe a group with at least $$n$$ distinct elements, and the relations themselves imply that that group is isomorphic to $$G$$.

Exercise 7.10.4 We solve this problem in three parts. First, we determine the normal closure $$N$$ of $$xyx^{-1}y^{-1}$$ in $$\mathcal{F}$$, the free group over two generators. Second, we characterize the group $$G = \mathcal{F}/N$$. Finally, we prove the universal property.

Define $$w : \mathcal{F} \to \mathbb{Z}^2$$ as the homomorphism with $$w(x) = (1, 0), w(y) = (0, 1)$$. By the universal property, this homomorphism exists and is unique, and simply counts occurrences of each generator minus its inverse in the reduced word. We claim that $$\ker w \subseteq N$$, that is, $$N$$ contains all balanced words, which are reduced words that have the same number of $$x$$'s as $$x^{-1}$$'s and the same number of $$y$$'s as $$y^{-1}$$'s. We prove the claim by induction on the length of the balanced word, which must obviously be even.

Base case: The only balanced word of length zero is the identity, which is of course in $$N$$. There are no balanced words of length two. We are given that $$xyx^{-1}y^{-1} \in N$$; successive conjugation by $$x^{-1}, y^{-1}, x$$ yields the elements $$yx^{-1}y^{-1}x, x^{-1}y^{-1}xy, y^{-1}xyx^{-1}$$. The inverses of these four elements must also be in $$N$$, namely $$yxy^{-1}x^{-1}, x^{-1}yxy^{-1}, y^{-1}x^{-1}yx, xy^{-1}x^{-1}y$$. These eight words exhaust the balanced words of length four.

Inductive case: Assume all balanced words of length less than $$2n$$ are in $$N$$. We will use this to prove that all balanced words of length $$2n$$ are in $$N$$ in the following steps:

1. Let $$a$$ denote $$x$$, $$y$$, $$x^{-1}$$, or $$y^{-1}$$. If a balanced word of length $$2n$$ has the form $$aWa^{-1}$$, then $$W$$ must also be balanced; by the induction hypothesis, $$W \in N$$; therefore $$aWa^{-1} \in N$$ also. That is, all balanced words of the form $$aWa^{-1}$$ of length $$2n$$ are in $$N$$.
2. If a balanced word of length $$2n$$ has the form $$a W_1 a^{-1} W_2$$ where the length of $$W_2$$ is at most $$n-2$$, then $$a W_1 a^{-1} W_2 = (a W_1 W_2 a^{-1})(a W_2^{-1} a^{-1} W_2)$$. By step 1, the left factor is in $$N$$. The right factor is balanced, and has length at most $$2(n-1)$$. By the induction hypothesis, the right factor is in $$N$$. Therefore all balanced $$a W_1 a^{-1} W_2 \in N$$.
3. If a balanced word of length $$2n$$ has the form $$a W_1 a^{-1} W_2$$ where the length of $$W_1$$ is at most $$n-2$$, then note that by step 2, the balanced word $$a^{-1} W_2 a W_1^{-1}$$ is in $$N$$, and $$a W_1 a^{-1} W_2$$ is the conjugate of that element by $$a W_1$$. Therefore $$a W_1 a^{-1} W_2 \in N$$.
4. If a balanced word of length $$2n$$ has the form $$W_1 a W_2 a^{-1} W_3$$, where the length of $$W_2$$ is not $$n-1$$, then note that by steps 1 and 2, the balanced word $$a W_2 a^{-1} W_3 W_1^{-1}$$ is in $$N$$, and the word $$W_1 a W_2 a^{-1} W_3$$ is the conjugate of that word by $$W_1$$. Therefore $$W_1 a W_2 a^{-1} W_3 \in N$$.
5. If a balanced word of length $$2n$$ is not covered by case 4, it can only take the form $$W = a_1 a_2 \ldots a_n a_1^{-1} a_2^{-1} \ldots a_n^{-1}$$. Note that the word $$W'$$ formed by swapping the initial $$a_1$$ and $$a_2$$ is then covered by case 4 so $$W' \in N$$; finally $$W = (a_1 a_2 a_1^{-1} a_2^{-1})W'$$, and $$a_1 a_2 a_1^{-1} a_2^{-1} \in N$$ by the base case $$n = 2$$, therefore $$W \in N$$.

We have proven that $$\ker w \in N$$. But $$\ker w$$ is normal, so $$N = \ker w$$.

Now let's turn our attention to the structure of $$G$$. By the first isomorphism theorem, $$G = \mathcal{F}/N$$ is isomorphic to $$\im w$$, which is simply $$\mathbb{Z}^2$$ itself. Explicitly, the elements of $$G$$ are fibers of $$w$$, so this isomorphism maps $$[xN]$$ to $$(1, 0)$$ and $$[yN]$$ to $$(0, 1)$$. In the group $$G$$, the symbol $$x$$ denotes the coset $$[xN]$$ and $$y$$ denotes $$[yN]$$. So we can also say that $$G$$ is isomorphic to $$\mathbb{Z}^2$$ with $$x$$ mapped to $$(1, 0)$$ and $$y$$ mapped to $$(0, 1)$$.

We can finally prove the universal property. Let an abelian group $$A$$ be given and let $$u, v \in A$$. We know that $$x, y$$ generate $$G$$ because $$(1, 0), (0, 1)$$ generate $$\mathbb{Z}^2$$. Therefore there is at most one homomorphism $$\varphi : G \to A$$ with $$\varphi(x) = u, \varphi(y) = v$$. To show that $$\varphi$$ actually exists and is a homomorphism, define it as $$\varphi = \gamma \circ \varphi'$$ where $$\varphi'$$ is the isomorphism from $$G$$ to $$\mathbb{Z}^2$$ and $$\gamma$$ is the homomorphism defined by $$\gamma((1, 0)) = u, \gamma((0, 1)) = v$$. It is obvious that $$\gamma$$ is a valid homomorphism, therefore $$\varphi$$ is also a valid homomorphism, as required.

Exercise 7.10.5 Idea: Let $$G$$ denote the group specified. Notice that the defining relation $$yxyz^{-2} = 1$$ is equivalent to $$x = y^{-1}z^2y^{-1}$$. Define a homomorphism $$\varphi : G \to \mathcal{F}(y, z)$$ as follows: to compute $$G(w)$$ for some reduced word $$w$$, replace every occurrence of $$x$$ with $$y^{-1}z^2 y^{-1}$$ and every occurrence of $$x^{-1}$$ with $$yz^{-2}y$$, and cancel to obtain a reduced word. Argue that the result doesn't depend on the order of substitutions and cancellations, and use this to show that $$\varphi$$ is a homomorphism. Argue that the normal closure $$N$$ of the defining relation $$x^{-1}y^{-1}z^2y^{-1}$$ is $$\ker \varphi$$, and by the first isomorphism theorem, $$G = \mathcal{F}(x, y, z)/N \cong \im \varphi = \mathcal{F}(y, z)$$. Details are left as an exercise.

Exercise 7.10.6

1. Let $$H$$ be a characteristic subgroup of $$G$$. Then $$H$$ is carried to itself by all inner automorphisms, which are simply conjugation by the various elements of $$G$$. Therefore $$H$$ is normal. To see that the centre is normal, observe that if $$x$$ is central and $$\varphi$$ is an automorphism, then for all $$y \in G$$, $$\varphi(x)y = \varphi(x)\varphi(\varphi^{-1}(y)) = \varphi(x\varphi^{-1}(y)) = \varphi(\varphi^{-1}(y)x) = y\varphi(x)$$, therefore $$\varphi(x)$$ is also central. Applying this logic using the inverse automorphism establishes that the converse is also true, so the centre must be carried to itself.
2. $$Q_8$$ has the trivial group and $$Q_8$$ itself as normal subgroups. We can classify the remaining normal subgroups based on whether they contain an element of order 4. If such a subgroup doesn't contain an element of order 4, it can only be the $$C_2$$ subgroup consisting of $$\pm 1$$. If it does contain an element of order 4, it must either be the group generated by that element or the entire group, by Lagrange's theorem. But all subgroups generated by an element of order 4 are normal since they have index 2 in $$Q_8$$. So all six subgroups of $$Q_8$$, namely the trivial group, $$Q_8$$, $$\langle -1\rangle$$, $$\langle i \rangle$$, $$\langle j \rangle$$, and $$\langle k \rangle$$, are normal.

Only three of those subgroups are characteristic. Obviously the trivial group and $$Q_8$$ itself are characteristic; $$\{1, -1\}$$ is characteristic because $$-1$$ is the only element of order 2. The subgroups of order 4 are not characteristic; for example, the automorphism that sends $$i, j, k$$ to $$j, k, i$$ respectively permutes those three subgroups in a 3-cycle.

Exercise 7.10.7 Let $$[x, y]$$ denote the commutator of $$x$$ and $$y$$. It is easy to see that if $$\varphi$$ is any homomorphism, then $$\varphi([x, y]) = [\varphi(x), \varphi(y)]$$. Also note that $$[x, y] = [y, x]^{-1}$$. Suppose now that $$\varphi$$ is an automorphism and that $$g \in C$$. Then we can write $$g$$ in the form $$[x_1, y_1][x_2, y_2] \ldots [x_n, y_n]$$ where all $$x_i$$'s and $$y_i$$'s are elements of $$G$$. (We don't need any inverses here since we can just exchange the two elements in the commutator, but the argument doesn't depend on this.) Then $$\varphi(g) = \varphi([x_1, y_1]) \ldots \varphi([x_n, y_n]) = [\varphi(x_1), \varphi(y_1)] \ldots [\varphi(x_n), \varphi(y_n)]$$, therefore $$\varphi(g) \in C$$ also. By applying this logic to the inverse automorphism $$\varphi^{-1}$$ we can establish the converse also. So $$C$$ is characteristic in $$G$$.

To show that $$G/C$$ is abelian, we will work with the cosets directly. Let $$[xC], [yC]$$ be two elements of $$G/C$$. Then $$[xC][yC] = [xyC]$$ and $$[yC][xC] = [yxC]$$. But $$C$$ contains the element $$[x^{-1}, y^{-1}] = x^{-1}y^{-1}xy$$, so $$[yxC]$$ contains $$yxx^{-1}y^{-1}xy = xy$$, so these two cosets are equal. Since $$[xC], [yC]$$ were arbitrary, $$G/C$$ is abelian.

Exercise 7.10.8

1. The group $$SO(2)$$ is abelian, so its derived subgroup is trivial.
2. Let $$x, y \in O(2)$$. The commutator $$[x, y]$$ must lie in the $$SO(2)$$ subgroup because in $$xyx^{-1}y^{-1}$$ any reversal of orientation caused by $$x$$ will be undone by $$x^{-1}$$ and likewise with $$y$$ and $$y^{-1}$$. If $$z = \rho_{\theta}$$, then a quick calculation shows that $$z = [\rho_{\theta/2}, r]$$. So $$[SO(2), SO(2)] = SO(2)$$.
3. As in part (b), it is clear that all commutators are orientation-preserving, and according to part (b), $$[M, M]$$ contains all rotations. Let $$t_a$$ be a rotation and let $$r$$ be the reflection that maps $$a$$ to $$-a$$, that is, a reflection across the line perpendicular to the vector $$a$$. A quick calculation shows that $$[t_a, r] = t_{2a}$$, so $$[M, M]$$ contains all translations too. Since the translations and rotations generate the orientation-preserving subgroup of $$M$$, that subgroup is $$[M, M]$$.
4. The cases $$n = 1$$ and $$n = 2$$ are trivial. By a similar argument to that of part (b), the commutators of $$S_n$$ must all be even permutations, so $$[S_n, S_n] \subseteq A_n$$. Let $$x = (1\ 2), y = (2\ 3)$$. Then $$[x, y] = (1\ 2)(2\ 3) (1\ 2)(2\ 3) = (1\ 2\ 3)(1\ 2\ 3) = (1\ 3\ 2)$$. It is clear by analogy that every 3-cycle is a commutator in $$S_n$$. By Lemma 7.5.5(a), $$A_n$$ is generated by the 3-cycles. So $$[S_n, S_n] = A_n$$.
5. Let $X = \begin{pmatrix} \cos \theta & -\sin\theta & 0 \\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix} \qquad Y = \begin{pmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$ That is, $$X$$ is a rotation of angle $$\theta$$ around the z-axis, while $$Y$$ is a rotation of 180 degrees around the x-axis. Calculation shows that $[X, Y] = \begin{pmatrix} \cos 2\theta & -\sin 2\theta & 0 \\ \sin 2\theta & \cos 2\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}$ So we see that $$[SO(3), SO(3)]$$ contains all rotations around the z-axis. But by Exercises 7.10.7 and 7.10.6(a), the derived subgroup is normal, so it contains all conjugates of rotations around the z-axis, and by Corollary 5.1.28(b), this includes all rotations in $$SO(3)$$. Therefore $$[SO(3), SO(3)] = SO(3)$$.

Exercise 7.10.9 We found in Exercise 2.12.2 that the centre of the group $$G$$ consists of all those matrices of the form $\begin{pmatrix} 1 & 0 & x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ where $$x$$ is an arbitrary element of the ground field; this did not depend on choice of field.

Direct calculation shows that $\left[\begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & d & e \\ 0 & 1 & f \\ 0 & 0 & 1 \end{pmatrix} \right] = \begin{pmatrix} 1 & 0 & af - cd \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$ so in fact $$[G, G] = Z(G)$$.

Finally, direct calculation shows, and induction can easily be used to prove, that $\begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}^n = \begin{pmatrix} 1 & na & nb + \binom{n}{2}ac \\ 0 & 1 & nc \\ 0 & 0 & 0 \end{pmatrix}$ For $$p > 2$$, the order of any element where either $$a$$ or $$c$$ is nonzero is at least $$p$$ since we need $$na = nc = 0$$, and $$p$$ divides $$\binom{p}{2}$$ so the condition that $$nb + \binom{n}{2}ac = 0$$ is then satisfied as well, and the order is exactly $$p$$. Also if $$a = c = 0$$ then it is clear the order is also $$p$$. So for $$p > 2$$, all elements have order $$p$$ except the identity. For $$p = 2$$, this argument fails since $$\binom{2}{2} = 1$$, which isn't a multiple of 2. This only matters in the case where $$ac \neq 0$$. So the two matrices $\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}, \qquad \begin{pmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}$ have order 4 while all other non-identity matrices have order 2.

Exercise 7.10.10

1. We showed in Exercise 7.10.4 that $$\mathcal{R}$$ consists of the balanced words; and $$x^2 y^2 x^{-2} y^{-2}$$ is a balanced word.
2. By Exercises 7.10.7 and 7.10.6(a), $$[\mathcal{F}, \mathcal{F}]$$ is a normal subgroup. Since it contains $$[x, y]$$, it must contain all of $$\mathcal{R}$$. Also, it is easy to see that a commutator is always a balanced word, so $$[\mathcal{F}, \mathcal{F}] \subseteq \mathcal{R}$$. Putting the two results together, we see that the derived subgroup is exactly $$\mathcal{R}$$.