Brian Bi

Return to table of contents for Brian's unofficial solutions to Artin's Algebra

Section 2.4. Cyclic Groups

Exercise 2.4.1 Since $a^7 = 1$, it follows that $ab = a^{15}b$. We can use $a^3 b = ba^3$ to exchange $b$ with 3 factors of $a$; doing this 5 times; we obtain $a^{15}b = ba^{15}$. So $ab = ba^{15} = ba$.

Exercise 2.4.2

The $n$^th roots of unity are the $n$ complex numbers $e^{2\pi ik/n}$ with $k = 0, 1, \ldots, n-1$. It is easily verified that this set contains 1 and is closed under multiplication and inverses, so it is a group. It is cyclic since it is generated by the element $e^{2\pi i/n}$.
If $n$ is odd, then the $n$^th roots of unity occur in $(n-1)/2$ complex conjugate pairs together with 1, so the product is 1. If $n$ is even, then there are $n/2-1$ complex conjugate pairs together with 1 and -1, so the product is -1. In general, the product is $(-1)^{n+1}$.

Exercise 2.4.3 Let $n$ be the order of $ab$. Then $1 = (ab)^n = a(ba)^{n-1}b$. Left-multiplying by $b$ right-multiplying by $b^{-1}$ yields $1 = ba(ba)^{n-1} = (ba)^n$. Therefore the order of $ba$ is at most $n$. By exchanging the roles of $a$ and $b$ in the above argument, we see that the order of $ab$ is also at most the order of $ba$. Therefore the two orders are equal.

Exercise 2.4.4 Suppose $G$ is a group that contains no proper subgroup. If $G$ is nontrivial, then $G$ must be cyclic; otherwise the subgroup generated by some element of $G$ other than the identity would be a proper subgroup. If $G$ is cyclic and finite with order greater than 1, the order must be prime, since $C_p$ is a proper subgroup of $C_{pq}$ whenever $p, q > 1$. If $G$ is infinite cyclic, then it has proper subgroups, such as the subgroup generated by $1 + 1$. Thus, the possible groups are the trivial group and the cyclic groups of prime order.

Exercise 2.4.5 Let $G$ be a cyclic group and $H$ a subgroup of $G$. Let $x$ be a generator of $G$, and write $H = \{x^i \mid i \in S\}$ where the set $S$ is possibly infinite. Let $d$ be the GCD of $S$, which is defined even if $S$ is infinite; see Exercise 2.3.3. By definition, $d$ is a linear combination of the elements of $S$, say, $d = \sum_{i \in S} c_i i$ where all $c_i$'s are integers and only finitely many are nonzero; and $d$ generates the set of such linear combinations. This implies that $x^d = \prod_{i \in S} (x^i)^{c_i}$, so $H$ contains $x^d$, and that $d$ divides all $i \in S$ so that all $x^i$'s are powers of $x^d$. It follows that $H$ is cyclic with generator $x^d$.

Exercise 2.4.6 In general, let $G = C_n$ and let $x$ be some generator of $C_n$. Observe that $m$ and $n$ are relatively prime if and only if $m$ and $-n$ are relatively prime, so by Corollary 2.3.6, there exist $r, s \in \mathbb{Z}$ such that $rm - sn = 1$ if and only if $m$ and $n$ are relatively prime. If such $r, s$ exist, then $(x^m)^r = x^{sn + 1} = x$, so $x^m$ generates $C_n$. Conversely, if $x \in \langle x^m \rangle$, then there is some $r$ with $x = (x^m)^r = x^{mr}$, so $x^{mr-1} = 1$, implying $mr - 1 = sn$ for some $s$, or $rm - sn = 1$; so when $m$ is not relatively prime to $n$, $x \notin \langle x^m \rangle$. Therefore the number of elements of $C_n$ that generate $C_n$ is precisely $\varphi(n)$, the number of integers from 0 to $n-1$, inclusive, that are relatively prime to $n$. Therefore $C_6$ has 2 generators, $C_5$ has 4, and $C_8$ has 4.

Exercise 2.4.7 Since all non-identity elements of $H$ have order 2, each such element is its own inverse, so each element of $H$ has an inverse in $H$. It remains to verify that $H$ is closed under multiplication. We only need to establish that $xy, x(xy), yx, y(xy), (xy)x$, and $(xy)y$ are in $H$; the products with 1 are obviously in $H$, and the squares of elements in $H$ have already been discussed. Now $x(xy) = (xx)y = 1y = y \in H$. Similar reasoning shows that $(xy)y \in H$. And $(yx)xy = y(xx)y = yy = 1$, so $yx$ is the inverse of $xy$, implying that $yx = xy \in H$. This then allows us to establish that $y(xy) = y(yx) = (yy)x = x \in H$ and likewise $(xy)x = (yx)x = y(xx) = y \in H$.

Exercise 2.4.8 I can't see how solving (a) first makes (b) easier, so instead I will give a standalone solution to (b) and reduce (a) to (b).

Suppose $A \in GL_n(\mathbb{R})$ is given. Let $c = \det A$. Then, $A = BC$ where $B = \diag(c, 1, 1, \ldots, 1)$ and $C$ is identical to $A$ except that each entry in the first row is equal to $1/c$ times the corresponding entry of $A$. By part (b), $C$ can be written as a product of elementary matrices of the first type. $B$ is an elementary matrix of the third type. Therefore $A$ is a product of elementary matrices of the first and third types.
First, some notation. Let $E_n(r, c, a)$ denote the $n \times n$ elementary matrix of the first type with the entry $a$ located at position $(r, c)$. Let $S_n(i_1, i_2, c)$ denote the diagonal matrix with the entry $c$ at $(i_1, i_1)$, the entry $1/c$ at $(i_2, i_2)$, and ones elsewhere on the diagonal. Note that left-multiplication by $S_n(i_1, i_2, c)$ has the effect of scaling rows $i_1$ and $i_2$ by a factor of $c$ and $1/c$, respectively, while leaving other rows unchanged. Let $T_n(i_1, i_2)$ denote the matrix whose entries are as follows: \begin{equation*} T_n(i_1, i_2)_{r,c} = \begin{cases} 1 & \text{if $r = c$, and $r \neq i_1$, and $r \neq i_2$} \\ -1 & \text{if $r = i_1$ and $c = i_2$} \\ 1 & \text{if $r = i_2$ and $c = i_1$} \\ 0 & \text{otherwise} \end{cases} \end{equation*} Note that left-multiplication by $T_n(i_1, i_2)$ has the effect of swapping rows $i_1$ and $i_2$, then changing the sign of the new row $i_2$.

As the text suggests, we first consider the case $n = 2$. If one can work out how to express $S_2$ and $T_2$ matrices in terms of $E_2$ matrices, the idea can be extended to arbitrary $n$. Here one can easily verify that \begin{align*} S_2(1, 2, c) &= \begin{pmatrix} c & 0 \\ 0 & 1/c \end{pmatrix} \\ &= \begin{pmatrix} 1 & 0 \\ -1/c & 1 \end{pmatrix} \begin{pmatrix} 1 & c-1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & \frac{1-c}{c} \\ 0 & 1 \end{pmatrix} \\ &= E_2(2, 1, -1/c)\, E_2(1, 2,c-1)\, E_2(2, 1, 1)\, E_2(1, 2, (1-c)/c) \\ T_2(1, 2) &= \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \\ &= \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & -1 \\ 0 & 1 \end{pmatrix} \\ &= E_2(1, 2, -1)\, E_2(2, 1, 1)\, E_2(1, 2, -1) \end{align*} This allows us to deduce a way of writing the $S_n$ and $T_n$ matrices as products of $E_n$ matrices in general: \begin{align*} S_n(i_1, i_2, c) &= E_n(i_2, i_1, -1/c)\, E_n(i_1, i_2, c-1)\, E_n(i_2, i_1, 1) E_n(i_1, i_2, (1-c)/c) \\ T_n(i_1, i_2) &= E_n(i_1, i_2, -1)\, E_n(i_2, i_1, 1)\, E_n(i_1, i_2, -1) \end{align*}

Now in general the idea is to use row reduction to write an arbitrary $A \in SL_n(\mathbb{R})$ as the product of $E_n$'s. The $S_n$ and $T_n$ matrices correspond naturally to certain operations done during row reduction. We claim that if the first $k$ columns of $A$ are the same as those of the $n \times n$ identity matrix, then $A$ can be written as a product of $E_n$'s. The result will be proven by induction on $n - k$.

Base case: If $n - k = 1$, then by expansion in minors, $\det A = A_{n,n}$, therefore $A_{n,n} = 1$. Therefore \begin{equation*} A = E_n(1,n,A_{1,n})\, E_n(2,n,A_{2,n}) \ldots E_n(n-1,n,A_{n-1,n}) \end{equation*}

Inductive case: Suppose $n - k \ge 2$ and the left $k$ columns of $A$ are the same as those of an identity matrix. We now work on column $k+1$. If $A_{k+1,k+1}$ is nonzero then let $B = A$. Otherwise, since $A$ has full rank, the $(1+k)$^th column can't be a linear combination of the first $k$ columns, so there must be some $m > k$ with $A_{m,k+1} \neq 0$. In the process of row reduction, we would swap rows $k+1$ and $m$, but since we are working with matrices of determinant 1, we also have to change the sign of one of the rows. Here, this is accomplished by writing $A = T_n(k+1, m) B$. Note that the first $k$ columns of $B$ are also the same as those of the identity matrix, but $B_{k+1,k+1} \neq 0$. So we have reduced the problem of $A$ to that of $B$.

In the process of row reduction, we would now scale row $k+1$ by $1/c$ to make the entry at $(k+1,k+1)$ equal to 1. Since we are working with matrices of determinant 1, we have to scale some higher-numbered row by $c$ as well in order to preserve the determinant. Since $n - k \ge 2$, we can write $B = S_n(k+1, k+2, B_{k+1,k+1})C$ where the first $k$ rows of $C$ are the same as those of the identity matrix, but $C_{k+1,k+1} = 1$. So we have reduced the problem of $B$ to that of $C$.

The final step of row reduction is to zero out all elements in column $k+1$ other than the pivot. This is done using the $E_n$'s themselves (see (1.2.5) in the text). The result is that $C = E_n(1, k+1, -C_{1,k+1})\, E_n(2, k+1, -C_{2,k+1})\ldots D$ where the first $k+1$ columns of $D$ are the same as those of the identity matrix. By the inductive hypothesis, $D$ is a product of elementary matrices of the first kind. This completes the induction.

By taking $k = 0$, we recover the desired result in full generality.

Exercise 2.4.9 A permutation $P$ that equals its own inverse can be expressed as a product of disjoint transpositions, where each transposition $(a\ b)$ is formed from a pair of distinct elements with $P(a) = b, P(b) = a$. Conversely, every product of disjoint transpositions is involutive. So the number of elements of order 2 in $S_4$ is equal to the number of ways to select some disjoint pairs of elements from a set of size 4, not counting the empty set of pairs (corresponding to the identity, which has order 1). There are 6 permutations containing one transposition and 3 more containing two tranpositions, for a total of 9.

Exercise 2.4.10 Define $A, B \in GL_2(\mathbb{R})$ as follows: \begin{align*} A &= \begin{pmatrix} -1 & 0 \\ 0 & 1 \end{pmatrix} \\ B &= \begin{pmatrix} -1 & 0 \\ 1 & 1 \end{pmatrix} \end{align*} Then $A^2 = B^2 = I$, but \begin{align*} AB &= \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix} \\ (AB)^n &= \begin{pmatrix} 1 & 0 \\ n & 1 \end{pmatrix} \end{align*} so $AB$ does not have finite order. But in an abelian group, if $a$ has order $m$ and $b$ has order $n$, then $(ab)^{mn} = (a^m)^n(b^n)^m = 1$, so $ab$ has finite order too.

Exercise 2.4.11

Write $P \in S_n$ as a product of disjoint cycles. It suffices to show that the tranpositions generate the cycles. Suppose $C = (a_1\ a_2 \ldots a_n)$. Then it can be written as the product of transpositions $C = (a_1\ a_2) (a_2\ a_3) \ldots (a_{n-1}\ a_n)$, and we are done.
Write $P \in A_n$ as a product of transpositions $P = \tau_1 \tau_2 \ldots \tau_{2k-1} \tau_{2k}$. The number of these transpositions must be even; we can pair them up like so: $P = (\tau_1\tau_2) (\tau_3\tau_4) \ldots (\tau_{2k-1}\tau_{2k})$. It suffices to write each parenthesized factor as a product of 3-cycles. We can do so using the fact that if $a, b, c, d$ are distinct, then $(a\ b)(c\ d) = (a\ d\ c)(a\ b\ c)$; otherwise WLOG assume $b = c$, and then $(a\ b)(c\ d) = (a\ b\ d)$. This completes the proof. Note that we didn't use the fact that $n \geq 3$, so the result holds for $n = 2$ as well; in this case $A_n$ is trivial so it is generated by the (empty) set of 3-cycles.