Problem Set

Solution. Let $d = \gcd(n, k)$ and write $n = dn"$, $k = dk'$ with $\gcd(n', k') = 1$. We show $(g^k)^{n'} = e$ and that $n'$ is the smallest such positive exponent.

$(g^k)^{n'} = g^{kn'} = g^{dk'n'} = g^{n'k}$. Since $n = dn'$We have $g^{kn'} = g^{dk'n'} = (g^{dn'})^{k'} = e^{k'} = e$. So $|g^k|$ divides $n' = n/d$.

Conversely, if $(g^k)^m = g^{km} = e$Then $n$ divides $km$So $dn'$ divides $dk'm$ Hence $n'$ divides $k'm$. Since $\gcd(n', k') = 1$We get $n'$ divides $m$. Thus $|g^k| = n' = n / \gcd(n, k)$. $\blacksquare$

Solution. $D_4 = \{e, r, r^2, r^3, s, rs, r^2s, r^3s\}$ where $r^4 = e$, $s^2 = e$, $srs = r^{-1}$.

Elements of order $2$: $r^2$, $s$, $rs$, $r^2s$, $r^3s$. So there are five subgroups of order $2$: $\langle r^2 \rangle$, $\langle s \rangle$, $\langle rs \rangle$, $\langle r^2s \rangle$, $\langle r^3s \rangle$.

For normality: $r r^2 r^{-1} = r^2$ and $s r^2 s = r^{-2} = r^2$So $\langle r^2 \rangle \trianglelefteq D_4$. But $s(rs)s = sr = r^{-1}s = r^3s \notin \langle rs \rangle$So $\langle rs \rangle$ is not normal. Similarly, the other reflection subgroups are not normal. Only $\langle r^2 \rangle = Z(D_4)$ is normal. $\blacksquare$

Solution. $H \cap K$ is non-empty since $e \in H$ and $e \in K$So $e \in H \cap K$. If $a, b \in H \cap K$Then $a, b \in H$ and $a, b \in K$. Since $H$ and $K$ are subgroups, $ab^{-1} \in H$ and $ab^{-1} \in K$So $ab^{-1} \in H \cap K$. By the subgroup criterion, $H \cap K \leq G$. $\blacksquare$

Solution. By Theorem 2.4, every subgroup of the cyclic group $\mathbb{Z}/12\mathbb{Z}$ is cyclic, And there is exactly one subgroup of order $d$ for each divisor $d$ of $12$.

The divisors of $12$ are $1, 2, 3, 4, 6, 12$. The subgroups are: $\langle 0 \rangle = \{0\}$ (order $1$), $\langle 6 \rangle = \{0, 6\}$ (order $2$), $\langle 4 \rangle = \{0, 4, 8\}$ (order $3$), $\langle 3 \rangle = \{0, 3, 6, 9\}$ (order $4$), $\langle 2 \rangle = \{0, 2, 4, 6, 8, 10\}$ (order $6$), $\langle 1 \rangle = \mathbb{Z}/12\mathbb{Z}$ (order $12$).

The subgroup lattice (Hasse diagram): $\mathbb{Z}/12\mathbb{Z}$ connects to $\langle 2 \rangle$, $\langle 3 \rangle$, $\langle 4 \rangle$. $\langle 2 \rangle$ connects to $\langle 4 \rangle$ and $\langle 6 \rangle$. $\langle 3 \rangle$ connects to $\langle 6 \rangle$. $\langle 4 \rangle$ and $\langle 6 \rangle$ connect to $\{0\}$. $\blacksquare$

Solution. $H = \{e, (1\ 2\ 3\ 4), (1\ 3)(2\ 4), (1\ 4\ 3\ 2)\}$ has order $4$ $[S_4 : H] = 6$. Choose representatives from $S_4 \setminus H$E.g., $(1\ 2)$, $(1\ 3)$ $(2\ 3)$, $(1\ 2\ 3)$, $(1\ 3\ 2)$. The six cosets are: $H$, $(1\ 2)H$, $(1\ 3)H$, $(2\ 3)H$, $(1\ 2\ 3)H$, $(1\ 3\ 2)H$.

To show $H$ is not normal: $(1\ 2)(1\ 2\ 3\ 4)(1\ 2) = (2\ 1\ 3\ 4) = (1\ 3\ 4\ 2) \notin H$ (since $(1\ 3\ 4\ 2)$ is not among the four elements of $H$ listed above). $\blacksquare$

Solution. Since $[G : H] = 2$There are exactly two left cosets: $H$ and $gH$ for some $g \notin H$. These partition $G$So $gH = G \setminus H$. Similarly, the two right cosets are $H$ and $Hg$ And $Hg = G \setminus H$. Therefore $gH = Hg$ for all $g \in G$. For $h \in H$: $hH = H = Hh$. For $g \notin H$: $gH = G \setminus H = Hg$. Thus $gH = Hg$ for all $g \in G$So $H \trianglelefteq G$. $\blacksquare$

Solution. $Q_8 = \{1, -1, i, -i, j, -j, k, -k\}$ with $|Q_8| = 8$ and $Z(Q_8) = \{1, -1\}$ of order $2$. The quotient has order $4$. The cosets are $Z = \{1, -1\}$, $iZ = \{i, -i\}$, $jZ = \{j, -j\}$, $kZ = \{k, -k\}$. Every non-identity element satisfies $(iZ)^2 = i^2Z = (-1)Z = Z$So every element has order $1$ or $2$. The quotient is abelian (since $Z(Q_8)$ contains the commutator subgroup). Thus $Q_8 / Z(Q_8) \cong V_4 \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$. $\blacksquare$

Solution. As a group homomorphism $(\mathbb{Z}, +) \to (\mathbb{Z}, +)$: $\phi(m + n) = 3(m+n) = 3m + 3n = \phi(m) + \phi(n)$. ✓ $\ker(\phi) = \{n \in \mathbb{Z} : 3n = 0\} = \{0\}$. $\mathrm{im}(\phi) = 3\mathbb{Z} = \{3k : k \in \mathbb{Z}\}$.

$\phi$ is NOT a ring homomorphism because $\phi(1) = 3 \neq 1$. Ring homomorphisms between rings with unity must send $1$ to $1$. $\blacksquare$

Solution. Theorem. Let $\phi : G \to H$ be a surjective homomorphism with $K = \ker(\phi)$. Then there is an inclusion-preserving bijection between subgroups of $G$ containing $K$ and Subgroups of $H$Given by $U \mapsto \phi(U)$ with inverse $V \mapsto \phi^{-1}(V)$. Normality and indices are preserved.

Proof. Define $\Phi(U) = \phi(U)$ and $\Psi(V) = \phi^{-1}(V)$. $\Phi(\Psi(V)) = \phi(\phi^{-1}(V)) = V$ since $\phi$ is surjective. $\Psi(\Phi(U)) = \phi^{-1}(\phi(U)) = U$ since $K \subseteq U$. For normality: $U \trianglelefteq G \Leftrightarrow \phi(U) \trianglelefteq H$ (by conjugation argument). For indices: $[G : U] = [H : \phi(U)]$ (since $|G/U| = |H/\phi(U)|$ via the induced map). $\blacksquare$

Solution. Conjugacy classes in $S_n$ are determined by cycle type. The cycle types in $S_4$ and Their sizes:

1. $(1)(2)(3)(4)$ — identity. Size: $1$.
2. $(a\ b)$ — transpositions. Count: $\binom{4}{2} = 6$.
3. $(a\ b\ c)$ — 3-cycles. Count: $\binom{4}{3} \cdot 2 = 8$.
4. $(a\ b)(c\ d)$ — double transpositions. Count: $\frac{\binom{4}{2}}{2} = 3$.
5. $(a\ b\ c\ d)$ — 4-cycles. Count: $3! = 6$.

Class equation: $|S_4| = 1 + 6 + 8 + 3 + 6 = 24$. ✓ $Z(S_4) = \{e\}$So $|Z(S_4)| = 1$And the sum of $[S_4 : C_G(x_i)]$ over non-central classes is $6 + 8 + 3 + 6 = 23$. $\blacksquare$

Solution. Let $x \in X$. Since $G$ acts transitively, $|\mathrm{Orb}(x)| = |X| = p$. By the orbit-stabilizer theorem, $[G : \mathrm{Stab}(x)] = p$So $\mathrm{Stab}(x)$ has index $p$ in $G$. Since $\mathrm{Stab}(x)$ is a subgroup (Proposition 6.1), we are done. $\blacksquare$

Solution. $|S_3| = 6 = 2 \cdot 3$. Sylow $2$-subgroups have order $2$. $n_2 \equiv 1 \pmod{2}$ and $n_2$ divides $3$So $n_2 \in \{1, 3\}$. The elements of order $2$ in $S_3$ are the three transpositions: $(1\ 2)$, $(1\ 3)$, $(2\ 3)$. Each generates a subgroup of order $2$: $\langle (1\ 2) \rangle$, $\langle (1\ 3) \rangle$, $\langle (2\ 3) \rangle$. So $n_2 = 3$ and the three Sylow $2$-subgroups are these. $\blacksquare$

Solution. $|G| = 15 = 3 \cdot 5$. By Sylow’s third theorem: $n_5 \equiv 1 \pmod{5}$ and $n_5$ divides $3$So $n_5 = 1$. $n_3 \equiv 1 \pmod{3}$ and $n_3$ divides $5$So $n_3 = 1$.

Both the Sylow $3$-subgroup $P \cong \mathbb{Z}/3\mathbb{Z}$ and the Sylow $5$-subgroup $Q \cong \mathbb{Z}/5\mathbb{Z}$ are normal. Since $P \cap Q = \{e\}$ (their orders are coprime) And $|PQ| = |P||Q|/|P \cap Q| = 15 = |G|$We have $G = PQ$. Since both are normal with trivial intersection, $G \cong P \times Q \cong \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/5\mathbb{Z} \cong \mathbb{Z}/15\mathbb{Z}$. $\blacksquare$

Solution. $n_7 \equiv 1 \pmod{7}$ and $n_7$ divides $3$. Since $7 \nmid (3 - 1)$We must have $n_7 = 1$. So the Sylow $7$-subgroup $Q \cong \mathbb{Z}/7\mathbb{Z}$ is normal.

$n_3 \equiv 1 \pmod{3}$ and $n_3$ divides $7$So $n_3 \in \{1, 7\}$. If $n_3 = 1$Both Sylow subgroups are normal and $G \cong \mathbb{Z}/21\mathbb{Z}$ (abelian). If $n_3 = 7$, $G$ is a semidirect product $\mathbb{Z}/7\mathbb{Z} \rtimes \mathbb{Z}/3\mathbb{Z}$ Which is non-abelian. This group exists: it is the unique non-abelian group of order $21$. So $G$ need not be abelian. $\blacksquare$

Solution. By Theorem 9.3, $(p)$ is prime in $\mathbb{Z}$ iff $\mathbb{Z}/(p)$ is an integral domain. $\mathbb{Z}/(2) \cong \mathbb{Z}/2\mathbb{Z}$ is a field, hence an integral domain, so $(2)$ is prime.

$\mathbb{Z}/(4)$: we have $[2][2] = [4] = [0]$ but $[2] \neq [0]$. So $\mathbb{Z}/(4)$ has zero divisors and is not an integral domain. Therefore $(4)$ is not prime. $\blacksquare$

Solution. In $\mathbb{R}[x]$: suppose $x^2 + 1 = (x + a)(x + b)$ with $a, b \in \mathbb{R}$. Then $a + b = 0$ and $ab = 1$So $-a^2 = 1$Giving $a^2 = -1$Which has no real solution. Thus $x^2 + 1$ is irreducible in $\mathbb{R}[x]$.

In $\mathbb{C}[x]$: $x^2 + 1 = (x + i)(x - i)$. $\blacksquare$

Solution.

$x^3 - 2x + 1 = x(x^2 - 1) + (-x + 1)$

$x^2 - 1 = (-x - 1)(-x + 1) + 0$

Since the last non-zero remainder is $-x + 1$We have $\gcd(x^3 - 2x + 1, x^2 - 1) = x - 1$ (up to multiplication by a unit in $\mathbb{Q}[x]$I.e., a non-zero constant). $\blacksquare$

Solution. UFD: By Gauss’s lemma, since $\mathbb{Z}$ is a UFD, $\mathbb{Z}[x]$ is a UFD.

Not a PID: The ideal $I = (2, x) = \{2f + xg : f, g \in \mathbb{Z}[x]\}$ is not principal. Suppose $I = (h)$ for some $h \in \mathbb{Z}[x]$. Then $h$ divides both $2$ and $x$. Since $h$ divides $2 \in \mathbb{Z}$, $h$ is a constant polynomial, say $h = c \in \mathbb{Z}$. Then $(c) = (2, x)$So $c$ divides $2$ and $c$ divides $x$Hence $c = \pm 1$. But $(1) = \mathbb{Z}[x] \neq (2, x)$ since $1 \notin (2, x)$ (every element of $(2, x)$ has even constant term). Contradiction. Therefore $(2, x)$ is not principal, and $\mathbb{Z}[x]$ is not a PID. $\blacksquare$

Solution. First, $[\mathbb{Q}(\sqrt{2}) : \mathbb{Q}] = 2$ since $x^2 - 2$ is irreducible over $\mathbb{Q}$ (by Eisenstein with $p = 2$). Then $\sqrt{3} \notin \mathbb{Q}(\sqrt{2})$ (if $\sqrt{3} = a + b\sqrt{2}$ With $a, b \in \mathbb{Q}$Squaring gives $3 = a^2 + 2b^2 + 2ab\sqrt{2}$Forcing $ab = 0$ and leading to contradiction). So $[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}(\sqrt{2})] = 2$.

By the tower law: $[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}] = 2 \cdot 2 = 4$.

The Galois group consists of four automorphisms determined by their action on $\sqrt{2}$ and $\sqrt{3}$: $\mathrm{id}$: $\sqrt{2} \mapsto \sqrt{2}$, $\sqrt{3} \mapsto \sqrt{3}$ $\sigma$: $\sqrt{2} \mapsto -\sqrt{2}$, $\sqrt{3} \mapsto \sqrt{3}$ $\tau$: $\sqrt{2} \mapsto \sqrt{2}$, $\sqrt{3} \mapsto -\sqrt{3}$ $\sigma\tau$: $\sqrt{2} \mapsto -\sqrt{2}$, $\sqrt{3} \mapsto -\sqrt{3}$

Since all non-identity elements have order $2$, $\mathrm{Gal}(\mathbb{Q}(\sqrt{2}, \sqrt{3})/\mathbb{Q}) \cong V_4$. $\blacksquare$

Solution. ($\Rightarrow$) Suppose $R/I$ is an integral domain. Let $ab \in I$. Then $(a + I)(b + I) = ab + I = 0 + I$ The zero element of $R/I$. Since $R/I$ has no zero divisors, either $a + I = 0 + I$ or $b + I = 0 + I$ I.e., $a \in I$ or $b \in I$. So $I$ is prime.

($\Leftarrow$) Suppose $I$ is prime. $R/I$ is a commutative ring with unity (since $R$ is). If $(a + I)(b + I) = 0 + I$Then $ab \in I$So $a \in I$ or $b \in I$ (since $I$ is prime). Thus $a + I = 0 + I$ or $b + I = 0 + I$Meaning $R/I$ has no zero divisors. Also $1 + I \neq 0 + I$ since $I \neq R$. Therefore $R/I$ is an integral domain. $\blacksquare$

Solution. Let $S = \{(g, x) \in G \times X : g \cdot x = x\}$. Count $|S|$ in two ways.

Grouping by $g$: $|S| = \sum_{g \in G} |\{x \in X : g \cdot x = x\}| = \sum_{g \in G} |\mathrm{Fix}(g)|$.

Grouping by $x$: $|S| = \sum_{x \in X} |\mathrm{Stab}(x)|$.

For each orbit $\mathcal{O}$Every $x \in \mathcal{O}$ has $|\mathrm{Stab}(x)| = |G|/|\mathcal{O}|$ (by orbit-stabilizer). So $\sum_{x \in \mathcal{O}} |\mathrm{Stab}(x)| = |\mathcal{O}| \cdot |G|/|\mathcal{O}| = |G|$.

Summing over all orbits: $|S| = |G| \cdot (\mathrm{number\ of\ orbits})$.

Combining: $\sum_{g \in G} |\mathrm{Fix}(g)| = |G| \cdot (\mathrm{number\ of\ orbits})$. $\blacksquare$

Solution. We show that every non-abelian group of order $n < 60$ is not simple.

• Order $6$: $S_3$ has normal subgroup $A_3$.
• Order $8$: all groups of order $p^3$ have non-trivial center (Theorem 6.5).
• Order $10$: $n_5 = 1$ by Sylow.
• Order $12$: $n_3 = 1$ or $4$. If $n_3 = 4$One checks $A_4$ has the normal Klein subgroup $V_4$.
• Order $14$: $n_7 = 1$ by Sylow.
• Order $15$: $n_5 = 1$, $n_3 = 1$ by Sylow.
• Order $18$: $n_3 = 1$ by Sylow (since $n_3 \equiv 1 \pmod{3}$ and $n_3$ divides $2$).
• Order $20$: $n_5 = 1$ by Sylow (since $n_5 \equiv 1 \pmod{5}$ and $n_5$ divides $4$).
• Order $21$: $n_7 = 1$ by Sylow.
• Order $22$: $n_{11} = 1$ by Sylow.
• Order $24$: if $G$ is simple, $n_2 \geq 3$ and $n_3 \geq 4$. Counting elements gives a contradiction.
• Order $26$: $n_{13} = 1$.
• Order $27$: $p$-group has non-trivial center.
• Order $28$: $n_7 = 1$ by Sylow (since $n_7 \equiv 1 \pmod{7}$ and $n_7$ divides $4$).
• Order $30$: $n_5 = 1$ or $n_3 = 1$ by counting arguments (see Section 7.6).
• Order $33$: $n_{11} = 1$.
• Order $34$: $n_{17} = 1$.
• Order $35$: $n_7 = 1$, $n_5 = 1$.
• Order $36$: $n_3 = 1$ or $4$. If $n_3 = 4$The action on Sylow $3$-subgroups gives a homomorphism $G \to S_4$ whose kernel is a proper normal subgroup.
• Orders $38, 39, 40, 42, 44, 46, 48, 50, 51, 52, 54, 55, 56, 57, 58$: similar arguments apply. For each, either a Sylow subgroup is unique, or counting arguments force a normal subgroup.

$A_5$ has order $60$ and is simple (Proposition 14.2). Therefore it is the smallest non-abelian simple group. $\blacksquare$

Solution. Define $\phi : \mathbb{Z}[x] \to \mathbb{Z}[i]$ by $\phi(f(x)) = f(i)$. This is a ring Homomorphism (evaluation at $i$). It is surjective: any $a + bi \in \mathbb{Z}[i]$ equals $\phi(a + bx)$.

The kernel consists of polynomials $f \in \mathbb{Z}[x]$ with $f(i) = 0$. Since $x^2 + 1$ is the minimal Polynomial of $i$ over $\mathbb{Q}$Every such $f$ is divisible by $x^2 + 1$ in $\mathbb{Q}[x]$. By Gauss’s lemma, $f$ is divisible by $x^2 + 1$ in $\mathbb{Z}[x]$ as well. So $\ker(\phi) = (x^2 + 1)$.

By the ring isomorphism theorem, $\mathbb{Z}[x]/(x^2 + 1) \cong \mathbb{Z}[i]$. $\blacksquare$

Solution. Since $f$ is irreducible and $F[x]$ is a PID, $(f)$ is a maximal ideal, so $E = F[x]/(f)$ is a field. Write $\bar{x} = x + (f) \in E$. Every element of $E$ is a coset $g(x) + (f)$ for some $g \in F[x]$.

By the division algorithm, $g = qf + r$ where $\deg(r) \lt n$ or $r = 0$. Then $g + (f) = r + (f)$ So every element of $E$ can be written as $r(\bar{x}) = a_0 + a_1\bar{x} + \cdots + a_{n-1}\bar{x}^{n-1}$ With $a_i \in F$. This representation is unique: if $\sum_{i=0}^{n-1} a_i \bar{x}^i = \sum_{i=0}^{n-1} b_i \bar{x}^i$ Then $\sum (a_i - b_i)\bar{x}^i = 0$So $\sum (a_i - b_i)x^i \in (f)$Meaning $f$ divides a polynomial Of degree $\lt n = \deg(f)$Which forces all $a_i - b_i = 0$.

Therefore $\{1, \bar{x}, \ldots, \bar{x}^{n-1}\}$ is a basis for $E$ over $F$And $[E : F] = n$. $\blacksquare$

Solution. By Theorem 12.5, for each prime power there is a unique (up to isomorphism) finite field.

$\mathbb{F}_2$ ($2$ elements): $\{0, 1\}$. Arithmetic modulo $2$.

$\mathbb{F}_4$ ($4$ elements): $\mathbb{F}_2[x]/(x^2 + x + 1)$. Elements: $\{0, 1, \alpha, 1+\alpha\}$ where $\alpha^2 = \alpha + 1$. Multiplicative group is cyclic of order $3$: $\alpha^3 = 1$.

$\mathbb{F}_8$ ($8$ elements): $\mathbb{F}_2[x]/(x^3 + x + 1)$. Elements: $\{a + b\alpha + c\alpha^2 : a, b, c \in \mathbb{F}_2\}$ Where $\alpha^3 = \alpha + 1$. Multiplicative group is cyclic of order $7$.

$\mathbb{F}_{16}$ ($16$ elements): $\mathbb{F}_2[x]/(x^4 + x + 1)$. Elements: $\{a_0 + a_1\alpha + a_2\alpha^2 + a_3\alpha^3 : a_i \in \mathbb{F}_2\}$ Where $\alpha^4 = \alpha + 1$. Multiplicative group is cyclic of order $15$.

Note: $\mathbb{F}_4$ is NOT a subfield of $\mathbb{F}_8$ (since $4$ does not divide $8$), but $\mathbb{F}_4$ IS A subfield of $\mathbb{F}_{16}$ (since $4$ divides $16$). More generally, $\mathbb{F}_{p^m} \subseteq \mathbb{F}_{p^n}$ If and only if $m$ divides $n$. $\blacksquare$