The Lumenary · Advanced undergraduate

Quotient groups and quotient rings

Name: ______________________

Date: ______________________

You will learn to

Construct $G/N$ and $R/I$ , and prove that the operations are well defined exactly when $N$ is normal, respectively when the additive subgroup $I$ is an ideal
State and prove the universal property of quotients and the first, second and third isomorphism theorems
Identify quotients such as $\Z/n\Z$ , $\R/\Z$ , $\mathrm{GL}_n(k)/\mathrm{SL}_n(k) \cong k^\times$ and $\R[x]/(x^2 + 1) \cong \C$ using the first isomorphism theorem
Prove the Chinese remainder theorem for rings
Show that $R/I$ is a field if and only if $I$ is maximal and a domain if and only if $I$ is prime, and use this to build finite fields such as $\mathbb{F}_4$ and $\mathbb{F}_9$

1.
The Klein four-group $V = \{e, (1\,2)(3\,4), (1\,3)(2\,4), (1\,4)(2\,3)\}$ is normal in $S_4$ . What is the order of the quotient group $S_4/V$ ?
2.
Let $H = \{0, 4, 8\}$ , a subgroup of $\Z/12\Z$ . What is the order of the element $5 + H$ in the quotient group $(\Z/12\Z)/H$ ?
3.
To show that $(aN)(bN) = abN$ is well defined, suppose $a^{-1}a' \in N$ and $b^{-1}b' \in N$ and write $(ab)^{-1}(a'b') = \big(b^{-1}(a^{-1}a')b\big)\big(b^{-1}b'\big)$ . At which point is the normality of $N$ needed?
- A.To conclude that $b^{-1}(a^{-1}a')b \in N$ .
- B.To conclude that $b^{-1}b' \in N$ .
- C.To multiply the two factors and stay in $N$ .
- D.To verify the identity $(ab)^{-1}(a'b') = (b^{-1}a^{-1}a'b)(b^{-1}b')$ .
4.
Which of these quotient rings are fields? (Select all that apply.)
- A. $\Z/(7)$
- B. $\Z/(6)$
- C. $\Q[x]/(x^2 - 2)$
- D. $\R[x]/(x^2 - 2)$
- E. $\mathbb{F}_2[x]/(x^2 + x + 1)$
- F. $\Z[x]/(x)$
5.
Which homomorphism, combined with the first isomorphism theorem, proves $\R[x]/(x^2 + 1) \cong \C$ ?
- A.Evaluation at $i$ : $f \mapsto f(i)$ , which is surjective with kernel $(x^2 + 1)$ .
- B.Evaluation at $0$ : $f \mapsto f(0)$ .
- C.The inclusion $\R[x] \to \C[x]$ .
- D.Differentiation $f \mapsto f'$ .
6.
How many residues $x \in \Z/35\Z$ satisfy $x^2 = x$ ? (Use the Chinese remainder theorem.)
7.
Which Lean statement formalises “for a commutative ring $R$ and an ideal $I$ , the quotient $R/I$ is a field if and only if $I$ is maximal”?
- A.theorem t {R} [CommRing R] (I : Ideal R) : I.IsMaximal ↔ IsField (R ⧸ I)
- B.theorem t {R} [CommRing R] (I : Ideal R) : I.IsPrime ↔ IsField (R ⧸ I)
- C.theorem t {R} [CommRing R] (I : Ideal R) : I.IsMaximal ↔ IsDomain (R ⧸ I)
- D.theorem t {R} [Field R] (I : Ideal R) : I.IsMaximal
8.
A student says: “ $\Z/2\Z$ is a subgroup of $\Z$ , namely $\{0, 1\}$ .” What is wrong?
- A. $\{0, 1\}$ is not closed under addition in $\Z$ ( $1 + 1 = 2$ ); the elements of $\Z/2\Z$ are the cosets $2\Z$ and $1 + 2\Z$ , not integers.
- B.Nothing: $\{0, 1\}$ with addition modulo $2$ is a subgroup of $\Z$ .
- C.The quotient should be written $2\Z/\Z$ .
- D. $2\Z$ is not normal in $\Z$ , so the quotient does not exist.
9.
Write cosets(n, k) returning the cosets of the subgroup $H = \langle k \rangle$ of $\Z/n\Z$ , as a list of sorted lists, ordered by their smallest element. Then write coset_sum(n, k, a, b) returning the coset (as a sorted list) of $a + b$ . Check that the coset of $a + b$ does not change when $a$ or $b$ is replaced by another element of its coset.
Computational — Practice

Answer key — Quotient groups and quotient rings

1.
$|S_4/V| = 24/4 = 6$ . In fact $S_4/V \cong S_3$ : $S_4$ permutes the three ways $\{\{1,2\},\{3,4\}\}$ , $\{\{1,3\},\{2,4\}\}$ , $\{\{1,4\},\{2,3\}\}$ of splitting $\{1,2,3,4\}$ into pairs, giving a surjective homomorphism $S_4 \to S_3$ whose kernel is $V$ ; the first isomorphism theorem finishes the job.
2.
The multiples of $5$ modulo $12$ are $5, 10, 3, 8, \dots$ , and the first one in $H = \{0, 4, 8\}$ is $4 \cdot 5 = 20 \equiv 8$ . So the order is $4$ . Conceptually, $\Z/12\Z \to \Z/4\Z$ , $x \mapsto x \bmod 4$ , is surjective with kernel $H$ , so $(\Z/12\Z)/H \cong \Z/4\Z$ by the first isomorphism theorem (or the third: $(\Z/12\Z)/(4\Z/12\Z) \cong \Z/4\Z$ ), and $5 \mapsto 1$ , a generator of order $4$ .
3.
The second factor $b^{-1}b'$ is in $N$ by hypothesis, and products of elements of $N$ stay in $N$ because $N$ is a subgroup. The first factor is the conjugate $b^{-1}(a^{-1}a')b$ of the element $a^{-1}a' \in N$ : it lies in $N$ for all choices exactly when $N$ is normal. The converse in the lesson shows normality is not just sufficient but necessary.
4.
Fields: $\Z/(7)$ , $\Q[x]/(x^2 - 2) \cong \Q(\sqrt2)$ and $\mathbb{F}_2[x]/(x^2 + x + 1) = \mathbb{F}_4$ . Not fields: $\Z/(6)$ has zero divisors; $\R[x]/(x^2 - 2) \cong \R \times \R$ because the polynomial factors over $\R$ ; $\Z[x]/(x) \cong \Z$ . The criterion used throughout is: $R/I$ is a field $\iff$ $I$ is maximal.
5.
Let $\varepsilon(f) = f(i)$ . It is a ring homomorphism, surjective since $\varepsilon(a + bx) = a + bi$ . Its kernel contains $x^2 + 1$ . Conversely, if $f(i) = 0$ , divide: $f = (x^2 + 1)q + (a + bx)$ with $a, b \in \R$ ; then $0 = f(i) = a + bi$ , so $a = b = 0$ and $f \in (x^2 + 1)$ . By the first isomorphism theorem $\R[x]/(x^2 + 1) \cong \C$ . Cauchy (1847) defined the complex numbers in exactly this way.
6.
By the Chinese remainder theorem, $x \mapsto (x \bmod 5, x \bmod 7)$ is a ring isomorphism $\Z/35 \to \Z/5 \times \Z/7$ . In each field the only solutions of $x^2 = x$ are $0$ and $1$ , so there are $2 \times 2 = 4$ solutions: $(0,0), (1,1), (1,0), (0,1)$ , which are $0, 1, 21, 15$ . Check: $21^2 = 441 = 12 \cdot 35 + 21$ and $15^2 = 225 = 6 \cdot 35 + 15$ .
7.
Option (a). With $R = \Z$ , $I = (0)$ : $I$ is prime but not maximal and $R/I \cong \Z$ is a domain but not a field, which refutes (b) and (c). (d) fails for $I = R$ . The two correct dictionaries are: field $\leftrightarrow$ maximal, domain $\leftrightarrow$ prime.
8.
The elements of $\Z/2\Z$ are the two cosets $2\Z$ (even numbers) and $1 + 2\Z$ (odd numbers). Writing them as $0, 1$ is a convenient labelling, not an inclusion into $\Z$ . And $\Z$ has no subgroup isomorphic to $\Z/2\Z$ , since no non-zero integer $n$ has $n + n = 0$ . Quotients are images of $G$ (under the quotient map), not subsets of $G$ .
9.
```
def cosets(n, k):
    H = sorted({k * m % n for m in range(n)})
    result = []
    for a in range(n):
        c = sorted({(a + h) % n for h in H})
        if c not in result:
            result.append(c)
    return sorted(result)

def coset_sum(n, k, a, b):
    H = sorted({k * m % n for m in range(n)})
    return sorted({(a + b + h) % n for h in H})
```
The last test checks well-definedness by brute force: every choice of representatives of two cosets gives the same coset of the sum. In an abelian group every subgroup is normal, so this always succeeds; the analogous test for $\{e, (1\,2)\}$ in $S_3$ fails (see the first experiment). Note that $\langle 4 \rangle = \langle \gcd(4, 10) \rangle = \langle 2 \rangle$ in $\Z/10\Z$ .