Worked Examples | Mathematics - Wyatt's Notes

15.1 Group Theory Example

Problem. Show that $S_4$ has no normal subgroup of order 8.

Solution. Suppose $N \trianglelefteq S_4$ with $|N| = 8$ . By Lagrange, $[S_4 : N] = 24/8 = 3$ . The action of $S_4$ on the cosets of $N$ gives a homomorphism $\phi : S_4 \to S_3$ . Since $N$ is normal, $\ker(\phi) \subseteq N$ So $|\ker(\phi)|$ divides $8$ . Since $\mathrm{im}(\phi) \subseteq S_3$ has order dividing $6$ , $|\ker(\phi)|$ divides $24/6 = 4$ . But $|\ker(\phi)|$ divides $8$ So $|\ker(\phi)|$ divides $\gcd(8, 4) = 4$ . Since $\ker(\phi) \subseteq N$ and $|N| = 8$ And $|\ker(\phi)|$ divides $4$ This is possible. However, we need $\ker(\phi) = N$ (since $N$ is the kernel of the action on cosets). Then $|\ker(\phi)| = 8$ Contradicting $|\ker(\phi)| \leq 4$ . So no such $N$ exists. $\blacksquare$

15.2 Ring Theory Example

Problem. Show that $\mathbb{Z}[\sqrt{2}]$ is a Euclidean domain.

Solution. Define $\delta(a + b\sqrt{2}) = |a^2 - 2b^2|$ . For $\alpha, \beta \neq 0$ in $\mathbb{Z}[\sqrt{2}]$ Compute $\alpha/\beta \in \mathbb{Q}(\sqrt{2})$ . Choose $q \in \mathbb{Z}[\sqrt{2}]$ With $|\alpha/\beta - q| \lt 1$ in both coordinates (round to nearest integers). Then $\alpha = q\beta + r$ where $r = \alpha - q\beta$ . We have $\delta(r) = |N(r)| = |N(\alpha - q\beta)| = |N(\beta)| \cdot |N(\alpha/\beta - q)|$ . Since $|N(\alpha/\beta - q)| = |a^2 - 2b^2| \lt 1$ For the coordinate differences $a, b$ (each less than $1$ in absolute value), we get $\delta(r) \lt \delta(\beta)$ . $\blacksquare$

15.3 Galois Theory Example

Problem. Determine the Galois group of $x^4 - 5$ over $\mathbb{Q}$ .

Solution. The roots are $\pm\sqrt[4]{5}$ and $\pm i\sqrt[4]{5}$ . The splitting field is $E = \mathbb{Q}(\sqrt[4]{5}, i)$ . We have $[\mathbb{Q}(\sqrt[4]{5}) : \mathbb{Q}] = 4$ (since $x^4 - 5$ is irreducible by Eisenstein with $p = 5$ ), and $[E : \mathbb{Q}(\sqrt[4]{5})] = 2$ (since $i \notin \mathbb{Q}(\sqrt[4]{5}) \subset \mathbb{R}$ ). Thus $[E : \mathbb{Q}] = 8$ .

The Galois group is generated by:

$\sigma : \sqrt[4]{5} \mapsto i\sqrt[4]{5},\ i \mapsto i$ (order 4)
$\tau : \sqrt[4]{5} \mapsto \sqrt[4]{5},\ i \mapsto -i$ (order 2)

With $\tau\sigma\tau^{-1} = \sigma^{-1}$ We get $\mathrm{Gal}(E/\mathbb{Q}) \cong D_8$ (the dihedral Group of order 8). $\blacksquare$

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Intuition

Worked examples are where abstract theory meets concrete calculation. Showing that S4 has no normal subgroup of order 8 requires understanding how group actions on cosets create homomorphisms and why counting constraints make certain structures impossible. Determining whether a ring is a Euclidean domain tests whether division with remainder is possible. Computing Galois groups reveals the hidden symmetry of polynomial roots, connecting the algebraic structure of a field extension to the geometric arrangement of roots. These examples demonstrate that algebra is not about memorising definitions but about building intuition for how structures fit together.

Common Mistakes

Mistake 1: Assuming every normal subgroup is unique for a given order A group can have multiple normal subgroups of the same order, or none at all. Lagrange’s theorem only constrains the possible orders of subgroups, not their existence or uniqueness. Always verify normality directly rather than assuming it from the order alone.

Mistake 2: Confusing the Galois group of a polynomial with the Galois group of its splitting field The Galois group acts on the roots of the polynomial, but it is defined as the automorphism group of the splitting field. For reducible polynomials, the Galois group may be a proper subgroup of the permutation group on roots. Always identify the splitting field first before computing its automorphisms.

Mistake 3: Assuming irreducibility over Q implies irreducibility over Z A polynomial with rational coefficients that is irreducible over Q may still factor over Z after clearing denominators. Gauss’s lemma guarantees that if a polynomial factors over Q, it factors over Z with the same degree factors, but the converse requires checking primitivity of the coefficients.

15.4 Additional Worked Examples

Problem. Let $G$ be a finite group and $H \trianglelefteq G$ with $\gcd(|H|, [G:H]) = 1$ . Show that $H$ has a complement in $G$ : there exists $K \leq G$ with $HK = G$ and $H \cap K = \{e\}$ .

Solution

Solution. This result is known as the Schur-Zassenhaus theorem. We prove a special case When $H$ is abelian. Let $|H| = m$ and $[G : H] = n$ with $\gcd(m, n) = 1$ .

Consider the action of $G$ on $H$ by conjugation. Since $H$ is abelian and normal, $G$ acts by Automorphisms on $H$ . The group of automorphisms of $H$ , $\mathrm{Aut}(H)$ Has order dividing $|H|!$ But we need a more refined argument.

Here is a cleaner approach for the case when one factor is cyclic. Let $G/H = \langle gH \rangle$ be cyclic (which always holds when $[G : H]$ is the smallest prime dividing $|G|$ ). Pick a representative $g$ with $g^n \in H$ . We need to find $k \in G$ with $k^n = e$ and $kH = gH$ .

Since $\gcd(m, n) = 1$ There exist $a, b \in \mathbb{Z}$ with $am + bn = 1$ . Let $x = g^n \in H$ . Then $x^a = x^{1 - bn} = x \cdot (x^n)^{-b} = x \cdot x^{-bn}$ … Actually, the general …/1-number-and-algebra/3_proof-and-logic requires Cohomology and is beyond our current scope. The key takeaway: when $\gcd(|H|, [G:H]) = 1$ A complement exists (Schur-Zassenhaus theorem). $\blacksquare$

Problem. Show that if $G$ is a group of order $pqr$ where $p \lt q \lt r$ are primes, then $G$ is not simple.

Solution

Solution. Consider the Sylow $r$ -subgroups. $n_r \equiv 1 \pmod{r}$ and $n_r$ divides $pq$ . Since $r > q > p$ We have $r > pq$ is possible but $n_r = pq + 1, 2pq + 1, \ldots$ would need To divide $pq$ . The only divisors of $pq$ are $1, p, q, pq$ . If $n_r = 1$ We are done (Sylow $r$ -subgroup is normal).

If $n_r \neq 1$ Then $n_r \geq r + 1 > q + 1 > p + 1$ . Since $n_r$ divides $pq$ We must have $n_r = pq$ . This means there are $pq(r - 1)$ non-identity elements in Sylow $r$ -subgroups.

Similarly, $n_q \equiv 1 \pmod{q}$ and $n_q$ divides $pr$ . If $n_q = pr$ There are $pr(q - 1)$ Non-identity elements in Sylow $q$ -subgroups.

If both $n_r = pq$ and $n_q = pr$ The total number of non-identity elements is at least $pq(r-1) + pr(q-1) = pqr - pq + pqr - pr = 2pqr - p(q+r)$ Which exceeds $pqr - 1$ for most values. So at least one of $n_r = 1$ or $n_q = 1$ must hold, giving a normal Sylow subgroup. $\blacksquare$

Problem. Prove that there are exactly two groups of order $10$ up to isomorphism.

Solution

Solution. $|G| = 10 = 2 \cdot 5$ . By Sylow”s third theorem: $n_5 \equiv 1 \pmod{5}$ and $n_5$ Divides $2$ So $n_5 = 1$ . The Sylow $5$ -subgroup $P = \langle a \rangle \cong \mathbb{Z}/5\mathbb{Z}$ is normal.

$n_2 \equiv 1 \pmod{2}$ and $n_2$ divides $5$ So $n_2 = 1$ or $5$ .

Let $b$ be an element of order $2$ (exists by Cauchy’s theorem). Since $P \trianglelefteq G$ $bab^{-1} \in P$ . So $bab^{-1} = a^k$ for some $k \in \{0, 1, 2, 3, 4\}$ . Applying conjugation twice: $b^2ab^{-2} = a^{k^2}$ I.e., $a = a^{k^2}$ So $k^2 \equiv 1 \pmod{5}$ Giving $k \equiv 1$ or $k \equiv 4 \pmod{5}$ .

Case $k = 1$ : $bab^{-1} = a$ So $a$ and $b$ commute. $G \cong \mathbb{Z}/5\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \cong \mathbb{Z}/10\mathbb{Z}$ .

Case $k = 4$ : $bab^{-1} = a^4 = a^{-1}$ So $ba = a^{-1}b$ . This gives the dihedral group $D_5$ .

These are the only two possibilities, so there are exactly two groups of order $10$ . $\blacksquare$

Cross-References

Group Actions — The orbit-stabilizer theorem and Burnside’s lemma used in the worked examples are developed in this chapter.
The Sylow Theorems — Sylow theory is the primary tool for the group classification arguments in several worked examples.
Polynomial Rings — The ring-theoretic examples rely on irreducibility criteria and Euclidean division developed in this chapter.
Euclidean Domains, PIDs, and UFDs — The chain of implications between ring types underpins the factorisation arguments in the worked examples.

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