Worked Examples

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). $\blacksquare$

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 $10$ . $\blacksquare$

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