For \(n = 5, 6\), either give an example of a nonabelian group of order \(n\), or prove that none exists.
Solution
Case \(n = 5\):
1. Every group of prime order is cyclic, hence abelian.
1.1. Let \(G\) be a group with \(|G| = 5\).
setup.
1.2. Choose an element \(g \in G \setminus \{e\}\).
\(|G| = 5 > 1\).
1.3. By Lagrange’s theorem, the order of the subgroup \(\langle g \rangle\) divides \(|G| = 5\).
Lagrange’s theorem.
1.4. Since \(g \neq e\), \(|\langle g \rangle| > 1\), so \(|\langle g \rangle| = 5\).
\(5\) is prime, so its only positive divisors are \(1\) and \(5\).
1.5. Therefore, \(G = \langle g \rangle \cong \mathbb{Z}/5\mathbb{Z}\), which is cyclic and abelian.
1.4.
1.6. No nonabelian group of order \(5\) exists.
5.
Case \(n = 6\):
2. The symmetric group \(S_3\) is a nonabelian group of order \(6\).
2.1. The order of \(S_3\) is \(3! = 6\).
order formula for symmetric groups on \(3\) elements.
2.2. Let \(\sigma = (1\ 2)\) and \(\tau = (2\ 3)\) in \(S_3\).
definitions of transpositions in \(S_3\).
2.3. \(\sigma \tau = (1\ 2)(2\ 3) = (1\ 2\ 3)\).
direct cycle composition: \(1 \mapsto 1 \mapsto 2\), \(2 \mapsto 3 \mapsto 3\), \(3 \mapsto 2 \mapsto 1\).
2.4. \(\tau \sigma = (2\ 3)(1\ 2) = (1\ 3\ 2)\).
direct cycle composition: \(1 \mapsto 2 \mapsto 3\), \(2 \mapsto 1 \mapsto 1\), \(3 \mapsto 3 \mapsto 2\).
2.5. Since \(\sigma \tau \neq \tau \sigma\), \(S_3\) is nonabelian.
\((1\ 2\ 3) \neq (1\ 3\ 2)\) (for example, they map \(1\) to different elements).
2.6. Hence \(S_3\) (or equivalently the dihedral group \(D_3\)) is a nonabelian group of order \(6\).
1 and 5.
3. Q.E.D.
1 and 2.