Classify all groups of order \(15\) and of order \(30\).
Solution
The only group of order \(15\) is \(C_{15}\). The groups of order \(30\), up to isomorphism, are \[ C_{30},\qquad C_3\times D_{10},\qquad C_5\times D_6,\qquad D_{30}, \] where \(D_{2m}\) denotes the dihedral group of order \(2m\).
1. Every group of order \(15\) is cyclic.
Sylow’s theorems give \(n_5\mid3\), \(n_5\equiv1\pmod5\), and \(n_3\mid5\), \(n_3\equiv1\pmod3\) [@DF04]. Thus \(n_5=n_3=1\). The respective Sylow subgroups \(Q,P\) are normal and have trivial intersection. Their commutators lie in that intersection, so they commute elementwise. Multiplication \(P\times Q\to G\) is an injective homomorphism: \(xy=1\) forces \(x=y^{-1}\in P\cap Q\). Both sides have order \(15\), so it is an isomorphism. Each prime-order factor is cyclic, and a pair of generators has order \(15\).
2. Every group \(G\) of order \(30\) is a semidirectproduct \(C_{15}\rtimes C_2\).
Let \(\lambda:G\to\operatorname{Sym}(G)\) be the action by left multiplication and put \(\varepsilon=\operatorname{sgn}\circ\lambda:G\to\{1,-1\}\). This is a homomorphism [@DF04]. Cauchy’s theorem provides \(t\in G\) of order two. Left multiplication by \(t\) has no fixed point, since \(tx=x\) would imply \(t=1\). Thus its cycle decomposition consists of fifteen transpositions and \(\varepsilon(t)=(-1)^{15}=-1\). The sign character is onto, so its kernel \(N\) is normal of order \(15\). Step 1 makes \(N\) cyclic.
Since \(t\notin N\) and \([G:N]=2\), every element has a unique form \(a^i t^e\), where \(a\) generates \(N\), \(i\in\mathbb Z/15\mathbb Z\), and \(e\in\{0,1\}\). This is the asserted semidirect product.
3. Exactly four conjugation actions can occur.
Write \(tat^{-1}=a^u\) for a unit \(u\) modulo \(15\). The equation \(t^2=1\) forces \(u^2\equiv1\pmod{15}\). Modulo each of \(3\) and \(5\), the product \((u-1)(u+1)\) vanishes in a field, so \(u=1\) or \(-1\). The Chinese remainder theorem gives exactly four solutions modulo \(15\), namely \(u=1,4,11,14\) [@DF04].
The multiplication in the coordinates of step 2 is \[ (i,e)(j,f)=(i+u^e j,e+f), \] with the coordinates reduced modulo \(15\) and \(2\). For each of the four values this defines a group: associativity follows from \(u^{e+f}=u^e u^f\), the identity is \((0,0)\), and the inverse of \((i,e)\) is \((-u^{-e}i,-e)\). Hence each action exists and determines the group up to isomorphism.
Under \(C_{15}\cong C_3\times C_5\), the respective actions are trivial on both factors, inversion on the \(C_5\) factor only, inversion on the \(C_3\) factor only, and inversion on both. These give the four groups displayed at the start, in the same order.
4. The four groups of order \(30\) are pairwise nonisomorphic.
For \(u=1\) the group is abelian and its center has order \(30\). For \(u\ne1\), every element outside \(N\) acts on \(N\) by the same nontrivial automorphism \(a\mapsto a^u\), so none is central. An element \(a^i\in N\) is central exactly when it commutes with \(t\), equivalently \((u-1)i\equiv0\pmod{15}\).
For \(u=4\), this holds precisely when \(5\mid i\), giving center order three. For \(u=11\), it holds precisely when \(3\mid i\), giving center order five. For \(u=14\), the integer \(13\) is coprime to \(15\), so only \(i=0\) qualifies, giving center order one. The distinct center orders \(30,3,5,1\) distinguish all four groups. Steps <1>1–<1>3 prove exhaustiveness.