Let \(G\) be a subgroup of \(S_5\), the group of all permutations of five objects. Prove that if \(G\) contains a \(5\)-cycle and a \(2\)-cycle, then \(G=S_5\).
Solution
Let \(\sigma\in G\) be a \(5\)-cycle and let \(\tau\in G\) be a transposition. Label the five objects by \(\ZZ/5\ZZ\) in the cyclic order of \(\sigma\), so that \[ \sigma(k)=k+1. \] Write \[ \tau=(r,s) \] and set \[ d=s-r\in\ZZ/5\ZZ. \]
1. For every \(k\in\ZZ/5\ZZ\), the group \(G\) contains the transposition\[ (r+k,s+k). \]
Since \(\sigma,\tau\in G\) and \(G\) is a subgroup, \[ \sigma^k\tau\sigma^{-k}\in G. \] Conjugation carries a transposition to the transposition of the images of its two entries, so \[ \sigma^k(r,s)\sigma^{-k} = (r+k,s+k). \]
2. The graph on \(\ZZ/5\ZZ\) whose edges are\[ \{r+k,s+k\}, \qquad k\in\ZZ/5\ZZ, \]is connected.
Since \(r\neq s\), one has \(d\neq0\) in \(\ZZ/5\ZZ\). Because \(5\) is prime, the elements \[ 0,d,2d,3d,4d \] are all distinct modulo \(5\). Hence \[ r,\ r+d,\ r+2d,\ r+3d,\ r+4d \] are all five vertices.
For each \(j\), the pair \[ \{r+jd,r+(j+1)d\} \] is one of the edges from step 1, obtained by taking \(k=jd\). Thus these edges give a path through all five vertices, so the graph is connected.
3. The group \(G\) contains every transposition of the five objects.
Let \(a\) and \(b\) be two distinct vertices. By step 2, there is a path \[ a=v_0,v_1,\ldots,v_m=b \] in the graph. Step 1 says that every edge transposition \[ (v_{i-1},v_i) \] lies in \(G\).
Induct on the length of the path. For length \(1\), \((a,b)\) is already an edge transposition. Suppose \(m>1\). By induction, \[ (v_0,v_{m-1})\in G, \] and the last edge gives \[ (v_{m-1},v_m)\in G. \] Conjugating the latter transposition by the former gives \[ (v_0,v_{m-1})(v_{m-1},v_m)(v_0,v_{m-1}) = (v_0,v_m). \] Hence \((a,b)\in G\).
4. One has\[ \boxed{G=S_5}. \]
By step 3, \(G\) contains every transposition of the five objects. Every permutation is a product of transpositions, so \(G\) contains every element of \(S_5\). Since \(G\leq S_5\), equality follows.
5. Q.E.D.
Step 4 is the required conclusion.