Suppose that \(G\) is a finite group of order 2013. Prove that \(G\) has a normal subgroup \(N\) of index 3 and that \(N\) is a cyclic group. Furthermore, prove that the center of \(G\) has order divisible by 11. (You will need the factorization \(2013=3\cdot11\cdot61\).)
Solution
1. The Sylow \(61\)-subgroup \(P_{61}\) of \(G\) is unique and hence normal.
By Sylow’s theorem, the number \(n_{61}\) of Sylow \(61\)-subgroups satisfies \[ n_{61}\mid 33, \qquad n_{61}\equiv1\pmod{61}. \] The divisors of \(33\) are \(1,3,11,33\), and only \(1\) is congruent to \(1\) modulo \(61\). Thus \(n_{61}=1\).
2. The Sylow \(11\)-subgroup \(P_{11}\) of \(G\) is unique and hence normal.
Again by Sylow’s theorem, \[ n_{11}\mid 3\cdot61=183, \qquad n_{11}\equiv1\pmod{11}. \] The divisors of \(183\) are \(1,3,61,183\), whose residues modulo \(11\) are \(1,3,6,7\), respectively. Hence \(n_{11}=1\).
3. The subgroup\[ N=P_{11}P_{61} \]is normal of order \(11\cdot61=671\).
Both factors are normal by 1 and 2, so their product is a normal subgroup. Their intersection is trivial because their orders are coprime. Therefore \[ |N|=|P_{11}||P_{61}|=11\cdot61=671. \] Since \(|G|=3\cdot671\), the index of \(N\) is \(3\).
4. The subgroup \(N\) is cyclic.
For normal subgroups \(A,B\trianglelefteq G\), every commutator \([a,b]\) lies in both \(A\) and \(B\). Hence \[ [P_{11},P_{61}]\subseteq P_{11}\cap P_{61}=1. \] Thus the two subgroups commute elementwise. Since each has prime order, they are cyclic, and therefore \[ N\cong C_{11}\times C_{61}\cong C_{671}, \] because \(11\) and \(61\) are coprime.
5. The Sylow \(11\)-subgroup lies in the center of \(G\).
Since \(P_{11}\trianglelefteq G\), conjugation gives a homomorphism \[ G\longrightarrow \operatorname{Aut}(P_{11}). \] Now \(P_{11}\cong C_{11}\), so \[ |\operatorname{Aut}(P_{11})|=10. \] The order of the image divides both \(|G|=2013\) and \(10\). Since \[ \gcd(2013,10)=1, \] the image is trivial. Thus every element of \(G\) centralizes \(P_{11}\), so \[ P_{11}\subseteq Z(G). \]
6. Consequently \(11\mid |Z(G)|\).
By 5, the center contains the subgroup \(P_{11}\) of order \(11\). Lagrange’s theorem therefore gives \(11\mid |Z(G)|\).