Classify all groups of order \(2009=7^2\times 41\).
Suppose that \(G\) is a group of order 2009. How many intermediate groups are there—that is, how many groups H are there with \(1\subsetneq H\subsetneq G\), where both inclusions are proper? (There may be several cases to consider.)
Solution
1. Every group \(G\) of order\[ 2009=7^2\cdot41 \]has a unique Sylow \(41\)-subgroup.
Let \(n_{41}\) be the number of Sylow \(41\)-subgroups. The Sylow theorems give \[ n_{41}\equiv1\pmod{41}, \qquad n_{41}\mid49. \] The divisors of \(49\) are \(1,7,49\), and neither \(7\) nor \(49\) is congruent to \(1\) modulo \(41\). Hence \[ n_{41}=1. \] Thus the Sylow \(41\)-subgroup \(Q\) is normal. Since \(|Q|=41\), it is cyclic: \[ Q\cong C_{41}. \]
2. Every such \(G\) also has a unique Sylow \(7\)-subgroup.
Let \(n_7\) be the number of Sylow \(7\)-subgroups. Then \[ n_7\equiv1\pmod7, \qquad n_7\mid41. \] Thus \(n_7\) is \(1\) or \(41\). But \[ 41\equiv6\pmod7, \] so \(41\) is impossible. Hence \[ n_7=1. \] Thus the Sylow \(7\)-subgroup \(P\) is normal and has order \(49\).
3. The group \(G\) is the direct product\[ G\cong P\times Q. \]
The normal subgroups \(P\) and \(Q\) have coprime orders, so \[ P\cap Q=1. \] For \(p\in P\) and \(q\in Q\), the commutator \([p,q]\) lies in \(P\) because \(P\trianglelefteq G\), and also lies in \(Q\) because \(Q\trianglelefteq G\). Hence \[ [p,q]\in P\cap Q=1. \] Thus \(P\) and \(Q\) commute elementwise. Moreover, \[ |PQ|=\frac{|P||Q|}{|P\cap Q|}=49\cdot41=|G|, \] so \(PQ=G\). Therefore \(G\) is the internal direct product \(P\times Q\).
4. Up to isomorphism, there are exactly two possibilities for \(P\):\[ P\cong C_{49} \qquad\text{or}\qquad P\cong C_7\times C_7. \]
Every group of order \(p^2\) is abelian. Indeed, a nontrivial finite \(p\)-group has nontrivial center. If \(|Z(P)|=p^2\), then \(P\) is abelian. If \(|Z(P)|=p\), then \(P/Z(P)\) has order \(p\) and is cyclic; a group whose quotient by its center is cyclic is abelian, again forcing \(Z(P)=P\). Thus \(P\) is abelian.
The classification of finite abelian groups of order \(7^2\) gives exactly the two possibilities \[ C_{49} \qquad\text{and}\qquad C_7\times C_7. \]
5. Consequently the two groups of order \(2009\) are\[ C_{49}\times C_{41}\cong C_{2009} \]and\[ (C_7\times C_7)\times C_{41}. \]
Combine <1>1–<1>4. Since \(49\) and \(41\) are coprime, the product of the cyclic groups \(C_{49}\) and \(C_{41}\) is cyclic of order \(2009\).
6. If \(G\cong C_{2009}\), then \(G\) has exactly \(4\) nontrivial proper subgroups.
A cyclic group has exactly one subgroup for each positive divisor of its order. Since \[ 2009=7^2\cdot41, \] the number of positive divisors is \[ (2+1)(1+1)=6. \] These correspond to subgroup orders \[ 1,\ 7,\ 49,\ 41,\ 7\cdot41,\ 2009. \] Removing the trivial subgroup and \(G\) itself leaves exactly \[ 6-2=4 \] nontrivial proper subgroups.
7. Let\[ G=(C_7\times C_7)\times C_{41}. \]Every subgroup \(H\le G\) has the form\[ H=A\times B \]for uniquely determined subgroups\[ A\le C_7\times C_7, \qquad B\le C_{41}. \]
Let \(E=C_7\times C_7\) and \(Q=C_{41}\). Any Sylow \(7\)-subgroup of \(H\) consists entirely of elements of \(7\)-power order, hence lies in \(E\times1\). Likewise any Sylow \(41\)-subgroup of \(H\) lies in \(1\times Q\).
Since \(H\) has order dividing \(7^2\cdot41\), its Sylow subgroups have coprime orders and are normal in \(H\); indeed the \(41\)-Sylow subgroup is unique, and the \(7\)-part is the set of all elements whose \(Q\)-component is trivial. Thus \(H\) is the direct product of its \(7\)-part and its \(41\)-part. Writing these as \(A\times1\) and \(1\times B\) gives \[ H=A\times B. \] The two factors are recovered as intersections with \(E\times1\) and \(1\times Q\), so they are unique.
8. The group \(E=C_7\times C_7\) has exactly \(10\) subgroups.
View \(E\) as the \(2\)-dimensional vector space \(\mathbb F_7^2\). Its subgroups are exactly its linear subspaces. There is one zero subspace, one whole space, and the \(1\)-dimensional subspaces. The number of \(1\)-dimensional subspaces is \[ \frac{7^2-1}{7-1}=8, \] because there are \(7^2-1=48\) nonzero vectors and each line contains \(7-1=6\) nonzero vectors. Hence the total number of subgroups is \[ 1+8+1=10. \]
9. The group \(C_{41}\) has exactly \(2\) subgroups.
A group of prime order has only the trivial subgroup and the whole group.
10. Therefore\[ (C_7\times C_7)\times C_{41} \]has exactly \(18\) nontrivial proper subgroups.
By <1>7–<1>9, subgroups correspond bijectively to pairs \((A,B)\) with \[ A\le C_7\times C_7, \qquad B\le C_{41}. \] Thus the total number of subgroups is \[ 10\cdot2=20. \] Removing the trivial subgroup and \(G\) itself leaves \[ 20-2=18 \] nontrivial proper subgroups.
11. Hence the complete answer is\[ \boxed{G\cong C_{2009}\text{ with }4\text{ intermediate subgroups}} \]or\[ \boxed{G\cong C_7\times C_7\times C_{41}\text{ with }18\text{ intermediate subgroups}}. \]