Let \(G\) be a group of order \[ 2010=2\cdot3\cdot5\cdot67. \] Determine how many subgroups of order \(67\) the group \(G\) can have, and give examples for each possible number.
Solution
Let \(n_{67}\) denote the number of subgroups of \(G\) of order \(67\).
1. Every subgroup of \(G\) of order \(67\) is a Sylow \(67\)-subgroup.
Since \[ \abs G = 2\cdot3\cdot5\cdot67, \] the largest power of \(67\) dividing \(\abs G\) is \(67^1\). Therefore the Sylow \(67\)-subgroups are exactly the subgroups of order \(67\).
2. One has\[ \boxed{n_{67}=1}. \]
By the Sylow theorems, \[ n_{67}\equiv1\pmod{67} \] and \[ n_{67}\mid\frac{2010}{67}=30. \] Every positive divisor of \(30\) is at most \(30\), while the next positive integer after \(1\) that is congruent to \(1\) modulo \(67\) is \(68\). Hence the only possibility is \[ n_{67}=1. \]
3. The cyclic group of order \(2010\) is an example realizing the uniquepossible value.
The cyclic group \[ C_{2010} \] has order \(2010\). A cyclic group has a unique subgroup of order \(d\) for each divisor \(d\) of its order, so it has exactly one subgroup of order \(67\), in agreement with step 2.
4. Q.E.D.
Step 2 determines the only possible number of order-\(67\) subgroups, and step 3 supplies the requested example.