Let \(G\) be a group of order \(2012\), which has prime factorization \(2^2 \cdot 503\).
Classify all such groups \(G\) in terms of semidirect products of known groups. (Explain what the possible semidirect products are, but you do not need to decide which of them are isomorphic.)
Show that any group of order \(2012\) has a non-trivial center.
Solution
Let \(N\) be a Sylow \(503\)-subgroup of \(G\).
1. The subgroup \(N\) is unique, normal, and cyclic of order \(503\).
The number \(n_{503}\) of Sylow \(503\)-subgroups satisfies \[ n_{503}\mid 4, \qquad n_{503}\equiv 1\pmod{503}. \] Hence \(n_{503}=1\). Since \(503\) is prime, \[ N\cong C_{503}. \]
2. If \(H\) is any Sylow \(2\)-subgroup, then\[ G\cong C_{503}\rtimes_\theta H, \]where \(H\cong C_4\) or \(H\cong C_2\times C_2\).
We have \(|H|=4\), so \(H\) is isomorphic to one of the two groups of order \(4\). Also \[ N\cap H=1 \] because their orders are coprime, and therefore \[ |NH|=\frac{|N||H|}{|N\cap H|}=503\cdot4=|G|. \] Thus \(G=NH\), with \(N\trianglelefteq G\), so \(G\) is the indicated semidirect product.
3. Every possible action\[ \theta:H\longrightarrow \operatorname{Aut}(C_{503}) \]has image of order at most \(2\).
Since \(C_{503}\) is cyclic of prime order, \[ \operatorname{Aut}(C_{503})\cong (\mathbb Z/503\mathbb Z)^\times\cong C_{502}. \] The image of \(\theta\) has order dividing both \(|H|=4\) and \(502=2\cdot251\). Hence \[ |\operatorname{im}\theta|\mid \gcd(4,502)=2. \] Thus the image is either trivial or the unique subgroup of order \(2\), generated by inversion on \(C_{503}\).
4. If \(H\cong C_4\), there are exactly two possible actions: the trivial action and the action in which a generator of \(C_4\) acts by inversion.
A homomorphism \(C_4\to C_{502}\) is determined by the image of a generator. By 3, that image is either the identity or the unique element of order \(2\). Hence the two possibilities are \[ C_{503}\times C_4 \] and \[ C_{503}\rtimes C_4, \] where a generator of \(C_4\) acts on \(C_{503}\) by \(x\mapsto x^{-1}\).
5. If \(H\cong C_2\times C_2\), there are four possible actions: the trivial action and the three nonzero homomorphisms\[ C_2\times C_2\longrightarrow C_2\leq \operatorname{Aut}(C_{503}). \]
Every nontrivial image must be the unique order-\(2\) subgroup of \(\operatorname{Aut}(C_{503})\). A nonzero homomorphism \[ C_2\times C_2\to C_2 \] has kernel an order-\(2\) subgroup, and \(C_2\times C_2\) has exactly three such subgroups. Conversely, each such kernel determines a unique nonzero homomorphism to \(C_2\). Thus there are three nontrivial actions, in addition to the trivial one.
6. Consequently, all groups of order \(2012\) occur among the semidirect products in 4 and
5.
By 2 every such group is \(C_{503}\rtimes H\) for a group \(H\) of order \(4\), and <1>4–<1>5 enumerate every possible action. The problem does not require deciding which listed semidirect products are mutually isomorphic.
7. Every group \(G\) of order \(2012\) has nontrivial center.
Write \[ G=C_{503}\rtimes_\theta H \] as above. If \(\theta\) is trivial, then \(C_{503}\leq Z(G)\), so the center is nontrivial.
Suppose \(\theta\) is nontrivial. By 3, \[ |\operatorname{im}\theta|=2, \] so \[ |\ker\theta|=\frac{|H|}{2}=2. \] Choose \(1\neq h\in\ker\theta\). Since \(H\) is abelian, \(h\) commutes with every element of \(H\). Since \(h\in\ker\theta\), it acts trivially on \(C_{503}\), so it also commutes with every element of \(C_{503}\). Because \(G=C_{503}H\), the element \(h\) commutes with every element of \(G\). Therefore \[ 1\neq h\in Z(G). \] Thus \(Z(G)\neq1\) in every case.