Groups of order 2023

Since the current year commonly features in math competition problems, it seems useful to begin the year by noting that {2023 = 7 \cdot 17^2}.

The past couple years around this time I have set myself the further puzzle of working out the groups of that order. Usually it is not difficult, but occasionally it will be essentially impossible (e.g., in {2048}).

This year is not very interesting actually, group-theoretically. This is an abelian year in the sense that all groups of order {2023} are abelian, and the only possibilities are {C_{2023}} and {C_7 \times C_{17} \times C_{17}}. This is an easy consequence of Sylow’s theorem. Let {G} be a group of order {2023}. The number of Sylow {7}-subgroups must divide {17^2} and be {\equiv 1 \pmod 7}, so it must be {1}, and similarly the number of Sylow {17}-subgroups must also be {1}, so {G \cong C_7 \times H} where {H} has order {17^2}. A group of order {p^2} must be abelian (standard exercise), so {H \cong C_{289}} or {C_{17} \times C_{17}}.

To make it more challenging, can we actually characterize the positive integers {n} such that all groups of order {n} are abelian? Call such an integer abelian.

Proposition 1 Let {n} be a positive integer. Then {n} is abelian if and only if
(1) {n} is cube-free,
(2) if {p, q} are distinct primes such that {pq \mid n} then {p} does not divide {q-1},
(3) if {p, q} are distinct primes such that {pq^2 \mid n} then {p} does not divide {q+1}.

Let us first show that the conditions are necessary. Suppose (1) fails. Then {p^3 \mid n} for some prime {p}. There is a nonabelian group {H} of order {p^3} (see extraspecial groups), so {H \times C_{n / p^3}} is a nonabelian group of order {n}. Similarly if (2) fails then there is a nonabelian semidirect product {C_q : C_p}, so we can again form a nonabelian group of order {n}. Finally if (3) fails then note that {\mathrm{Aut}(C_q^2) \cong \mathrm{GL}_2(q)}, and {\mathrm{GL}_2(q)} contains an element of order {q^2 - 1} (since the finite field of order {q^2} has a primitive root), so we have another nonabelian semidirect product {(C_q \times C_q) : C_p}.

Now we show that the conditions are sufficient. We use induction on {n}. Let {G} be a group of order {n}, where {n} satisfies conditions (1), (2), and (3). Assume first {G} is not simple. Let {N} be a maximal normal subgroup of {G}. Then {|N|} and {|G/N|} are both integers smaller than {n} satisfying (1), (2), and (3), so {N} and {G/N} are abelian by induction. Since all subgroups of {G/N} are normal and {N} was chosen to be maximal normal, {G/N} must be cyclic of prime order, say {p}. Let {P \le G} be a Sylow {p}-subgroup and let {H \le N} be the product of the Sylow {q}-subgroups for {q \ne p}. Then it follows that {G} is the semidirect product {G = H : P}.

Next consider the conjugation action of {P} on {H}. Let {Q \le H} be a Sylow {q}-subgroup. Then {Q} is isomorphic to one of {C_q, C_{q^2}, C_q \times C_q}, and {\mathrm{Aut}(Q)} is respectively isomorphic to one of {C_{q-1}, C_{q(q-1)}, \mathrm{GL}_2(q)}. Note that {|\mathrm{GL}_2(q)| = q (q-1)^2(q+1)}. By the given conditions, none of these has order divisible by {p}, so {P} acts trivially on {Q}. Hence {P} acts trivially on {H} and {G = H \times P} is abelian, as claimed.

To finish we must rule out the possibility that {G} is simple. This seems to require some nontrivial group theory. One lazy method is to note that condition (2) implies that {n} is odd (or a power of {2}), so we are done by the Feit–Thompson theorem. An easier argument is to use Burnside’s transfer theorem, which states that if {P} is a Sylow {p}-subgroup and {C_G(P) = N_G(P)} then {G} has a normal subgroup of order {|G:P|} (in particular {G} is not simple). In our case, let {p} be the smallest prime dividing {n} and let {P} be a Sylow {p}-subgroup. Let {C = C_G(P)} and {N = N_G(P)}. Then {N / C} is isomorphic to a subgroup of {\mathrm{Aut}(P)}. As before {\mathrm{Aut}(P)} is one of {C_{p-1}}, {C_{p(p-1)}}, or {\mathrm{GL}_2(p)}. The orders of these groups have no prime factor larger than {p+1}. Since {p} is the smallest prime dividing {n} and {p > 2}, it follows that {N / C} is trivial, so we are done.

It is also possible to characterize nilpotent numbers in general, and even solvable numbers. See Pakianathan, J., & Shankar, K. (2000). Nilpotent Numbers. The American Mathematical Monthly, 107(7), 631–634. https://doi.org/10.2307/2589118.

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