Subgroups & centralizers

Goal

Test the subgroup relation, compute a centralizer, and scan all subgroup classes with ConjugacyClassesSubgroups / Representative. The tools for showing "this factor centralises that one" and "this group has no subgroup of order n."

Concept

Centralizer(G, x) is everything in G commuting with x (an element or a subgroup); it's the algebraic form of "what leaves this structure alone." The set of all subgroups up to conjugacy is ConjugacyClassesSubgroups(G). That's a list of classes, each with a Representative. That lets you ask existence questions ("is there a subgroup of order 60?") by scanning representatives, which is how you rule out a wrong guess about a group's structure.

Commands to try

S4 := SymmetricGroup(4);
H := Subgroup(S4, [ (1,2,3,4) ]);
Size(H);
IsSubgroup(S4, H);

Expect 4 and true. Subgroup(G, gens) builds a subgroup from generators that already live in G; IsSubgroup(G, H) tests containment.

c := Centralizer(S4, (1,2,3,4) );
Size(c);
Elements(c);

Predict the size: what commutes with a 4-cycle in S4? (A 4-cycle generates a cyclic group of order 4, and in S4 little else commutes with it.) Then check that the centralizer here is the cyclic group ⟨(1,2,3,4)⟩ itself, order 4.

Centralizer(S4, Subgroup(S4, [(1,2)(3,4), (1,3)(2,4)]) );

The centralizer of a whole subgroup (the Klein four V). Same function, subgroup argument: it returns everything commuting with all of V.

Scanning subgroup classes

ccs := ConjugacyClassesSubgroups(A4);;   # A4 first; it's small
Length(ccs);
List(ccs, c -> Size(Representative(c)));

where A4 := AlternatingGroup(4);;. Each class gives one Representative subgroup; mapping Size over the representatives shows which subgroup orders occur. Predict: A4 has order 12; does an order-6 subgroup appear? (Famously, A4 has no subgroup of order 6 - watch 6 be absent from the list.)

ForAny(ConjugacyClassesSubgroups(A4), c -> Size(Representative(c)) = 6);

Expect false. The GAP one-liner for "no order-6 subgroup," the same shape as any "does a subgroup of order n exist?" check.

Predict-then-check

Build the direct-product-style centralizer reading on a toy:

G := SymmetricGroup(5);;
z := (1,2,3,4,5);;
Centralizer(G, z) = Group(z);
Size(Centralizer(G, z));

Predict: in S5, what commutes with a 5-cycle? Is the centralizer exactly the cyclic group it generates, and what is its order?

Exercise

1. In S4, compute Centralizer(S4, (1,2)). Predict its size (what commutes with a single transposition?) before checking.
2. Using ConjugacyClassesSubgroups(SymmetricGroup(4)), list the distinct subgroup orders that occur (use Set + List + Representative + Size). Compare against the divisors of 24 - which divisors are missing?
3. Write the one-liner that answers "does A5 have a subgroup of order 30?" (Predict the truth value: A5 is simple. Then verify.)
4. Foreshadow the capstone: confirm ForAny(ConjugacyClassesSubgroups(SymmetricGroup(4)), c -> Size(Representative(c))=12) and identify which subgroup of order 12 that is.

Pitfalls

• Centralizer(G, x) accepts an element or a subgroup as x. The centre is the special case Centralizer(G, G) = Center(G).
• ConjugacyClassesSubgroups is exponential in general - fine for A4, S4, and other small groups, but do not call it on a group with millions of elements. Existence questions on big groups need other tools instead.
• Representative(c) gives one subgroup from the class; other members are its conjugates. Two different classes can share the same order - Set of sizes collapses that, which is what you usually want for an existence scan.
• Subgroup(G, gens) requires the generators to already be elements of G; Group(gens) builds a standalone group instead. Use Subgroup when you want the containment relationship tracked.
• A centralizer is itself a group object. You can take its Size, Center, feed it back into IsSubgroup, etc.