给定乘法表的群G，如何高效判断元素a是否属于中心C(G)？

Great question! The brute-force approach of checking commutativity with every element works, but there are much smarter methods that cut down the number of operations significantly, depending on what you know about the group (G):

1. Verify Commutativity Only with Generators #

Every group (G) can be generated by a subset (S = {g_1, g_2, ..., g_k}) — meaning every element of (G) is a product (and inverses) of elements from (S). Instead of checking (a \cdot b = b \cdot a) for all (b \in G), you only need to verify this equality for each (g_i \in S).

Why this works: #

If (a) commutes with every generator, it commutes with every product of generators (and their inverses). For example, if (b = g_1 g_2^{-1} g_3), then:
[
a \cdot b = a g_1 g_2^{-1} g_3 = g_1 a g_2^{-1} g_3 = g_1 g_2^{-1} a g_3 = g_1 g_2^{-1} g_3 a = b \cdot a
]
You can formalize this with induction on the length of the product.

Efficiency win: #

If (k) (the number of generators) is much smaller than (n = |G|), this reduces your operations from (n-1) to (k). For cyclic groups, (k=1) — you only need one check!

2. Leverage Known Group Structure #

If you already know the type of group (G) is, you can skip computations entirely using structural properties:

• Abelian groups: The center (C(G)) is the entire group, so every element is in the center automatically.
• Symmetric groups (S_n) (n ≥ 3): The center only contains the identity element. So if (a) isn't the identity, it's not in (C(G)).
• p-groups: While the center is non-trivial, you can use properties of p-group subgroups (like the fact that every maximal subgroup contains the center) to narrow down checks instead of verifying all elements.

3. Check the Size of the Conjugacy Class #

An element (a) is in the center if and only if its conjugacy class (Cl(a) = {g a g^{-1} | g \in G}) has size 1 (since only elements that commute with everyone have no non-trivial conjugates).

Instead of computing the entire conjugacy class, you can use the orbit-stabilizer theorem: (|Cl(a)| = [G : C_G(a)]), where (C_G(a)) is the centralizer of (a) (the set of elements that commute with (a)). If (|Cl(a)| = 1), then (C_G(a) = G), so (a \in C(G)).

For many groups (like symmetric groups), conjugacy classes are determined by easy-to-check properties (e.g., cycle type for permutations), so you can instantly tell if a class has size 1.

4. Compute the Order of the Centralizer #

Since (a \in C(G)) exactly when (C_G(a) = G), you can compute the order of (C_G(a)). If (|C_G(a)| = |G|), then (a) is in the center.

To compute (|C_G(a)|) efficiently:

• Start with elements you know commute with (a) (like generators that commute with (a), or powers of (a) itself).
• Generate the subgroup they span — if its order equals (|G|), you're done.
• Use the orbit-stabilizer theorem alongside conjugacy class size (as above) to calculate (|C_G(a)|) without enumerating all elements.

内容的提问来源于stack exchange，提问作者user275490