11.2.1 Actions of a Group on Itself
A group acts on itself in two basic ways. Under left multiplication, , the action is transitive, every stabilizer is trivial, and no has a fixed point. In particular the action is faithful, which gives Cayley’s theorem (Theorem 3.2.14): a finite group is isomorphic to a subgroup of , where . This embedding tells us little about the structure of . The second action is far more informative.
The group also acts on itself by conjugation, . Each map is an automorphism of , and is a homomorphism , so this is indeed an action. Recall from Definition 3.2.12 that its orbits are the conjugacy classes , and that the stabilizer of is its centralizer , the subgroup of elements that commute with ; we write when the group must be named. The kernel of the action is the center (Definition 3.1.13).
Note. The conjugation action is trivial (every element is fixed) if and only if is abelian, and it is faithful if and only if . A subgroup is normal if and only if it is a union of conjugacy classes (Proposition 3.2.13); it contains the class , and by Lagrange’s theorem its order divides . We use this repeatedly to find normal subgroups.
11.2.2 The Class Equation
By Proposition 3.2.13, the conjugacy class of has elements, a divisor of when is finite, and if and only if .
Theorem 11.2.1 (The Class Equation). Let be a finite group, and let be representatives of the conjugacy classes of with more than one element. Then where each divides and is greater than 1.
Proof. The conjugacy classes partition . The classes with one element are the sets with , and together they contain elements. The other classes are , and by the Orbit–Stabilizer Theorem. ∎
Example 11.2.2 (The class equation of ). By Theorem 3.4.6, the conjugacy classes of are , the three transpositions, and the two 3-cycles. So and the class equation is . We compute more class equations in Section 11.4.
11.2.3 -Groups
Definition 11.2.3 (-group). Let be a prime. A finite group is a -group if for some .
Theorem 11.2.4 (-groups have nontrivial center). If is a -group and , then divides . In particular .
Proof. Let with . In the class equation each divides and is greater than 1, so it is divisible by . Hence is divisible by . Since , , and therefore . ∎
Corollary 11.2.5 (Groups of order ). If with prime, then is abelian, and or .
Proof. By Theorem 11.2.4, divides , and divides by Lagrange’s theorem, so is or . Suppose , and pick . The centralizer contains and , so ; as a subgroup of its order divides , so . Then commutes with every element, so , a contradiction. Hence and is abelian. By the classification of finitely generated abelian groups (Theorem 10.2.1), is a product of cyclic groups whose orders multiply to , so or . These are not isomorphic, since only the first has an element of order . ∎
The next lemma is elementary, and we use it several times in this chapter.
Lemma 11.2.6 (Subgroups of index 2). Let be a subgroup of index 2 in a group . Then is normal in , and for every . In particular, contains every element of of odd order.
Proof. The subgroup is normal by Proposition 3.1.11. The quotient has order 2, so , that is, . If has odd order , then . ∎
Proposition 11.2.7 (Groups of order 8). Up to isomorphism there are exactly five groups of order 8: the abelian groups , and , the dihedral group of symmetries of the square, and the quaternion group , in which commutes with every element, , multiplication by changes signs, and
Proof. First, is a group. In , the matrices and satisfy , , and . So the eight distinct matrices multiply by the rules above, with replaced by ; they form a subgroup of , and is a group isomorphic to it.
The three abelian groups listed are pairwise non-isomorphic (the largest orders of their elements are 8, 4 and 2), and by Theorem 10.2.1 every abelian group of order 8 is one of them. The groups and are not abelian, and they are not isomorphic: has five elements of order 2 (the four reflections and the rotation by ), while is the only element of order 2 in .
Now let be a non-abelian group of order 8. No element of has order 8, since is not cyclic. If every element satisfied , then for all , and would be abelian. So some has order 4. The subgroup has index 2, so it is normal by Lemma 11.2.6. Fix . Then , and every element of can be written uniquely as with and .
Since is normal, , and it has order 4, so . If , then and commute and , whose elements are the , is abelian. Hence , and so for all . By Lemma 11.2.6, , say ; since commutes with , we get , so and . Now so the multiplication of is completely determined by whether or .
- If , the same rules hold in for the rotation by and a reflection, whose products are the eight elements of . So is an isomorphism .
- If , the same rules hold in for and : has order 4, , and . The elements are , so .
∎
Note. The class equation gives another way into this classification. If is non-abelian of order 8, then by Theorem 11.2.4, and the argument of Corollary 11.2.5 rules out 4: for , contains and , so it would have more than 4 elements and be all of . So . For each , the centralizer lies strictly between and , so it has order 4 and . The class equation is therefore : two central elements and three classes of size 2. One can then work out the possible multiplications from there.