Conjugacy classes in
Last time: every element of can be expressed as a product of a unique collection of disjoint cycles. Conjugacy classes in correspond to partitions of , i.e. ways to express as a sum of positive integers (the lengths of the cycles).
The alternating group is a nontrivial normal subgroup of . We’ll see that for it’s the only one.
Proposition 25.1. If is a conjugacy class, then either or . In the latter case, either is a conjugacy class in , or it splits into 2 conjugacy classes in .
is a single conjugacy class in iff, given , there exists an odd permutation that commutes with .
Proof.
- All elements of have the same cycle lengths, hence the same sign. So or . (Or: is a normal subgroup of , hence a union of conjugacy classes.)
- Assume . Then split into the 2 (right) cosets of : for any with . Then
These 2 conjugacy classes are either equal or disjoint; they are equal iff is in the latter conjugacy class, i.e. there exists (odd) such that , i.e. .
∎
In other terms: for , let be the centralizer. If , then the conjugates of by odd permutations are different from the conjugates by even permutations, and they form two conjugacy classes in ; if , then all conjugates of in are conjugates by elements of .
Example 25.2. :
The 3-cycles still form a single conjugacy class in (because ); so do the ’s (because ). But the 5-cycles split into 2 conjugacy classes in . So the class equation of is
More generally:
Proposition 25.3. For , splits into 2 conjugacy classes in iff the cycle lengths of are all odd and distinct.
Proof.
- commutes with the cycles in its own cycle decomposition. So any even-length cycle in gives an odd permutation in , and then doesn’t split.
- If two odd cycles and of the same length appear in the cycle decomposition of , then is odd. (This includes the case ! We can’t have 2 fixed points.)
- If the cycle lengths are all distinct, then an element of must preserve each of the corresponding subsets of ; now, on a -element subset,
So .
∎
Simplicity of
Now: the class equation of is . We can now look for normal subgroups of . A normal subgroup is a union of conjugacy classes including , and its order divides 60; we can’t reach a divisor of 60 in any nontrivial way as such a union, except by taking all of them. Hence:
Proposition 25.4. is simple, i.e. its only normal subgroups are and itself.
Theorem 25.5. is simple for all .
Corollary 25.6. The only normal subgroups of () are , , and .
Proof. normal subgroup is a normal subgroup of , hence of , hence it is or . So either (then , so or ), or . In the latter case, there can’t be 2 distinct odd permutations in ( would be in ), so ; but a subgroup with can’t be normal. ∎
Proof of the theorem? we’ve just seen; is a similar argument using the class equation (on HW 9)! However, the result is false for ( is normal). The general case relies on:
Lemma 25.7. is generated by 3-cycles.
Proof. Induction on : this is true for .
Now assume is generated by 3-cycles. Let . If , then it belongs to the subgroup , so it’s a product of 3-cycles by the induction hypothesis. Else, let and any element distinct from and ; then and , so by the induction hypothesis , and so is . ∎
Moreover: for , the 3-cycles form a single conjugacy class in , since and are conjugate by any permutation , and some of these are in . So: to prove that a normal subgroup , , is all of , it suffices to show that it contains a 3-cycle.
Proof of the theorem. Let , , be a normal subgroup. As just noted, it’s enough to show that contains a 3-cycle (hence all 3-cycles by conjugation, hence ).
Let , . Replacing by some power of , we can assume that it has prime order: let and a prime with ; then has order . Since the order of is the lcm of its cycle lengths, this implies is a product of disjoint -cycles. We now look at cases depending on .
- If : , where fixes and permutes the remaining elements. Let ; then normal and . Now
takes
so , and contains a 3-cycle.
- : if is a 3-cycle we’re done. Else it is a product of at least two disjoint 3-cycles: write and let . We find , a 5-cycle in ; this reduces to the previous case. ✓
- , and is a product of only 2 transpositions (a single !): . Let and . Then ; back to the first case.
- , and is a product of at least 3 transpositions (in fact ): . Again let ; then has order 3, which reduces to case 2.
∎
The Sylow theorems
Our next topic, still very much related to understanding finite groups, is the Sylow theorems.
Notes not reached in this lecture, covered at the start of {{LINK:L26}}
If and , then in general there is no reason for to contain an element of order , or even a subgroup of order : the “converse to Lagrange’s theorem” fails.
Example 25.8. (resp. ) has no subgroup of order 6 (resp. 30): such a subgroup would be normal.
The first Sylow theorem says: if , prime, , then there exist subgroups of order .
Fix a prime (which divides ) and write , .
Definition 25.9. A subgroup of order is called a Sylow -subgroup of .
Theorem 25.10 (Sylow theorems (Sylow, 1872)).
- For every prime , a Sylow -subgroup of exists.
- All Sylow -subgroups are conjugates of each other: if are Sylow -subgroups, then there exists such that . Moreover, any subgroup with a power of is contained in a Sylow -subgroup.
- Let be the number of Sylow -subgroups of . Then , and (or equivalently, ).
Example 25.11 (Classify groups of order 15). Show solution
If , then there exist Sylow subgroups with , . The numbers of such Sylow subgroups:
This implies and are normal! (Since their conjugates , are also Sylow subgroups, but and are the unique such.) Using a criterion for direct products coming up next time, this implies
Every group of order 15 is cyclic!
Digression: extensions and semidirect products
If is a normal subgroup, then we have an exact sequence
This does not imply that , or even that contains a subgroup isomorphic to !
Example 25.12. is a subgroup, , but has no subgroup .
If can in fact be identified with a subgroup of containing one element of each right coset of , then every element of is uniquely for some , . Yet need not be :
i.e. the bijection , , isn’t a homomorphism. We’ll study this further next time.