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Cycle types and conjugacy in Sn |
Recall
that every permutation in Sn can be expressed as a product
of disjoint cycles, which can be written in any order. So suppose a1Sn
has cycle lengths of k1, k2, ..., kr,
where k1 >= k2 >= ... > = kr and
k1 + k2 + ... + kr = n, so we have included all the singleton (length 1) cycles. The r-tuple (k1, k2, ..., kr) is called the cycle type of a, and the expression k1 + k2 + ... + kr = n with k1 > or = k2 > or = ... > or = kr is called a partition of n so there is a 1 - 1 correspondence between cycle types in Sn and partitions of n. Now suppose a = (a1, a2, ..., ak) is a cycle in Sn and s is any element of Sn. s-1.a.s is the cycle (a1s, a2s, ..., aks) , because s -1.a.s moves a1s to a2s, a2s to a3s and so on, so that the conjugate of a cycle is another cycle of the same length, and hence the conjugate of any a in Sn is a permutation of the same cycle type. Conversely, suppose a and b are two elements of Sn with the same cycle types. Define s in Sn to be that permutation which maps each element a in a cycle of a to the corresponding b in the corresponding cycle of b. Check for yourself that sigma is a permutation of 1,2,...,n. Then by the previous paragraph, b = s-1.a.s. Thus every pair of elements with the same cycle type are conjugate! We can state this as: Theorem 6.1 There is a 1-1 correspondence
between conjugacy classes in Sn and partitions of n.
If k1 + k2 + ... + kr = n is a partition
of n, the corresponding conjugacy class is the set of permutations in Sn
whose cycle type is (k1, k2, ... , kr).
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Conjugacy in An |
Now
let's consider An. If a
is conjugate to b
in An, say b = s-1.a.s,
with s in An,
then certainly a
is conjugate to b
in Sn, so a
and b have the
same cycle type. But it is not true that if a
and b have the
same cycle type then they are conjugate in An. For example,
recall that the group of rigid motions of the tetrahedron is isomorphic
to A4 acting on the corners. Then (234) represents rotation
by (2p)/3 anticlockwise
round the axis through 1 and the median of the opposite face. This symmetry
is conjugate to rotation by (2p)/3
anticlockwise round each of the other such axes, but it is not conjugate
to rotation by (4p)/3
round any axis, so (234) is not conjugate to its inverse (243). To
see why, note that (243) = (34)(234)(34), but (34) is not in A4, so no conjugate of (234) by an element of A4 can equal (243).
This situation occurs also with An: each conjugacy class in Sn which contains an element of An is either a conjugacy class in An or the union of two conjugacy classes in An. More generally, we have: Lemma 6.2 Let H be a normal subgroup of a group G. If c is in H, then the conjugacy class of c in G is a disjoint union of conjugacy classes of c in H. Proof If d in G is conjugate to c, then since H is normal, d is in H. Thus the complete conjugacy class (in G) of c is in H; call it C. Then the conjugacy class of c in H, call it C1, is contained in C. If it is all of C we are done. If not, let c2 be an element of C not in C1, and call its conjugacy class in H C2. Continuing in this way, every element of C is in some conjugacy class in H, as required. Another general fact about conjugates we shall need later is: Lemma 6.3 Let G be any group and H a subgroup. Let x be in G. Then x-1Hx is a subgroup of G isomorphic to H. Proof Just check for yourself that the map y |->
x-1.y.x is an automorphism of G. It therefore maps the
subgroup H to an isomorphic copy of H in G.
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To continue, click here. To return to Page 1, click here . Last update Nov 24 1999 Author: Phill Schultz, schultz@maths.uwa.edu.au |
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