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§ 3.6 Permutation Groups: Solved problems

In the problems in this section, we will use the following description of the dihedral group Dn of order 2n.

Dn = { ai bj | 0 i < n, 0 j < 2, o(a) = n, o(b) = 2, ba = a-1b }

22. In the dihedral group Dn, show that bai = an-i b, for all 0 i < n.

Solution: For i = 1, the equation bai = an-i b is just the relation that defines the group. If we assume that the result holds for i = k, then for i = k+1 we have
bak+1 = (bak)a = (an-kb)a = an-k(ba) = an-ka-1b = an-(k+1)b .
This implies that the result must hold for all i with 0 i < n.

Comment: This is similar to a proof by induction, but for each given n we only need to worry about a finite number of equations.

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23. In the dihedral group Dn, show that each element of the form ai b has order 2.

Solution: Using the result from Problem 22, we have

(ai b)2 = (ai b)(ai b) = ai (bai) b = ai (an-i b) b = (ai an-i)( b2 ) = an e = e.

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24. In S4, find the subgroup H generated by (1,2,3) and (1,2).

Solution: Let a = (1,2,3) and b = (1,2). Then H must contain
a2 = (1,3,2), ab = (1,3) and a2b = (2,3),
and this set of elements is closed under multiplication.
(We have just listed the elements of S3.) Thus
H = { (1), a, a2, b, ab, a2b } = { (1), (1,2,3), (1,3,2), (1,2), (1,3), (2,3) }.

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25. For the subgroup H of S4 defined in Problem 24, find the corresponding subgroup H -1, for = (1,4).

Solution: We need to compute -1, for each in H. Since
(1,4)-1 = (1,4), we have (1,4) (1) (1,4) = (1), and (1,4) (1,2,3) (1,4) = (2,3,4).
As a shortcut, we can use Exercise 2.3.10, which shows that
(1,2,3) -1 = ((1), (2), (3)) = (4,2,3).
Then we can quickly do the other computations:

(1,4) (1,3,2) (1,4)-1 = (4,3,2)
(1,4) (1,2) (1,4)-1 = (4,2)
(1,4) (1,3) (1,4)-1 = (4,3)
(1,4) (2,3) (1,4)-1 = (2,3) .

Thus (1,4) H (1,4)-1 = { (1), (2,3,4), (2,4,3), (2,3), (2,4), (3,4) }.

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26. Show that each element in A4 can be written as a product of 3-cycles.

Solution: We first list the 3-cycles:

(1,2,3),
(1,2,4),
(1,3,2),
(1,3,4),
(1,4,2),
(1,4,3),
(2,3,4),
(2,4,3).

Rather than starting with each of the other elements and then trying to write them as a product of 3-cycles, it is easier to just look at the possible products of 3-cycles. We have

(1,2,3)(1,2,4) = (1,3)(2,4),
(1,2,4)(1,2,3) = (1,4)(2,3),
(1,2,3)(2,3,4) = (1,2)(3,4).

Including the 3-cycles themselves, and the identity, this accounts for all 12 of the elements in A4.

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27. In the dihedral group Dn, find the centralizer of a.

Solution: The centralizer C(a) contains all powers of a, so we have < a > C(a). This shows that C(a) has at least n elements. On the other hand, C(a) Dn, since by definition b does not belong to C(a). Since < a > contains exactly half of the elements in Dn, Lagrange's Theorem shows that there is no subgroup that lies strictly between < a > and Dn, so
< a > C(a) Dn and C(a) Dn together imply that C(a) = < a >.

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28. Find the centralizer of (1,2,3) in S3, in S4, and in A4.

Solution: Since any power of an element a commutes with a, the centralizer C(a) always contains the cyclic subgroup < a > generated by a. Thus the centralizer of (1,2,3) always contains the subgroup
{ (1), (1,2,3), (1,3,2) }.

In S3, the centralizer of (1,2,3) is equal to < (1,2,3) >, since it is easy to check that (1,2) does not belong to the centralizer, and by Lagrange's Theorem a proper subgroup of a group with 6 elements can have at most 3 elements.

To find the centralizer of (1,2,3) in S4 we have to work a bit harder. It helps to have some shortcuts when doing the necessary computations. To see that x belongs to C(a), we need to check that xa = ax, or that a x a-1 = x. Exercise 2.3.10 provides a quick way to do this in a group of permutations. That exercise shows that if (1,2,. . . ,k) is a cycle of length k and is any permutation, then

(1,2,. . . ,k) -1 = ((1), (2), . . . , (k)).

Let a = (1,2,3). From the computations in S3, we know that
(1,2), (1,3), and (2,3) do not commute with a. The remaining transpositions in S4 are
(1,4), (2,4), and (3,4). Using Exercise 2.3.10, we have

a (1,4) a-1 = (2,4),
a (2,4) a-1 = (3,4), and
a (3,4) a-1 = (1,4),

so no transposition in S4 commutes with a.
For the products of the transposition, we have

a (1,2)(3,4) a-1 = (2,3)(1,4),
a (1,3)(2,4) a-1 = (2,1)(3,4), and
a (1,4)(2,3) a-1 = (2,4)(3,1),

and so no product of transpositions belongs to C(a).

If we do a similar computation with a 4-cycle, we will have
a (x,y,z,4) a-1 = (u,v,w,4),
since a just permutes the numbers x, y, and z. This means that
w z, so (u,v,w,4) (x,y,z,4).
Without doing all of the calculations, we can conclude that no 4-cycle belongs to C(a). This accounts for an additional 6 elements. A similar argument shows that no 3-cycle that includes the number 4 as one of its entries can belong to C(a). Since there are 6 elements of this form, we now have a total of 21 elements that are not in C(a), and therefore C(a) = < a >. Finally, in A4 we must get the same answer: C(a) = < a >.


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