D_{n}
= { a^{i} b^{j} |
0 i < n,
0 j < 2,
*o*(a) = n,
*o*(b) = 2,
ba = a^{-1}b }

*Solution:*
For i = 1, the equation
ba^{i} = a^{n-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

ba^{k+1}
= (ba^{k})a
= (a^{n-k}b)a
= a^{n-k}(ba)
= a^{n-k}a^{-1}b
= a^{n-(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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*Solution:*
Using the result from Problem 22, we have

(a^{i} b)^{2}
= (a^{i} b)(a^{i} b)
= a^{i} (ba^{i}) b
= a^{i} (a^{n-i} b) b
= (a^{i} a^{n-i})( b^{2} )
= a^{n} e
= e.

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*Solution:*
Let a = (1,2,3) and b = (1,2).
Then H must contain

a^{2} = (1,3,2), ab = (1,3) and a^{2}b = (2,3),

and this set of elements is closed under multiplication.

(We have just listed the elements of S_{3}.) Thus

H = { (1), a, a^{2}, b, ab, a^{2}b }
= { (1), (1,2,3), (1,3,2), (1,2), (1,3), (2,3) }.

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*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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*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 A_{4}.

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*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) D_{n},
since by definition b does not belong to C(a).
Since < a > contains exactly half of the elements in D_{n},
Lagrange's Theorem
shows that there is no subgroup
that lies strictly between < a > and D_{n}, so

< a > C(a)
D_{n}
and C(a) D_{n}
together imply that C(a) = < a >.

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*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 S_{3}, 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 S_{4}
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 S_{3}, we know that

(1,2), (1,3), and (2,3) do not commute with a.
The remaining transpositions in S_{4} 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 S_{4} 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 A_{4} we must get the same answer:
C(a) = < a >.

Forward to §3.7 | Forward to §3.7 problems | Back to §3.5 problems | Up | Table of Contents

Forward to §3.7 | Forward to §3.7 problems | Back to §3.5 problems | Up | Table of Contents