6,7, and 8 please, this is abstract algebra 4. I t-(12345). Express oduct of tra

Question

6,7, and 8 please, this is abstract algebra

4. I t-(12345). Express oduct of transpositions in two different ways. Briefly justify why you are confident as a t that both are correct. 2 3 2 3 4 3451 1 2 3 4 5 6 (41 5.Expressthe permutation as a product of transpositions 234S6 41 6 235 5) 6 2 3 (14)(92) (35%5) Definition 25.16 Let n E Z+. A permutation E S is even if can be written as a product of an even number of transpositions. A permutation is odd if is not even. ired (why?) Note: In order for this definition to make sense, the following theorem is requ Theorem 25.14 Let the number of transpositions in the product will have the same parity [That is, the number of transpositions may be different for different representations, but a single permutation can never have both an even and odd representation- n E Z+ and let permutation E Sn. No matter how is written as a product of transpositions, see pp. 339-340 in your book for a proof of this theorem) Definition 25.19 For n 2 2, the alternating group A, is the subgroup of S, consisting of the even perutations of Sn- 6, Find the parity of the permutations and from problems 4 and 5 above. 7. Show that An as defined above is a subgroup of S, S. Find the elements of As. Then find the elements of As Hint: how does the order of A, compare to the order of S.?

Explanation / Answer

4.The simplest permutations are cyclic, such as (1; 2; 3; 4; 5) is
(5; 1; 2; 3; 4) : Thinking of the numbers 1; 2; 3; 4; 5 arranged around
a circle clockwise, with 1 at the top, they all moved 1 place
clockwise. But this notation is clumsy. A more efficient nota-
tion is to think of the permutation as a function, say p; and,
in this example, we have p (1) = 5; p (2) = 1; p (3) = 2;
p (4) = 3; and p (5) = 4: We will then denote p by (15432) :
p (1) = 5; p (5) = 4; p (4) = 3; p (3) = 2; p (2) = 1As a second cyclic permutation, think of
(1; 2; 3; 4; 5) ! (3; 2; 5; 4; 1) :
Buried in here is (1; 3; 5) ! (3; 5; 1) ; an obvious cycle. Denoting
this by q; we have q (1) = 3; q (3) = 5; q (5) = 1; and 2 and
4 are Öxed, so that q (2) = 2; q (4) = 4: In this case, a simpler
notation is used. We write
q = (135)
to mean exactly what is stated in the deÖnition of q:
The simplest kind of cycle has two elements, such as (24) : These
are called ìtranspositionsî.
But it must be understood that in the formulas for p and q above,
the numbers 1,2,3,4,5 here are ìplace holders". One way to look
at this is to say that (15432) is the cycle which takes (a; b; c; d; e)
to (e; a; b; c; d) ; and (135) is the cycle which takes (a; b; c; d; e)
into (c; b; e; d; a) ; whatever a; b; c; d; e are. (They should all be
di§erent from each other.)
The notations (15432) or (135) are used only for cycles.Consider the following two steps: p : (1; 2; 3; 4; 5) ! (5; 1; 2; 3; 4)
and q : (1; 2; 3; 4; 5) ! (3; 2; 5; 4; 1) : (Here p and q are as be-
fore.) If we start with (1; 2; 3; 4; 5) and Örst apply p and then
apply q; we get
qp : (1; 2; 3; 4; 5) ! (2; 1; 4; 3; 5)
Both p and q are cycles, but they are not independent cycles.
We can express qp as the product of independent cycles as qp =
(12) (34) ; with 5 unchanged. So we have the equation
(135) (15432) = (12) (34) = (34) (12) :
Note that the last product commutes because the two cycles are
independent of each other. It doesnít matter which you do Örst.
It should be noted that this particular example does not have a
geometrical interpretation, because (135) does not represent a
permutation of the vertices by a rotation or a reáection. In a
rotation, all vertices move. In a reáection, four vertices move.

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