![]() ![]() See Lagrange's theorem, in wikipedia:Ī consequence of the theorem is that the order of any element $a$ of aįinite group (i.e. That number is not a multiple of the prime $11$. ![]() $S_7$ has $7 \times 6 \times 5 \times 4\times 3 \times 2$ elements. You should be able to complete this list and find the orders of the elements of each type. The last entry is the identity permutation. The possible patterns when you write an element of $S_7$ as a product of disjoint cycles (which you can always do) are (xxxxxxx) More help, in response to the OP's comment. Examples of permutation groups include the symmetric group (of order ), the alternating group (of order for ), the cyclic group (of order ), and the dihedral group (of order ). Your guess that all orders up to $12$ would occur can't be right since the order of any element must divide the order of the group, and $11$ does not divide $7!$. Conjugacy classes of elements which are interchanged in a permutation group are called permutation cycles. ![]() So all you need to do is write down the possible cycle patterns for permutations in $S_7$. You already know the crucial pieces of information: the order of a cycle is its length, and the order of two group elements that commute is the least common multiple of their orders. Please don't make it too advance because I am just a beginner in studying abstract algebra. So I wonder how I can also include the elements formed by the joint cycles in the consideration toward the answer(Ps: not just the case I mention for the transposition, but also like in some general case such as $(134)(235)$)and conduct it properly?Īnd I want to know the rigorous proof towards this problem of finding orders of elements for the permutation group and also if it is possible tell me some general method that I can use for finding orders not just in the case of $S_7$ and $A_7$, but also in all the other cases. ![]() However, I am not sure if I am correct or not.Īs a matter of fact, I also find out the elements formed by all the transpositions which share a common number has a higher order than the element formed by the disjoint cycle in the case when for example $|(12)(32)|>|(23)(14)|$. The term also refers to the combination of the two. And since $A_7$ which takes all even permutation of $S_7$ is a subgroup of $S_7$, so $A_7$ should take all elements of odd orders, such as $1$, $3$, $5$, $\dots$, $11$. In mathematics, the term permutation representation of a (typically finite) group can refer to either of two closely related notions: a representation of as a group of permutations, or as a group of permutation matrices. At first, I thought $S_7$ should take all elements from order $1$ to order $12$, since the maximum order of element formed by disjoint cycles is $lcm(3,4)=12$ and the least order of element it can form is the single cycle $(1)$. This permutation group is known, as an abstract group, as the dihedral group of order 8.I was working on a problem that is about finding all possible orders of elements in $S_7$ and $A_7$. Through the ubiquity of group actions and the concrete representations which they afford. The only remaining symmetry is the identity (1)(2)(3)(4). Permutation Groups form one of the oldest parts of group theory. The reflection about the 1,3−diagonal line is (24) and reflection about the 2,4−diagonal is (13). The reflection about the horizontal line through the center is given by (12)(34) and the corresponding vertical line reflection is (14)(23). The rotation by 90° (counterclockwise) about the center of the square is described by the permutation (1234). The symmetries are determined by the images of the vertices, that can, in turn, be described by permutations. Let the vertices of a square be labeled 1, 2, 3 and 4 (counterclockwise around the square starting with 1 in the top left corner). This permutation group is, as an abstract group, the Klein group V 4.Īs another example consider the group of symmetries of a square. G 1 forms a group, since aa = bb = e, ba = ab, and abab = e. This permutation, which is the composition of the previous two, exchanges simultaneously 1 with 2, and 3 with 4.Like the previous one, but exchanging 3 and 4, and fixing the others.This permutation interchanges 1 and 2, and fixes 3 and 4.This is the identity, the trivial permutation which fixes each element.The term permutation group thus means a subgroup of the symmetric group. The group of all permutations of a set M is the symmetric group of M, often written as Sym( M). In mathematics, a permutation group is a group G whose elements are permutations of a given set M and whose group operation is the composition of permutations in G (which are thought of as bijective functions from the set M to itself). ![]()
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