![]() 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 map S n GL(n, Z 2) that sends a permutation to its column representation is a faithful representation. ![]() By the formulas above, the n × n permutation matrices form a group under matrix multiplication with the identity matrix as the identity element. The possible patterns when you write an element of $S_7$ as a product of disjoint cycles (which you can always do) are (xxxxxxx) Since there are n permutations, there are n permutation matrices. More help, in response to the OP's comment. 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 $7!$. 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 $|(12)(32)|>|(23)(14)|$. 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 $1$, $3$, $5$, $\dots$, $11$. 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 $(1)$. I think Cayley's Theorem has more historical interest than practical interest these days, but your mileage may vary.I was working on a problem that is about finding all possible orders of elements in $S_7$ and $A_7$. Having both viewpoints is better than having just one. ![]() But, as he pointed out, it is sometimes more convenient or useful to consider the group abstractly, sometimes to consider it as a group of permutations. Cayley was trying to abstract the notion of group he then pointed out that his more abstract definition certainly included all the things that people were already considering, and that in fact it did not introduce any new ones in the sense that every abstract group could be considered as a permutation group. The reason for Cayley's Theorem is that, historically, people only considered permutation groups: collections of functions that acted on sets (the sets of roots of a polynomial, the points on the plane via symmetries, etc). Mathematicsa mathematical group whose elements are permutations and in which the product of two permutations is the same permutation. You usually get more information if the set you are acting on is "small"-ish. But this gives you an embedding of $G$ into a very large symmetric group, because the set on which it is acting is large. Cayley's Theorem tells you that every group $G$ can be thought of as a permutation group, by taking $X$ to be the underlying set of $G$, and $\sigma$ to multiplication. You think of a permutation group as a group $G$, together with a faithful action $\sigma\colon G\times X\to X$ on a set $X$ (faithful here means that if $gx=x$ for all $x$, then $g=e$). For example, Jordan proved that the only finite sharply five transitive groups are $A_7$, $S_6$, $S_5$, and the Mathieu group $M_$ see. "Permutation group" usually refers to a group that is acting (faithfully) on a set this includes the symmetric groups (which are the groups of all permutations of the set), but also every subgroup of a symmetric group.Īlthough all groups can be realized as permutation groups (by acting on themselves), this kind of action does not usually help in studying the group special kinds of actions (irreducible, faithful, transitive, doubly transitive, etc), on the other hand, can give you a lot of information about a group. ![]()
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