group proof, permutations

[jrwwest]

I've got this question to complete and i'm not even sure how to start,
closure proof of a group of permutations confuses me somewhat.

A set G contains permutations in S6 and is given by
G={e , (12)(36)(45), (14)(25)(36), (15)(24)}.

proove that G forms a set under compostion of permutation


and find a subgroup of this group which contains fewer than 4 elements.

 

[daon]

You haven't said where you're having trouble. To prove G is a group you need: (1) The identity permuation is contained. I guess thats the 'e' you see there. (2) The set is closed under the operation (composition). For each pair of non-idenity elements, compute their product in both orders. There are (3 choose 2) = 3 pairs, so there are six computations that have to be done ((g o h) and (h o g) are two different products). If you know in your class that disjoint cyles commute, you don't need to do both. (3) Every element in G has an inverse (it must be both a right and left inverse to be an inverse). In doing step (2) you should be able to find the inverses. Note that the identity element e here is the permutation that sends every position to itself. In cycle notation it is (1)(2)(3)(4)(5)(6).

SInce G has only 3 non-denity elements, you are looking for a set of one or two of these that together with the idenity form a group. Theres very few possibilities and in doing the first part you should see which elements you may remove. In particular, look for either an element that is its own inverse (then that with e will form a group), or look for two elements that are inverses of eachother, then those two with e will form a group.