That's what I thought you meant- so I don't think that you follow my answer:
If you take the group which consists of ALL permutations of elements of R(3) (which is going to be a huge set, of cardinality larger than that of the real numbers), then you will clearly get a group, but what I'm saying is that it won't be the Euclidean group.
The Euclidean group consists of all Euclidean transformations of R(3)- think of taking your copy of R(3), rotating it a bit, reflecting it and/or moving it about by translation. All such transformations will give you the Euclidean group- elements in this group of have the property, for example, that all elements remain the same distance from each other after the transformation.
This clearly isn't so for ALL permutations. Although all Euclidean transformations do describe permutations, I can imagine a permutation e.g. one which just switches (1,0,0) and (0,0,0) and fixes the rest which won't be in the Euclidean group. I suppose you can even see this from the cardinality of the groups- the cardinality of the Euclidean group will only be of order the same as the real numbers (I think), where as ALL permutations will be larger.
A subgroup of the Euclidean group could be where you force the origin to remain fixed. This will give you the orthogonal group O(3). This group is now compact (in a sense, doesn't go off to infinity) because you don't have infinite translations. There is a subgroup of this which is connected, called SO(3), the special orthogonal group. This one doesn't allow "flips", or orientation reversing transformations. All of these things are just permutations, but of a special sort. So, being groups themselves, we could say that they are subgroups of the group of ALL permutations of R(3). e.g. the Euclidean group, I imagine, is just all permutation of R(3) which preserves distances between points.
I hope this longer answer is more clear!