Set Permutation

Show that the set {I, (12)(34),(13)(24),(14)(23)} of permutations is a group.

Now is this set above equivalent to the symmetrical set of permutations for n=4, i.e.. is the above Sym(S_4)? If so then I imagine i only have to show it is equivalent as we know that the set of permutations of a finite set is a group (under composition).

If not, then is there a less cumbersome way than going through the 4 group axioms. I mean it is easy to show that I is the identity map and that each permutation is its own inverse. But then would i have to go through each possible composition to show that closure and associativity hold? That seems quite cumbersome which makes me think Im going about this the wrong way and am perhaps missing something.

Any help please

Whoa, I don't really want to think about what $\text{Sym}(S_4)$ looks like - do you mean $\text{Sym}(\{1,2,3,4 \})$? That's of order 4! = 24, which is far too big.

Anyway, there is a much less cumbersome way. What is the order of each element?

(There is also some theory you can use: that a conjugacy class in S_n is precisely the set of things with the same cycle type. Also, a normal subgroup of a group is precisely a union of conjugacy classes. Hence the set you have specified is a normal subgroup of S_4, and in particular is a group. That's a bit theory-heavy for what I had in mind, though.)

Ah yes I meant $\text{Sym}(\{1,2,3,4 \})$.

I'm not quite sure the book has covered 'order' yet. But I think that would mean the length of each cycle which is 2 because each permutation applied 2 times would be the identity map. Is that right?

Fair enough - yes, the order of those elements is all 2, apart from I which has order 1. (That's because of two facts: transpositions like (1 2) are self-inverse, and disjoint cycles commute.)
The group is isomorphic to the Klein 4-group. A subset of a group, containing I, in which all elements are self-inverse, is a group iff it is closed. (That's quite easy to prove - you should do it.) You can fairly quickly prove that the group is closed, just by hammering out a couple of operations.

More generally, notice that subsets of a group, using the same operation, containing the identity, are groups themselves iff the group is closed and each element has an inverse in the group.

Thanks. Also, if transpositions like (12) are self inverse, then how can you use this to find the inverse of products of disjoint transpositions? Like how would I go about finding the inverses of (12)(34) or (12)(34)(56).

Im finding some proof questions I can do fine, but others I get really stuck at. Like the one youve said to do which is easy, I just have no idea where to start. (Ive only just started uni maths so Im finding it a bit difficult for some problems to see what direction to take when writing a proof, though some i seem to manage fine)

Because transpositions are self-inverse, and because disjoint cycles commute, we have that (12)(34)(12)(34) = (12)(34)(34)(12) by commuting; (34)(34) = id, so LHS = (12)(12), which is id. Hence the inverse of a product of disjoint transpositions is itself.

Suppose you have a subset of a group, together with the parent operation, which contains I and where all the elements are self-inverse, and where the product of two elements of the subset is also in the subset. Prove that it is a group. (What axioms must it satisfy to be a group?)

Oh ok thats what i thought, thankyou- i was getting confused thinking everything is selfinverse as all cycles can be expressed as a product of transpositions. But the transpositions have to be disjoint right, so that commutativity holds?

Ok. Well it already satisfies the identity and inverse axioms. Is what youve said enough to explain that in a proof? And then we need to show that associativity and closure hold. Though we dont have to show associativity because it is a subset of the klein 4 group under which associativity must hold because that is a group so for a subset with the same operation, associativity will follow. Is it enough to just state this. Finally, we have to consider closure: Youve said the product of two elements in the subset is also contained in the subset - but how do we know this, i dont see why it might not appear that it is in the parent set but not the subset.

Yes, the transpositions have to be disjoint for an element to be self-inverse. This is because (132) can be written as (12)(23), which is of order three and so not self-inverse - despite being written as two transpositions.

Note, the group is the Klein-4 (V4) group, not a subset/subgroup of it. You can argue that it's a subset of the group S4 and so must satisfy the associativity property, but if you're going to do that you might as well just use the subgroup test and prove it's a subgroup of S4, and hence a group in its own right, just by showing the product of any two elements in the set is also contained in the set (because all non-identity elements are self-inverse). The question sounds more like an exercise in axiom grubbing anyway, so I'd avoid both arguments.

Yes, you have to consider closure; it's one of the main aspects of a group/subgroup, and group theory would pretty quickly fall apart if closure wasn't an axiom of a group. Checking closure is fairly straight-forward given that there are only four elements and one of them is the identity. All you have to do is show that six possible products of elements gives you another element in the set (since the job is trivial for any product involving the identity, and there's little point considering the product of two of the same non-identity element as they're all self-inverse). You can argue closure from a geometric standpoint because if you consider a rectangle with the corners numbered from 1 through to 4, your set (V4) represents the identity, horizontal and vertical reflection and 180 degree rotation. You can then argue closure by arguing that if you combine two of the four you're guaranteed to get another one (i.e. if you reflect the rectangle horizontally and vertically you end up with 180 degree rotation) and then the property follows from fairly simple geometry.

Oh ok thats what i thought, thankyou- i was getting confused thinking everything is selfinverse as all cycles can be expressed as a product of transpositions. But the transpositions have to be disjoint right, so that commutativity holds?