Use the left regular representation of to produce two elements of which generate a subgroup isomorphic to .
Recall that .
Now , , , , , , , and . Thus .
Similarly, , , , , , , , and . Thus .
With the labeling
we have .
Consider .
To save effort we will perform this computation in and then use the labeling bijections to produce subgroups of . Let be the left regular representation.
Now via the labelling
we have the isomorphic image
Similarly, via the labeling
we have the isomorphic image
It is easy to see that these two subgroups are different.
List the elements of as , , , , , and label these with the integers , respectively. Exhibit the image of each element of under the left regular representation of into .
We use the notation .
Let be the Klein 4-group.
The multiplication table for is as follows.
1 | a | b | c | |
1 | 1 | a | b | c |
a | a | 1 | c | b |
b | b | c | 1 | a |
c | c | b | a | 1 |
We can see that with labelled as , respectively, since , , , and we have , , , and . Thus .
Similarly, , , , and , so that . Likewise .
Now we relabel the elements as , respectively. Now , , , and . Thus . Now , , , and , so that , and , , , and , so that . Clearly this is the same subgroup as the image of under the first labelling.
Let be a group and subgroups. For each , define the double coset of by .
Let . Then for some , so that for some . Then for some . So .
Let . Then for some , so that for some . Then for some . So .
Note that if , then .
Now suppose such that . Then we have for some and . Then , so that . Similarly . Thus two double cosets are either disjoint or equal. Thus the double cosets form a partition of .
Lemma 1: . Proof: Suppose . Then , and we have . So , hence . Suppose . Then , so that , thus .
We saw in part 1 above that ; moreover, this union is disjoint because the are distinct left cosets of , each of order . Thus , using Lemma 1.
Lemma 2: . Proof: Suppose . Then , and we have . So , hence . Suppose . Then , so that , thus .
We saw in part 2 above that ; moreover, this union is disjoint. Thus .
Suppose acts transitively on and let be normal. Let be the distinct orbits of on .
Now let be arbitrary; since the action of on is transitive, there exists such that . Then ; thus the action of on the orbits is transitive.
Let be arbitrary, and let such that . Because is normal, for every there exists a unique such that . We have a mapping given by . Because , this mapping is bijective, and . Because the action of is transitive on the orbits, all orbits have the same cardinality.
We claim that . Proof of claim: Suppose . Now , so that in particular for some . Then , so that . Thus . Now suppose . Then for some and . Now , so that . Thus .
A transitive permutation group acting on is called doubly transitive if for all , the subgroup is transitive on .
We saw that the action of on the four vertices of a square is not primitive in the previous exercise. Thus this action is not doubly transitive.
Let act transitively on the set . A block is a nonempty subset such that for all either or .
Now suppose . We have , so that . Hence is nonempty, and we have since is a block. Thus . By a previous exercise, .
Now suppose . Then there exist such that . Now , so that is not empty. Thus , and we have . So elements of are pairwise disjoint; hence is a partition of .
So that is a nontrivial block; hence this action is not primitive.
Lemma 1: Let act transitively on and let be a block under the action. Then if and only if . Proof: The direction is clear. Suppose is a proper subset and let , . Now and , so that . Since is a block, we have . But since and , a contradiction. So is not proper, and we have .
Now we move to the main result.
Suppose is primitive on ; then the only blocks of are and the singletons. Now let and let be a subgroup with . We consider the set .
We claim that is a block. Proof of claim: is not empty since . Now let . If , then . If , then . Suppose that ; say that for some . Then , so that . But then , a contradiction. Thus if , then . Hence is a block.
Next we claim that . Proof of claim: If , then . Suppose . Then for some , hence . Then , so that .
Since is primitive on and , we have or . If , we have , so that . If , we have . Thus is a maximal subgroup of .
Suppose that for all , is a maximal subgroup in . Let be a block with . Now by part 1 above. Since is maximal, there are two cases.
If and such that , then since acts transitively there exists such that . Now , a contradiction. Thus .
If , then by Lemma 1 we have .
Thus is primitive on .
As in this previous exercise, let act on the set of all polynomials with integer coefficients in the independent variables by permuting the indices: .
Repeat the previous example with each of the following actions.
This exhausts the elements of .
. |
Now under the permutation representation , we have
Each remaining orbit has order 6. Thus, if , , then ; hence .
Under the permutation representation ,
If , we similarly have . In particular, since , we have .
Now , so that .