Let be a group of order , let , and form . Let act by left multiplication on the left cosets of in and let be the induced permutation representation.
- Prove that the elements of are coset representatives for the left cosets of in and that with this choice of representatives, is the left regular representation of .
- Prove that is the normalizer in of . Deduce that under the left regular representation of a finite group of order , the noramlizer in of the image of is isomorphic to . [Hint: Show that using §5.5 #1 and §5.5 #2.
- Deduce that the normalizer of the group generated by an -cycle in is isomorphic to and has order , where denotes the Euler totient function.
- Recall that . It suffices to show that if such that , then . To that end, if , then . There are left cosets of , and maps to these injectively. Thus is a set of coset representatives of . Define by , and let be the left regular representation of . Then . Hence ; thus is the left regular representation of up to an isomorphism.
- First, let . Now . Thus .
Now we prove a lemma.
Lemma [Adapted from Wikipedia]: If and , then . Proof: Since normalizes , for every there exists such that . Applying this mapping to 1, we see that . Thus, if , then for all we have . Now for all , we have , which is equal (since is a homomorphism) to . Since is invertible and is injective, we have ; hence is a homomorphism . Since is also bijective, .
Now let . By the argument above, also normalizes . Since the normalizer is a subgroup, we have . Moreover, ; by the lemma, is an automorphism of . Finally, note that , so that . Thus we have . Hence .
Since is injective, we conclude that .
- Let be an -cycle in . Up to a renaming, is the image of in via the permutation representation used above. Now , and . We know that , where denotes the Euler totient. Thus .