# Prove If a subgroup H of Sn contains an odd permutation, then H is

Content type
User Generated
Rating
Showing Page:
1/1
Prove: If a subgroup H of Sn contains an odd permutation,
then |H| is even and exactly half the elements of H are odd
permutations.
Solution
Take any subgroup H of S_n which is _not_
contained in A_n then since A_n is a normal subgroup of
S_n it follows that (H A_n)/A_n is isomorphic to H/(H
intersection A_n). Since H is not contained in A_n then H
contains an odd permutation so the product HA_n is in fact
equal to S_n. Therefore [H Int A_n: H] = 2 so H has even
order. Consider HnA(n) and HA(n). Use the fact that A(n)
is an index 2 normal subgroup of S(n). Even simpler, if H
contains an odd permutation \'o\', then H*o=H. But
multiplying by o turns even permutations into odd, and vice
versa. Thus exactly half elements are even and half are
odd permutations

Sign up to view the full document!

Unformatted Attachment Preview
Prove: If a subgroup H of Sn contains an odd permutation, then |H| is even and exactly half the elements of H are odd permutations. Solution Take any subgroup H of S_n which is _not_ contained in A_n then since A_n is a normal sub group of S_n it follows that (H A_n)/A_n is isomorphic to H/(H inte ...
Purchase document to see full attachment
User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service.

### Review

Anonymous
Really useful study material!

Studypool
4.7
Trustpilot
4.5
Sitejabber
4.4