Wilkins D.R.'s Course 311: Michaelmas Term 2001. Topics in Group Theory PDF

By Wilkins D.R.

Show description

Read Online or Download Course 311: Michaelmas Term 2001. Topics in Group Theory PDF

Best symmetry and group books

Additional resources for Course 311: Michaelmas Term 2001. Topics in Group Theory

Sample text

Now if Ki and Kj are two distinct subgroups of order q then Ki ∩ Kj is a proper subgroup of both Ki and Kj , and its order is a proper divisor of the order q of Ki and Kj , by Lagrange’s Theorem. But q is a prime number. It follows that Ki ∩ Kj = {e}, where e is the identity element of G. We deduce from this that no element of G of order q can belong to more than one subgroup of order q. But each subgroup of G of order q contains q − 1 elements of order q (namely all elements of that subgroup with the exception of the identity element).

Let d be a divisor of p. Then xd is an element of order p/d, since p/d is the smallest positive integer k for which xdk = e. It follows that either d = 1 or d = p (since the group G contains no element whose order is greater than 1 but less than p). It follows that the order p of G is a prime number, as required. We have already proved that if n ≥ 5 then the alternating group An has no normal subgroups other than the trivial subgroup and the whole of An . Thus An is a simple group when n ≥ 5. /2.

Moreover H and K are subgroups of J, and therefore |J| is divisible by 9, by Lagrange’s Theorem. But J is a subgroup of G, and hence |J| divides 36. Also |J| > 9, since H (and K) are proper groups of J. It follows that either |J| = 18 or 36. If |J| = 36 then J = G and H ∩ K is a normal subgroup of G of order 3. If |J| = 18 then J is a subgroup of G of index 2, and is therefore a normal subgroup of order 18. We conclude that any group of order 36 contains at least one non-trivial normal subgroup.

Download PDF sample

Course 311: Michaelmas Term 2001. Topics in Group Theory by Wilkins D.R.


by Donald
4.0

Rated 4.34 of 5 – based on 8 votes