rev1 - Math 5125 Friday, September 23 First Test Review The...

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Unformatted text preview: Math 5125 Friday, September 23 First Test Review The test will cover sections 3.1, 3.2, 3.3, 3.5, 4.14.6, 5.1, 5.2, 5.4, 5.5, 6.3. Topics will include If Z is a central subgroup of G and G / Z is cyclic, then G is abelian. Fundamental homomorphism theorem and related isomorphism theorems. Group actions, | G x || G x | = | G | If G has a subgroup H of index n , then there exists K G such that K H and | G / K | n !. Class equation The center of a nontrivial p-group is nontrivial. Groups of order p 2 are abelian. Sylow theorems. If G is a simple group which contains a subgroup of index n and | G | 6 = 2, then G is isomorphic to a subgroup of A n . Proving groups of certain orders cannot be simple. A n is simple for n 5 Fundamental structure theorem for finitely generated abelian groups; elementary di- visors and invariant factors Semidirect products Free groups, universal property of free groups Solution to Problem 1 on Sample Test no. 1 We consider the conjugation action of G on G given by g x = gxg- 1 for g , x G . Since the order of x is the same as the order of gxg- 1 , elements of order two are sent to elements of order two under this action and therefore the size of an orbit containing an element of order two is 1, 2 or 3. Lets consider separately each of these possibilities (its worth noting at this stage that G must have at least 4 elements). Suppose { x } is an orbit of size one. This means that gxg- 1 = x for all g G and thus x is in the center of G . It follows easily that h x i is a nontrivial normal subgroup of G and so G is not simple. Suppose there is an orbit of size two. Then the stabilizer of an element in this orbit will have index two in G . Since subgroups of index two are always normal, G has a nontrivial normal subgroup (because...
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rev1 - Math 5125 Friday, September 23 First Test Review The...

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