Given a group , embed in a larger group such that any automorphism of is the restriction of an inner automorphism of .
Exhibit a group , a set of generators for and a subgroup such that for all and yet is not normal in .
Finite subgroups of the group of units of a field are cyclic.
Every morphism of abelian groups vanishing on is identically zero.
Use this fact to prove that is not a free abelian group.
Let be a finite group such that , with prime. Then divides the order of .
Let . Prove that the group of homeomorphisms of the -dimensional ball that fix its boundary contains a copy of the free group on generators.
Recall that a left order on a group is an order of the underlying set such that left multiplication by any element of the groups is monotonous (for very simple examples see Some orders on groups). Define an element of an ordered group to be positive if it is greater than the unit of the group. Define an element to be negative if its inverse is positive. Clearly an element of an ordered group is either positive, negative or it is the unit of the group.
We now restrict our atention to orders on the group . The goal is the following: Suppose given an order on , prove that there exists a straight line passing through such that the elements strictly above this line have all the same sign, and the elements strictly below this line have also the same sign. That is, the elements strictly above are either all positive or all negative, and the same holds for the elements strictly below the line.