Written Assignment No. 1

due September 28, 2005

General Directions: Written assignments should be submitted typeset. What you submit must represent your own work.

Example of a Solved Exercise

Please note that the directions for this solved exercise differ from those for the exercises in the present assignment.

Prove the following statement: If the number of elements in a finite group G with identity e is even, show that there is at least one element g in G such that g ≠ e but g ∗ g = e.

Proof. Let 2n be the number of elements of the given finite group G. The assertion is that there is at least one element of G other than e for which g ∗ g = e, i.e., g = g^{−1}. If this were not the case then for every g ≠ e in G one would have g ≠ g^{−1}, i.e., g and g^{−1} would be different elements. So the set G − {e} would be the disjoint union of two element subsets of the form {g, g^{−1}}, and, therefore, the number |G − {e}| of elements of G − {e} would be even. Since G is the disjoint union of {e} and G − {e},

|G| = 1 + |G − {e}| ,

and, therefore, the number of elements of G would be odd. Hence, if the number of elements of G is even, there must be at least one element of G {e} for which g ∗ g = e.

Assigned Exercises

Read these directions carefully: for each of the following statements either provide a proof that the statement is true or label the statement as false and provide justification.

1. The multiplicative group of the integers mod 11 is a cyclic group.

2. If G is an abelian group with identity e, then the set T of all elements t ∈ G such that t^{2} = e is a subgroup of G.