Here are a two exercises.  

1) Show that the set of even integers is a group.

2) Another group is a permutation group, also known as a symmetric group on n elements, denoted as `S_n`.  This group permutes the elements in a list to some different order.  Symmetric groups are important because it turns out that **ALL** groups can be realized as the subgroup of a symmetric group.  This result is known as Cayley's Theorem.  An example of a symmetric group  is`S_3`.  Consider the natural ordering of a list as {1, 2, 3}.  Now, we want to know how we can re-arrange this list.  We can have {1, 2, 3}, {1, 3, 2}, {2, 1, 3}, {2, 3, 1}, {3, 1, 2}, and {3, 2, 1}.  How do we go from {1, 2, 3} to {1, 3, 2} ?  We *permute* the elements 2 and 3 in the ordered list.  This permutation is denoted as (2 3).  Notation-wise, we start from the right and say that 3 goes to 2 and then 2 wraps back to the beginning of 3. The group operation is taken as performing permutations in order starting from right to left.  That is to say with (1 2)(1 3)(1 2)(2 3), we see that 3 goes to 2 in the right most permutation, and then 2 goes to 1, and then 1 goes back to 3.  So 3 stays fixed.  Similarly, 2 goes to 3 which goes to 1 which goes to 2.  And so 2 stays fixed.  Since we have only 3 elements, and 2 of them are fixed, the third also must be fixed.  So (1 2)(1 3)(1 2)(2 3) is actually the identity permutation (written in a more complicated form).  We denote the identity of any symmetric group as (1), and we see that the identity never permutes any of the elements.  Question: what group permutation element moves {1, 2, 3} to {3, 2, 1} ?