is abelian since every element is it's own inverse (I proved this in the last homework I had).Let be the group of all permutations of .
The subset of is a normal subgroup of .
By proposition 3.5 of the notes, since / ,either (a cyclic group of order 6) or .
Decide which and explain your answer.
is not abelian. This is the group of isometries of a triangle. If we label the vertices 1,2 and 3 respectively, the permutations are a=(1,2,3) and b=(2,3). ab sends 2 to 1 but ba sends 2 back to 2. Hence it isn't abelian.
For two groups to be isomorphic, they must be structurally similar (ie. must be isomorphic to another abelian group).
I think this is right, but one of my friends has got it isomorphic to . Which one of us is right? (I can't see a flaw in my reasoning).