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).
Therefore .
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).
So...every (non-trivial) element has order 2? This is another problem. If you haven't already, you will encounter a theorem called Sylow's theorem. If you have encountered it, apply it to your comment I just quoted. If not, what can you tell me about the generating elements of ?
2 is prime.
Therefore Given a prime number .
As I said before, and .
is not a proper subgroup of any other p-subgroup of . This is because so the only possible primes that divide 6 are 3 and 2.
Let be the number of sylow-2 subgroups. Then [tex]n_2|3[/MATh] and . Then we have that or since these are the only numbers that satisfy these constraints.<--- I'm not sure where to go from here, i'm trying to show that G/N is not a proper subgroup of any other p-subgroup of G. I think want to show that ?
As for your other question, there must be more than generator for . Therefore it cannot be cyclic.
A nice corollary of Sylow's theorem (it may be a corollary of an earlier theorem, but it very obviously falls out of Sylow's theorem) is that for every prime dividing the order of your group you have a subgroup of order .
Looking for a contradiction to what you said about every element having order 2, apply this to your case.