An is the subgroup of all even permutations of Sn. to prove/disprove the statement, one usually defines what is meant by the sign or parity of a permutation,
and that sgn:Sn-->({-1,1},*) or parity:Sn--->({0,1},+) is a homomorphism from Sn to a cyclic group of order 2.
in other words, An is/is not normal because (even)(even)(even) = ____ and (odd)(even)(odd) = _____.
the real "meat" of this, is in proving that the sign or parity of a permutation is "well-defined", because a permutation can be written as a product
of transpositions in many, many different ways.
the second problem is a bit easier. for a homomorphism φ:G-->G', ker(φ) is a normal subgroup of G, so |ker(φ)| divides |G|, by lagrange.
by the first isomorphism theorem, φ(G) ≅ G/(ker(φ)). so we have 6 = 20/d, where d is an integer that divides 20.
proving that if [G:H] is the smallest prime p dividing |G|, then H is normal, is kind of involved, and usually requires that one knows about group actions.
the special case p = 2, is a bit more straight-forward, and can be proved directly from the definition.
Thanks Deveno! I get the second problem now! However for the first one, I've never read anything on the sign/parity of a permutation actually, but I have come across even/odd permutations, I was just wondering if you could explain it a bit further for me, I don't think my knowledge in this area is that strong yet haha.
The sign of a permutation is how many parts it is made up of. For instance, (123)=(12)(23), so it is even (as it is made up of an even number of parts) while (1234)=(12)(23)(34) is odd. Similarly, (123)(4567)=(12)(23)(45)(56)(67) is odd.
From here it is easy to see that odd times odd is even, even times odd is odd, and even times even is even. And just to confuse you, an odd-cycle is even while and even-cycle is odd.
Now, as even times even is even (and the identity is even), these form a subgroup. This is your alternating subgroup, .
Wikipedia covers this, although it may not cover it well.
the "parts" Swlabr refers to, are called 2-cycles, or transpositions. in point of fact, every permutation can be achieved by "swapping two numbers at a time". for example, the permutation (1 2 3) the mapping:
1-->2
2-->3
3-->1
we can get this same mapping first by switching 2 and 3:
1-->1
2-->3
3-->2
and then by switching 1 and 2:
1-->1-->2
2-->3-->3
3-->2-->1 so if f = (2 3) and g = (1 2), (1 2)(2 3) = gof = (1 2 3). since it takes two transpostions to achieve the 3-cycle (1 2 3), and 2 is an even number, a 3-cycle is even.
a single transpostion is odd, because 1 (the number of transpostions it takes to form a transpostion) is an odd number.
by the same reasoning, a "double swap", such as (1 2)(3 4) (where the numbers switched have nothing to do with each other), is an even permutation.
there is a nice explanation of permutations and parity here.
Ahh okay thanks for that guys, I managed to come up with another soln to the first Q without the use of signs/parities, is it correct though? I just used the definition of a normal subgroup.
So suppose that and . Suppose we have written (product of m cycles). Then .
Thus:
So you see, has been written as a product of 2m+(number of cycles in h) cycles which is an even number, and so is closed under conjugation. Assuming that we already know A_n is a subgroup of S_n, then we have shown A_n is in fact a normal subgroup of S_n.
yes, almost. you don't want to say "cycles" though, but "transpositions" or 2-cycles. there is a difference. for example, even though a 3-cycle is even,
we can write (1 2 3) = (1 2 3)(1 3 2)(1 2 3), as a product of 3 3-cycles.
that's why the evenness/oddness of a permutation is such an important concept. no matter HOW we write a permutation as a product of transpositions,
the evenness/oddness (parity is just a fancy word for this) is invariant. (1 2 3) can NEVER be written as the product of an odd number of transpositions.