- #1

logarithmic

- 107

- 0

1. In a proof for the fact that if a finite factor group G/N has 2 elements, then N is a normal subgroup of G, it says:

"For each a in G but not in H, both the left coset aH, and right coset Ha, must consist of all elements in G that are not in H".

Why is this so?

2. For the alternating group A_4, let H be a subgroup of order 6 (the point is to show there's no subgroup of order 6), it says that A_4/H = {H, sH} for some s in A_4 but not in H. Also, (H)(H) = H, (sH)(sH) = H. It then says, a in H implies a^2 in H, and b in sH implies b^2 in H. That is, the square of any element in A_4 must be in H.

Where does the last sentence come from? The line before that says the square of anything in H or sH is in H, so why does it then conclude this is also true for any element of A_4?

3. In an example, it says that (Z_4 x Z_6)/<(2,3)> has order 12 (just use lagrange's theorem), it then concludes that this factor group must be isomorphic to Z_4 x Z_3 or Z_2 x Z_2 x Z_3. (Z_n is the group {0, 1, ..., n-1}.)

I think it used the fundamental theorem of finitely generated abelian groups here. But that would require (Z_4 x Z_6)/<(2,3)> to be finitely generated. Is there any obvious way to see this? (It is obviously abelian).

We can see that (Z_4 x Z_6) = <(1,0) , (0,1)> so is finitely generated, but is there a theorem that says if G is finitely generated then so is its factor group G/N? (My textbook has a theorem that if G is cyclic then G/N is cyclic, but (Z_4 x Z_6) is not cyclic.) Or is every finite group finitely generated?