Let H = { (1), (12)(34) } in A_4.......(Abstract Algebra)?

Question

Part I)
How do you show that H is not normal in A_4?

Part II)
Referring to the Multiplication table for A_4 as a set of even permutations (http://www.math.niu.edu/~beachy/aaol/grouptables2.html)
how do you show that, althought, X_6H = X_7H and X_9H = X_11H, it is not true that X_6X_9H = X_7X_11H. Why does this prove that the left cosets of H do not form a group under coset multiplication??

(For easier typing, for this problem I am letting alpha = X and the even permutations in the chart are labeled as follows:
(1) = X_1
(12)(34)= X_2
(13)(24)= X_3
(14)(23)= X_4
(123)= X_5
(243)= X_6
(142)= X_7
(134)= X_8
(132)= X_9
(143)= X_10
(234)= X_11
(124)= X_12

Answer 1

I) In order for H to be normal then for any h in H and k in A_4 then h^(-1)kh is in H.
Pick h = (12)(34), k = (123), then k^(-1) = (321) and
h^(-1)kh = (13)(24) which is not in H so H is not normal
II) x6H= {(243)(1), (243)(12)(34)} = {(243), (142)} =
x7H = {(142)(1), (142)(12)(34)}.
I hope you get the idea how these multiplications go and are able to show x9H = x11H and X_6X_9H is not equal to X_7X_11H.
If the left cosets of H formed a group and since x6H =x7H and x9H = x11H then it would follow:
x6Hx9H =x6x9H = x7x11H =x7Hx11H, which is not true.
Basically it's saying the operation is not well defined because mult equals by equals doesn't give us equals.