then
is 2-sylow subgroup is normal in it?
is 3-sylow subgroup is normal in it?
is a subgroup of order 72 is normal?
i think all are possible.
since order of A =144
the order of A is 24*6 = 144.
a subgroup of order 72 must be normal. why? why is it that ANY subgroup of index 2 is ALWAYS normal? (think of how many cosets there are, and then compare left and right cosets).
a sylow 3-subgroup is of order 9. sylow theory tells us that the number of sylow 3-subgroups must be 1,4,or 16. since sylow subgroups are all conjugate, the only way a sylow subgroup can be normal is if it is the only sylow subgroup (for that prime divisor).
since <((1 2 3),e),(e,(1 2 3))> is one such subgroup, and <((2 3 4),e),(e,(1 2 3))> another, we can conclude that the sylow 3-subgroups are not normal.
a sylow 2-subgroup is of order 16. again we have either 1,3 or 9 of them, and a sylow 2-subgroup will be normal only if there is only one. again <((1 2 3 4),e),(e,(1 2))> is one such subgroup, and <((1 2 3 4), e), (e,(2 3))> is another, so the sylow 2-subgroups are not normal.
<a,b> means the subgroup generated by the set {a,b}, where a and b are elements of our the group G.
in this case, G consists of pairs, the first element of the pair is from S_{4}, and the second element of the pair is a member of S_{3}.
for example, <((1 2 3),e),(e,(1 2 3))> is the 9-element subgroup:
{(e,e), (e,(1 2 3)), (e,(1 3 2)), ((1 2 3),e), ((1 2 3),(1 2 3)), ((1 2 3),(1 3 2)), ((1 3 2),e), ((1 3 2),(1 2 3)), ((1 3 2),(1 3 2))}
(this is isomorphic to Z_{3} x Z_{3}, being generated by a 3-cycle in S_{4} x {e}, and a 3-cycle in {e} x S_{3}).
it is easy to check that in S_{4}, (2 3 4) is conjugate to (1 2 3):
(2 3 4) = (1 2 3 4)(1 2 3)(1 2 3 4)^{-1}:
1→4→4→1
2→1→2→3
3→2→3→4
4→3→1→2
therefore, in G, we have ((1 2 3 4),e)((1 2 3),e)((1 2 3 4), e)^{-1} = ((2 3 4),e), which shows that <((1 2 3),e),(e,(1 2 3))> is not normal in G.