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

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- June 19th 2012, 06:13 PMsaravananbsgroups
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 - June 19th 2012, 08:15 PMDevenoRe: groups
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. - June 21st 2012, 05:45 PMsaravananbsRe: groups
- June 21st 2012, 11:09 PMDevenoRe: groups
<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.