Is it true or false that a group of order 12 always has a normal 2-sylow subgroup? I have a hunch it is false..
Group of order 12
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$\begingroup$
group-theory
finite-groups
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5Dear Gargle, Write down a 2-Sylow subgroup in your example, and start conjugating it. (There are only 12 elements to conjugate by!) If you have trouble "seeing" things, then just compute. (This is an advantage of algebra --- one can always just start computing and see what happens.) – 2010-12-18
3 Answers
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Hint : Use the second of the Sylow theorems. All the $p$-Sylow subgroups are conjugate to each other and a group is normal iff it is equal to each of its conjugates.
Some semidirect product would give you an explicit example.
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The 5 groups of order 12 are $C_{12}$, $C_6 \times C_2$ in the abelian case, and $A_4$ (group of all even permutations of length 4), $D_6$ (group of all symmetries of the regular hexagon), $C_3 \rtimes C_4$ in the nonabelian case.
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$D_{12}$, the dihedral group of order 12, which can also be described as the direct product of the symmetric group $S_3$ and the cyclic group of order 2, has a non-normal 2-Sylow subgroup.
This follows from the fact that $S_3$ itself has a non-normal 2-Sylow subgroup.