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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..

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Do you know examples of nonabelian groups of order 12? – Jonas Meyer Dec 18 '10 at 6:17
Hint: Try to find a non-abelian group of order 12 with a normal 3-Sylow. To do this it will be helpful to understand the semidirect product construction (see – Noah Snyder Dec 18 '10 at 6:17
I do not know how to classify them, but I know there is either a normal 3-subgroup or a normal 2-subgroup. – Gargle Dec 18 '10 at 6:19
You don't need to classify them, just an example. One way to get them is to take products or semidirect products of smaller groups (whose orders multiply to 12). – Jonas Meyer Dec 18 '10 at 6:22
Dear 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.) – Matt E Dec 18 '10 at 7:13

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.

You can check this page for more references. See also here

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