Re: Sylow subgroups of Alt(4)

On 27 Jan., 19:55, qsymmetry <qsymme...@xxxxxxxxx> wrote:
On Jan 27, 9:33 am, qsymmetry <qsymme...@xxxxxxxxx>
wrote:
How can we determine all the Sylow subgroups of
Alt(4) [or A_4] explicitly?

The number of Sylow 2-subgroups is either

n_2 = 1 or 3,

the order of any of which is 4.

Also, n_3 = 1 or 4.

I also know that A_4 contains 3 elements of order
2, and 8 elements of order 3; in particular, no Sylow
2-subgroup is cyclic.

If P were the unique Sylow 2-subgroup, then

P ~ Z/2 x Z/2, so that P contains 3 elements of
order 2.

Do we need to rule out n_2 = 3?   For the sake of
this discussion, let's suppose n_2 = 3;  then
how many elements do 3 distinct subgroups
isomorphic to Z/2 x Z/2  have?  Must they intersect
trivially?

Finally, since n_3 = 1 implies that there are only
2 elements of order 3,   we must have n_3 = 4.

Since there are only 4 elements in A_4 that can
possibly be in a Sylow
2-subgroup, namely the 3 elements of order 2 together
with the
identity element, then this subgroup has to be the
unique Sylow 2-
subgroup since any other such subgroup would have to
contain exactly
the same 4 elements.

Or, is it enough to show that those 3 elements of order 2 together identity constitute  a  subgroup of order 4; and that this is the only subgroup of order 4 in A_4?

So let  P  = {e, (12)(34), (13)(24), (14)(23)}

P is a group of order 4 and hence Sylow.
It is normal because a conjugate of e.g. (12)(34) again looks like (ab)
(cd) -- that is
a product of two disjoint transpositions.
Since P is normal, there is only one 2-Sylow group.


n=3:
If there were only one 3-Sylow group, it would contain all elements of
order 3,
esp. (123) and (124) and their product (13)(24) -- which is
impossible.
Therefore n3=4
.

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