Compute all Sylow subgroups of Dih(12) and the direct square of Sym(3)

Exhibit all Sylow 2- and Sylow 3-subgroups of D_{12} and S_3 \times S_3.


The Sylow 2-subgroups of D_{12} have order 4. By Sylow’s Theorem, the number n_2 of Sylow 2-subgroups is odd and divides 3; hence n_2 \in \{1,3\}. Note that

  • \langle r^3, s \rangle = \{ 1, r^3, s, sr^3 \},
  • \langle r^3, sr \rangle = \{ 1, r^3, sr, sr^4 \}, and
  • \langle r^3, sr^5 \rangle = \{ 1, r^3, sr^5, sr^2 \}

are distinct subgroups of order 4. Thus n_2 = 3, and we have exhausted the list of Sylow 2-subgroups of D_{12}.

The Sylow 3-subgroups of D_{12} have order 3, hence are cyclic. Note that any element of the form sr^a$ has order 2, and so is not in a Sylow 3-subgroup. Thus the Sylow 3-subgroups are contained in \langle r \rangle; this group is isomorphic to Z_6, and thus D_{12} has a unique subgroup of order 3, namely \langle r^2 \rangle.

The Sylow 2-subgroups of S_3 \times S_3 have order 4. Now by Sylow’s Theorem, the number n_2 of Sylow 2-subgroups is odd and divides 9; thus n_p \in \{1,3,9\}. Note that the following are subgroups of order 4.

  1. \langle ((1\ 2), 1), (1, (1\ 2)) \rangle
  2. \langle ((1\ 2), 1), (1, (1\ 3)) \rangle
  3. \langle ((1\ 2), 1), (1, (2\ 3)) \rangle
  4. \langle ((1\ 3), 1), (1, (1\ 2)) \rangle
  5. \langle ((1\ 3), 1), (1, (1\ 3)) \rangle
  6. \langle ((1\ 3), 1), (1, (2\ 3)) \rangle
  7. \langle ((2\ 3), 1), (1, (1\ 2)) \rangle
  8. \langle ((2\ 3), 1), (1, (1\ 3)) \rangle
  9. \langle ((2\ 3), 1), (1, (2\ 3)) \rangle

Moreover, these groups are all distinct since for any pair, some generator of one is not contained in the other. Thus n_2 = 9 and we have exhausted the Sylow 2-subgroups of S_3 \times S_3.

The Sylow 3-subgroups of S_3 \times S_3 have order 9. By Sylow’s Theorem, the number n_3 of Sylow 3-subgroups divides 4 and is congruent to 1 mod 3. Thus n_3 \in \{1,4\}. Recall that S_3 has one element of order 1, three elements of order 2, and two elements of order 3, and that the order of (a,b) \in S_3 \times S_3 is the least common multiple of the orders of a and b. So every element of S_3 \times S_3 has order 1, 2, 3, or 6, and moreover we can count that there are 15 elements of order 2, 8 of order 3, and 12 of order 6. Since every element in a Sylow 3-subgroup must have order 3, the elements of order 3 together with the identity comprise the only possible Sylow 3-subgroup. So n_3 = 1.

Now \langle ((1\ 2\ 3), 1), (1, (1\ 2\ 3)) \rangle \cong Z_3 \times Z_3 is a subgroup of order 9, thus is the unique Sylow 3-subgroup of S_3 \times S_3.

Post a comment or leave a trackback: Trackback URL.

Comments

  • Gobi Ree  On December 19, 2011 at 8:31 pm

    How can you conclude that n_3=1 in the last paragraph? It can be n_3=1,4, but after checking the actual subgroup we can deduce that it is unique.

    • nbloomf  On January 19, 2012 at 10:22 am

      You’re right. Thanks!

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: