Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.5 Exercise 4.5.1
Solution: 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. $$\langle ((1\ 2), 1), (1, (1\ 2)) \rangle $$ $$\langle ((1\ 2), 1), (1, (1\ 3)) \rangle$$ $$\langle ((1\ 2), 1), (1, (2\ 3)) \rangle$$ $$\langle ((1\ 3), 1), (1, (1\ 2)) \rangle$$ $$\langle ((1\ 3), 1), (1, (1\ 3)) \rangle$$ $$\langle ((1\ 3), 1), (1, (2\ 3)) \rangle$$ $$\langle ((2\ 3), 1), (1, (1\ 2)) \rangle$$ $$\langle ((2\ 3), 1), (1, (1\ 3)) \rangle$$ $$\langle ((2\ 3), 1), (1, (2\ 3)) \rangle$$ 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 $1\pmod 3$. 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 $S_3 \times S_3$.
