Finding a 2-Sylow subgroup of $\left(\mathbb Z^\times_{11} \times \mathbb Z^\times_{13} , \cdot \right)$

Question

so I'm trying to work through this problem:

How many elements does a 2-Sylolw subgroup of $\left(\mathbb Z^\times_{11} \times \mathbb Z^\times_{13} , \cdot \right)$ have (i.e. what is the order) and how many 2-Sylow subgroups there are in $\left(\mathbb Z^\times_{11} \times \mathbb Z^\times_{13} , \cdot \right)$ ? Find at least one 2-sylow subgroup and list all of its elements. (just in case $\mathbb Z^\times_n$ denotes the set of all $k \in \mathbb N$ that are relatively prime to n)

Now i think i can answer some of it so here it goes:

so the order of $\mathbb Z^\times_{11} \times \mathbb Z^\times_{13}$ is (or not?) $\left|\mathbb Z^\times_{11} \times \mathbb Z^\times_{13}\right|$ =$\phi(11)\phi(13) = 10\cdot12=120=2^3\cdot15$. So if that is correct then i know they are of order $2^3=8$ and i also know that if $s$ is the total number of 2-Sylow subgroups then the following must hold: $s\equiv 1\pmod 2$ and $s\mid15$. So from that i can say that $s \in \{1,3,5,15\}$.

Now that's as far as i can go. I don't know how to construct such subgroups i know that the elements of those subgroups must be of order that divides 8 that is of order 1,2,4 or 8, i know that there must be an element of order 2 (by Cauchy theorem i think). But i don't have an efficient way of finding those elements and i don't know how to specify exactly how many 2-Sylow subgroups there are other than saying it's either 1,3,5 or 15.

Any help will be greatly appreciated.

Answer 1

$\mathbb{Z}_{11}^\times \times \mathbb{Z}_{13}^\times$ is abelian, and all Sylow subgroups of a given order are conjugate, so there is exactly one Sylow 2-subgroup. It consists of all elements of order a power of two. Since $ord(a,b) = lcm(ord(a),ord(b))$, it is generated by $(-1,1)$ and $(1,x)$ for $x$ a square root of $-1$ in $\mathbb{Z}_{13}$.

Answer 2

We have $\mathbb{Z}_{11}^\times \times \mathbb{Z}_{13}^\times \cong C_{10} \times C_{12}$, whose $2$-Sylow subgroup is $C_{2} \times C_{4}$.

Therefore, the $2$-Sylow subgroup of $\mathbb{Z}_{11}^\times \times \mathbb{Z}_{13}^\times$ is $\langle g_{11}^5 \rangle \times \langle g_{13}^3 \rangle$, where $g_{p}$ is a primitive root mod $p$.

Note that $g_{11}^5$ has order $\frac{10}{5}=2$ and $g_{13}^3$ has order $\frac{12}{3}=4$, as needed.

We can take $g_{11}=2$ and $g_{13}=2$ and then the $2$-Sylow subgroup is $\langle 10 \rangle \times \langle 8 \rangle$.