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.


$\mathbb{Z}_{11}^\times \times \mathbb{Z}_{13}^\times$ 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 $\mathbb{Z}_{13}$.

  • $\begingroup$ thank you very much for your response.However, I'm not sure i completely understand your answer. Can you please elaborete a little more? why is there only one sylow 2-subgroup if the group is abelian? and how does one quickly finds all elements that are of order power of two? and i also don't see how ord(a,b)=lcm(ord(a),ord(b)) tells me that it is generated by (−1,0) and (0,x) $\endgroup$ – strangeattractor Jun 16 '18 at 11:57
  • $\begingroup$ If the group is abelian, then it acts trivially on itself by conjugation, so conjugate subgroups are equal. The $lcm$ formula tells you that it suffices to find elements of order a power of two in the factors of the product. The unique element of order two in $\mathbb{Z}_n^\times$ is $-1$, and elements of order 4 are square roots of $-1$. No higher power of two is an order of an element of the factors since their orders are $10$ and $12$. $\endgroup$ – Joshua Mundinger Jun 16 '18 at 12:01
  • $\begingroup$ thank you again. I was suspecting it has something to do with action. I see it more clearly now. However there are still thing that are not clear to me.This might be a silly question, but Is -1 always the only element of order two even if i'm not working in $\mathbb Z^\times _{prime}$ ? And why is there not an element of order 8 again? does it have to do something with 2*4=8 because i have elements of order 4? I really have to revise modular arithmetic it seems $\endgroup$ – strangeattractor Jun 16 '18 at 12:09
  • $\begingroup$ $(-1)^2 = 1$ holds in any ring. There is no element of order $8$ in $\mathbb{Z}_{11}^\times$ or $\mathbb{Z}_{13}^\times$ by Langrange's theorem since neither of them have order divisible by 8. $\endgroup$ – Joshua Mundinger Jun 16 '18 at 12:11
  • $\begingroup$ oh Lagrange's theorem! Thank you. But is it possible that i can find elements of order two other than -1 in some $\mathbb Z^\times _{not-prime}$ ? $\endgroup$ – strangeattractor Jun 16 '18 at 12:14

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 $\langle 10 \rangle \times \langle 8 \rangle$.

  • $\begingroup$ $C_{n}$ denotes cyclic group of n elements? $\endgroup$ – strangeattractor Jun 16 '18 at 14:27
  • $\begingroup$ I'm afraid i don't understand the notation. Can you please explain what you mean by $g^5_{11}$ and $g^3_{13}$? in particular what does the 5 and the 3 in the exponent mean and where do they come from? $\endgroup$ – strangeattractor Jun 16 '18 at 17:42
  • $\begingroup$ thank you very much now i understand. $\endgroup$ – strangeattractor Jun 16 '18 at 17:58

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