4
$\begingroup$

Please check my proof:

Consider the order of subgroup that can exist in H and K are For H are $1,2,3,6,12$ For K are $1,5$

Because the set of subgroup order n for H and K has no common member that has the same order except $1$

then only subgroup that in $H\cap K$ is $1$ or ${e}$

$\endgroup$
5
  • 2
    $\begingroup$ Correct. You're using Lagrange's Theorem which also projects on the orders of elements in finite groups. $\endgroup$ – DonAntonio Dec 19 '16 at 11:22
  • 1
    $\begingroup$ You are right, just by Lagrange's theorem. $\endgroup$ – m-agag2016 Dec 19 '16 at 11:22
  • $\begingroup$ @DonAntonio thank you ^ ^ $\endgroup$ – Lingnoi401 Dec 19 '16 at 11:23
  • 2
    $\begingroup$ For $H$, the order can also be $4$. Of course that doesn't change your proof. $\endgroup$ – Bernard Dec 19 '16 at 11:49
  • 1
    $\begingroup$ It was sufficient to calculate the $\operatorname{gcd}(|H|,|K|)$. $\endgroup$ – Marc Bogaerts Dec 19 '16 at 17:32
3
$\begingroup$

By Lagrange's theorem we have that $H\cap K \subseteq K$ and $H\cap K \subseteq H$, so $|H\cap K| \bigm| |K|$ and $|H \cap K| \bigm| |H|$ , therefore $|H\cap K| \bigm| \gcd(|K|,|H|)=1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.