# If $H,K$ be subgroup of $G$ such that $|H|=12$ and $|K|=5$ prove that $H\cap K ={e}$

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 ${e}$

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

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 $|H\cap K| \bigm| \gcd(|K|,|H|)=1$.