Let the group, $G$
There is a unique subgroup $H (\leq G) s.t. \vert H \vert =n$ then, $H \lhd G$
I believe that the above statement is true. Let me allow telling my proof idea to you.
If we take the $\phi_g : G \to G $ by $H \to gHg^{-1}, \forall g \in G$ ($\phi_g$ is Inner automorphism)
Then $\phi(H) \leq G$ and $\vert \phi(H) \vert = n$. By the way, there is a unique subgroup whose order is $n$. This means $\phi(H) = H$.
Hence $gHg^{-1} = H$(I.e. $H \lhd G$)
I have a little confidence my proof and statement is true. But, Can't totally ensure myself all the things right. Are my statement and proof right?
Thanks.