# Relationship between "uniqueness of the subgroup and normal"

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.

• This is correct. Well done. $H$ is even characteristic, meaning fixed by any automorphism of $G$. Mar 7, 2020 at 9:29
• Your proof is 'correct' but the definition for the conjugation automorphism should be $\phi_g:G\to G$ where $x\mapsto g^{-1}xg$. Mar 7, 2020 at 9:31
• Huh, @Sebastian Cor, why would that be? The author only shows that conjugation leads to another subgroup of the same order. Mar 7, 2020 at 9:55
• I dont think defined that way it defines a morphism since the product of two subgroups need not be a subgroup. Mar 7, 2020 at 9:58

By contrapositive, suppose $$H$$ is not normal in $$G$$; then, $$\exists a\in G \mid Ha \ne aH$$; but then $$aHa^{-1}$$ is a subgroup of order $$n^{(*)}$$ distinct from $$H$$, contradiction. So, your statement is true.
$$^{(*)}$$Because the map $$H \to aHa^{-1}$$, defined by $$h \mapsto aha^{-1}$$, is bijective.