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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.

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

1 Answer 1

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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.

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