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What we do in this proof is :

We say let $G$ acting on the left cosets of $H$ in $G$ where $H$ is a subgroup of $G$ , with $ | G : H | = p$

We consider $K$ as the kernel of the action such that $ | H : K | = k$,

Then we make a statement that $ G/K $ is isomorphic to a "subgroup" of $S_p$ ,

I know the first theorem of Isomorphism is applied here , but I am a little confused ,

According to the first isomorphism theorem , $ G/K $ should be isomorphic to $S_p$ and hence , $ |G/K| = |S_p|$ , i.e $ |G/K| = p!$ ..

From where does that "subgroup" term comes from ?

I know , only that "subgroup" helps in completing the proof as , that implies $|G/K|$ divides $p!$ , but the actual statement of the theorem confuses me , could anyone help me with this?

User9523
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    $G$ acts on $G/H$, which means there is a group homomorphism $G\to S_p$. Then if $K$ is the kernel of this homomorphism, $G/K$ is isomorphic to its image, which is a subgroup of $S_p$ but is not necessarily the whole $S_p$. – Arnaud D. Nov 16 '15 at 12:43
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    You should read the First Isomorphism Theorem more carefully. If $f:G \to H$ is a homomorphism, then $$G /\ker f \cong Im(f) \le H.$$ In a special case when $f$ is ONTO we have $G/\ker f \cong H$. – Amin Nov 16 '15 at 22:49
  • @User9523: https://math.stackexchange.com/questions/164244 – Watson Jun 17 '16 at 09:46

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There is a much simpler way to prove this:

Proposition Let $p$ be the smallest prime dividing the order of the finite group $G$, and assume that the subgroup $H$ has index $p$. Then $H \lhd G$.

Lemma Let $H \leq G$ with $G=HH^g$ for some $g \in G$. Then $G=H$.

Proof Apparently $g=hk^g$, for some $h,k \in H$. Hence $g=hg^{-1}kg$, from which one derives that $g=kh \in H$. Hence $H^g=H$, so $G=HH^g=HH=H$.

Let us proceed with a proof of the Proposition: assume that $H$ is not normal, so there exists a $g \in G$ with $H^g \ne H$. Since $H \ne G$, we have (using the Lemma) $|G| \gt |HH^g|$. But $|HH^g|=\frac{|H| \cdot |H^g|}{|H \cap H^g|}$. This shows that $|G:H|=p \gt |H^g:H \cap H^g|$. We conclude that $|H^g:H \cap H^g|=1$, as $p$ is the smallest prime dividing $|G|$. This means that $H^g=H \cap H^g$, so $H^g \subseteq H$, implying $H=H^g$, since $H^g$ and $H$ have the same order. We arrive at a contradiction.

Nicky Hekster
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