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Given that $H\leq G$ of index $p$ in $G$, where $p$ is the smallest prime integer such that $p \mid |G|$, then $H \trianglelefteq G$. I would appreciate some hints, as I don't even know where to begin. We can use a group action of $G$ on the left cosets of $H$ by left multiplication, but I don't know how to move further, and, particularly, how to use the fact that $p$ is the smallest prime dividing the order of $G$.

Michael Hardy
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sequence
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    see http://math.stackexchange.com/questions/164244/normal-subgroup-of-prime-index – St Vincent Nov 27 '15 at 05:18
  • Thanks for the link, but I didn't get in the top answer how it follows that: "But it must also have order dividing $|G|$, and since $p$ is the smallest prime that divides $|G|$, it follows that $|G/K|=p$". – sequence Nov 27 '15 at 05:47
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    @sequence ... Lagrange's Theorem gives you $|K| |G/K| = |G|$. In the number $p!$, all primes $\leq p$ occur ("quite often"). But knowing that $|G / K|$ divides $p!$ means that $|G / K|$ is itself build up of primes $\leq p$, but by the above only $p$ is possible and so $|G / K| = p$. – M.U. Nov 27 '15 at 07:54
  • @M.U.: I still don't understand. If $|G/K| \rvert p!$ and $p$ is the smallest prime dividing $|G|$, why can't the order of $|G/K|$ be $p-11$, for example? Suppose that $p-11$ is another prime that divides $|G|$, and $|G/K|$ divides $p!$, and $p-11$ is another prime in the prime factorization of $p!$ which divides $|G|$. Why can't this be true? – sequence Dec 01 '15 at 02:52
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    @sequence.: $p - 11$ would be a prime smaller than $p$, but by assumption (which is essential as you now hopefully see) $p$ is the smallest prime dividing $|G|$. – M.U. Dec 02 '15 at 10:23

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