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I am struggling to understand this proof.

Suppose that $p$ is the smallest prime that divides $|G|$ show that any subgroup of index $p$ is normal in $G$.

Proof

Let $\phi \rightarrow S_p$ be the homomorphism given by the generalized Cayley theorem. We claim $ker(\phi)=H$ which implies $H$ is normal. By first isomorphism theorem $|G/Ker \phi|$ divides $|S_p|=p!$ON the other hand $|G/Ker \phi|$ divides $|G|$ by lagrange's theorem. Since $ker\phi \leq H$ and $p$ is the smallest prime factor of $|G|$, $ker\phi=H$. I am struggling to understand this if anyone could help explain this?

Scott Frazier
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1 Answers1

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You don't actually detail your difficulties, so here's a description based on a guess...

$|G / \ker \phi|$ divides $|S_p| = p!$

means the factorization of $|G / \ker \phi|$ uses only powers of primes in the list $1, 2, \dots, p$.

$|G / \ker \phi|$ divides $|G|$

means the factorization of $|G / \ker \phi|$ uses only power of primes listed in the factorization of $|G|$, the smallest of which is $p$.

The only prime on both lists is $p$.

Going back to the first quote, this says the index of $H$ in $G$ is a multiple of $p$. Going back to the second quote, this says the index of $H$ in $G$ is a divisor of $p$. So the index of $H$ in $G$ is $p$.

Eric Towers
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  • We're given that the index of $H$ is $p$. –  Dec 14 '20 at 07:45
  • @ChrisCuster : "You don't actually detail your difficulties, so here's a description based on a guess..." Interesting that the only thing you read was the last sentence. – Eric Towers Dec 14 '20 at 13:22
  • It looks like you maybe meant that the index of $\rm{ker}\phi$ is $p$. –  Dec 14 '20 at 17:55
  • @ChrisCuster : Maybe. But OP still hasn't given any hint about what he does or doesn't understand, so there is no coherent way to respond to the Question. – Eric Towers Dec 14 '20 at 17:59
  • I fully agree. We are left having to explain the whole proof. –  Dec 14 '20 at 18:00