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This proof is from Dummit & Foote text.

Suppose by way of contradiction that $\vert G \vert=60$ and $n_5$>1 but that there exists $H$ a normal subgroup of $G$ with $H$ $\neq$ $1$ or $G$. By Sylow's theorem the only possibility for $n_5$ is $6$. Let $P \in Syl_5(G)$,so that $N_G(P)$=$10$ since its index is $n_5$.

If $5 $ $\vert $ $\vert H \vert $ then $H $ contains a Sylow 5-subgroup of $G$ and since $H$ is normal, it contains all $6$ conjugates of this subgroup. $\color{red}{(*)}$** In particular, $\vert H \vert \ge 1+6\cdot4=25$, and the only possibility is $H=30$**. This leads to a contradiction since any group of order $30$ has a normal (hence unique) Sylow 5-subgroup. This argument shows that $5$ does not divide $\vert H \vert$ for any proper normal subgroup $H$ of $G$.

If $\vert H \vert =6$ or $12$, $H$ has a normal, hence characteristic, Sylow subgroup which is therefore also normal in $G$. Replacing $H$ by this subgroup if necessary, we may assume $\vert H \vert=2,3$ or $4$. Let $G=G/H$, so $\vert \bar G \vert $=$30,20$ or $15$. In each case, $\bar G$ has a normal subgroup $\bar P$ of order $5$. $\color{red}{(**)}$** If we let $H_1$ be the complete preimage of $\bar P$ in $G$, then $H_1$ is normal in $G$,$H_1 \neq G$ and $5 $ $\vert $ $\vert H \vert $**. This contradicts the preceding paragraph and hence the proof.

NOTIONS WHICH I AM NOT GETTING

$\color{red}{(*)}$ I'm not getting from where does $\vert H \vert \ge 1+6.4=25$ comes (why he is manipulating so much, if it can be simply be done as). Since $H $ contains $6$ conjugates of $P$ & $\vert P \vert =5$, this implies $\vert H \vert=6\cdot5=30$

$\color{red}{(**)}$ If $H_1$ is the complete preimage of $\bar P$ in $G$, then how $H_1$ is normal in $G$, $H_1 \neq G$ and $5 $ $\vert $ $\vert H_1 \vert $.

Help me in understanding these points.

thank you!!

Styles
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1 Answers1

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(*) All six subgroups contain $1$ (and must intersect only in $1$, since they're cyclic of prime order).

(**) The preimage of the normal group $P$ under the quotient $\pi$ is normal, since $\pi(h^{-1}gh) = \pi(h)^{-1} \pi(g) \pi(h)\in P$ for any $g$ with $\pi(g)\in P$.

anomaly
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