This is a part of a proof that there is no simple group of order $120 = 2^{3} \cdot 3 \cdot 5$.

Suppose there is such $G$. Then checking $n_{5}(G) = 6$ and considering action on right cosets of $N_{G}(P)$ with $P \in \text{Syl}_{5}(G)$, we can inject this group $G \leqslant S_{6}$.

How do we show $G \leqslant A_{6}$? And in general, what assumptions are used when a subgroup of $S_{n}$ is in $A_{n}$? For the second question any examples will be appreciated whether they are trivial or not.

  • $\begingroup$ What is $n_5(G)$ ? $\endgroup$ – Amr Jan 6 '13 at 3:36
  • $\begingroup$ $n_{5}(G) := |\text{Syl}_{5}(G)|$. $\endgroup$ – user123454321 Jan 6 '13 at 3:38

Consider the map $$G\hookrightarrow S_6\rightarrow \lbrace -1,+1\rbrace$$ where the last morphism is the signature morphism. By simplicity of $G$, this has to be trivial, so the image of $G$ under te embedding in $S_6$ lies in the kernel of the signature, which happens to be the alternating group $A_6$.

It's the same idea as in this answer.


Well if $G\not\subset A_6,$ then $G\cap A_6$ would be a proper normal subgroup.

  • 1
    $\begingroup$ Could the intersection be just the identity? $\endgroup$ – gnometorule Jan 6 '13 at 3:29
  • $\begingroup$ @gnometorule It can't; if it were then any non-identity $x \in G$ would satisfy $x^2=1$, so the order $G$ would be a power of 2. $\endgroup$ – Ted Jan 6 '13 at 3:35
  • $\begingroup$ Not a power of 2. Order exactly 2 (or 1). $\endgroup$ – jspecter Jan 6 '13 at 3:45

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