# A question on $S_{6}$.

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.

• What is $n_5(G)$ ? – Amr Jan 6 '13 at 3:36
• $n_{5}(G) := |\text{Syl}_{5}(G)|$. – 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 $A_6$.
Well if $G\not\subset A_6,$ then $G\cap A_6$ would be a proper normal subgroup.
• @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. – Ted Jan 6 '13 at 3:35