Every group of order $4n+2$ has a subgroup of index $2$.
Here is what I have.
Let $G$ acts on itself with left shifts:
$$(\forall g\in G)(\forall x\in G) \quad g(x) = gx, \quad \phi: G \rightarrow S(G).$$
Then $\ker\phi = \{{\rm id}\}, \phi(G) \cong G \cong H \leq S_n.$
Let $\forall h \in H \quad\psi(h) = {\rm sign}(h), \phi(H) \cong \mathbb{Z}_2 \cong H/\ker\psi \implies |H/\ker\psi| = |H : \ker\psi| = 2$
And for isomorphism $\theta : G \rightarrow H \quad K = \theta^{-1}(\ker\psi) \leq G, |G : K| = 2$.
But I never used the fact that order $G$ is $4n+2$. What's wrong? Thanks.
Best Answer
You write $\phi(H)\cong \mathbb{Z}_2$ whereas I think you should write $\psi(H)\cong \mathbb{Z}_2$.
But why $\psi(H)\cong \mathbb{Z}_2$? Why not $\psi(H)\cong 1$ i.e. $\ker(\psi)=H$ ?
Use the fact that $2\mid |G|$. So $G$ contains an element of order $2$ which is a product of $2n+1$ transpositions in $H$. Hence has sign $-1$ ....