Group of order $2m$ where $m$ odd has a subgroup of index 2. Show that a group $G$ of order $2m$, where $m$ is odd, has a subgroup of index $2$.
I am feeling a little dubious about my proof. 
Let $G$ act on itself by left multiplication to induce the homomorphism $\varphi: G \longrightarrow S_{2m}$ defined by $g \mapsto \sigma_{g}$ where $\sigma_{g}: S_{2m} \longrightarrow S_{2m}$ is defined by $\sigma_{g}(x)=gx$ for all $x \in G$. Let $\psi: S_{2m}\longrightarrow \{\pm1\}$ be the sign map homomorphism. Then $\Phi = \psi \circ \varphi$ is a homomorphism from $G \longrightarrow \{\pm1\}$. 
If we can show $\Phi$ is surjective then $G/\ker\Phi \approx \{\pm1\} \implies |G:\ker\Phi|=2$. 
By Cauchy's theorem there exists an element $g$ of order $2$ in $G$.

(This is the part I'm questioning) We have $\sigma_{g}$ corresponds to a product of transpositions in $S_{2m}$, namely $(x_{1},gx_{1})(x_{2},gx_{2})\cdots(x_{m},gx_{m})$. Therefore $\psi(\sigma_{g})=-1$ because $\sigma_{g}$ is a product of an odd number of transpositions.

Therefore (provided highlighted part is correct) $\Phi$ is surjective and the kernel is the desired subgroup.
Does this work? 
Thanks.
 A: Since we know that every group of order $n$ is isomorphic to a subgroup of $S_n$.
So suppose $\phi:G\to S_{2m}$ is that homomorphism. 
Now since $|G|=2m~(even)$ hence $\exists$ an element $a$ whoes order is $2$.
Define $a_l:G\to G$ as $$a_l (x)=ax$$then clearly $a_l$ is a bijection so  $a_l\in sym(G)=S_{2m}$ 
Notice that $a_l$ gives a transposition with two symbols. since if $a_l(x)=y$ then $a_l(y)=a_l(ax)=a^2(x)=x$. So we get a transposition $ (x~y)$. In other words $a_l$ is product of $m$ transpositions but $m$ is odd so $a_l$ is an odd permutation.
Also we know that alternating map $\psi:S_{2m}\to \{1,-1\}$ is a group homomorphism. So composition map $\psi\circ\phi:G\to\{1,-1\}$ is also group homomorphism. Notice that this group homomorphism is also surjective since  $\psi\circ\phi(a)=\psi(\phi
(a))=\psi(a_l)=-1$ and  $\psi\circ\phi(e)=1$. And from fundamental theorem of group homomorphism $$G/ker(\psi\circ\phi)\approx \{1,-1\}$$
And hence it shows that $G$ has a subgroup of index 2
