Show group of order $4n + 2$ has a subgroup of index 2. Let $n$ be a positive integer. Show that any group of order $4n + 2$ has a subgroup of index 2. (Hint: Use left regular representation and Cauchy's Theorem to get an odd permutation.)
I can easily observe that $\vert G \vert = 2(2n + 1)$ so $2 \mid \vert G \vert$ and 2 is prime. We have satisfied the hypothesis of Cauchy's Theorem so we can say $G$ contains an element of order 2.
This is where I am stuck
I am confused about how the left regular representation relates to the group. So my understanding at this point is that every group is isomorphic to a subgroup of some symmetric group. My question: is the left regular representation $\varphi : G \to S_G$ an isomorphism? where $G \cong S_G$ or is $S_G$ the same thing as $S_{\vert G \vert}$ and $\varphi$ is only an injection? I'm using Dummit and Foote for definitions.
I saw an argument online that said that since we have an element of order 2, there is a $\sigma \in S_G$ of order 2, but it is a product of $2n + 1$ disjoint 2-cycles. I don't understand how they could claim this and tried working it out on my own but didn't get there. -- They then went on to use the parity mapping $\varepsilon : S_G \to \{\pm1\}$ and since we have an odd permutation $\sigma$, we have $[S_G:\text{ker }\varepsilon] = 2$. I understood their computation but not how that directly shows that $G$ has a subgroup of order 2? unless $G \cong S_G$ because of how left regular representation is defined. (but again, I'm not understanding that concept very well yet.)
So, to be clear about my questions:


*

*What is meant by left regular representation, is it an isomorphism or just an injection? and how would it be used here.

*If it is an isomorphism, the argument online starts to make more sense, but how can they say that since $\sigma$ is even, it is made up of $2n + 1$ disjoint transpositions?

*If you have a full, proof, I'd appreciate it, but good hints are just as good!


Thanks you!
 A: Consider the left regular representation of $G$ -- $\varphi : G \to S_G$. So by Cayley's Theorem we know $\text{Im }\varphi \cong G$.  Now, observe that $\vert G \vert = 2(2n + 1)$ so $2 \mid \vert G \vert$ and 2 is prime. We have satisfied the hypothesis of Cauchy's Theorem so we can say $G$ contains an element of order 2, call it $g$. Now, by property of isomorphism, we can say there exists $\sigma \in \text{Im }(\varphi)$ of order 2. Moreover $\sigma$ is a product of $2n + 1$ disjoint 2-cycles, an odd permutation, by property of $\sigma_g$ having no fixed points, since $\sigma \neq 1$. Now consider the sign mapping $\varepsilon : \text{Im } (\varphi) \to \{\pm 1\}$. This mapping is surjective sice $\sigma \in \text{Im }(\varphi)$ and $\varphi$ is odd. Now, by the Fundamental Theorem of Homomorphisms $\text{Im }(\varphi) /\text{ker }\varepsilon \cong \{\pm 1\} \cong \mathbb Z_2$. It follows immediately that $[\text{Im }(\varphi) : \text{ker }\varepsilon] = 2$ or equivalently $[G: \text{ker } \varepsilon] = 2$. Thus we can conclude that $G$ has a subgroup of index 2.
This argument can be generalized for any group of order $2k$ with $k$ odd.
A: Let $G=\{g_1,\ldots,g_{4n+2}\}$. For any $g\in G$, define $\sigma_g\in S_{|G|}$ as the permutation for which $\sigma_g(g_i) = g\cdot g_i$.
Step1. The set of elements $g\in G$ such that $\operatorname{sign}(\sigma_g)=+1$ is a subgroup of $G$, say $H$.
Step2. The only possibilities for $|H|$ are $|H|=|G|$ or $|H|=|G|/2$, since the "sign" is a homomorphism.
Step3. By Cauchy's theorem there is a $g\in G$ with order $2$ (that is $g=g^{-1}$, so $\sigma_{g}=\sigma_{g}^{-1}$). Since $g\neq id$, $\sigma_g$ has no fixed points, but $\sigma_g$ has order $2$ in $S_{|G|}$, so it is a product of $(2n+1)$ disjoint transpositions. It follows that $\operatorname{sign}(\sigma_g)=-1$ and we cannot have $|H|=|G|$.
A: Here is a more general theorem.
Theorem: Let $G$ be a group of order $ 2^n m$ (where $m,n \in \mathbb{N}$ ). If $2 \nmid m$ and $G$ has an element of order $2^n$, then there exists a normal subgroup of $G$ of order $m$. 
Proof: We start with the following facts.
Fact 1: If $G$ is a finite group and $\pi : G \to S_G$ be permutation representation of the group action such that $\pi_g (h) = gh$ for $g,h \in G$. If $t$ is an element of even order and $|G|/ \text{ord} (t)$ is odd then $ (\pi_ t)$ is an odd permutation.
Sketch of proof: Observe that $(\pi_t)$ is a product of $|G|/\text{ord}(t)$ number of $\text{ord}(t)$-cycles.So $\text{sgn}(\pi_t) = -1$.
Fact 2: If there exists such an element $t$ then $G$ has a subgroup of index $2$.
Sketch of proof: We use a result. Let $G$ is a finite group and $H$ is a subgroup of $G$ of index $p$ , where $p$ is a prime number. If $K \le G$, then either $K \le H$ or $[K : K \cap H] = p $. Since $G \cong \pi(G)$, to prove the above fact it suffices to show that $\pi(G)$ has a subgroup of index $2$. In the mentioned result let $H = A_G$ ($A_G$ is the subgroup of even permutations) and $K = \pi(G)$. Since $\pi(G)$ contains an odd permutation we can not have $\pi(G) \le A_G$. So we get $[\pi(G) : \pi(G) \cap A_G] = 2$. So $\pi(G) \cap A_G$ is our required subgroup of $\pi(G)$ of index $2$ and the fact is proved.
Now we are ready to prove the theorem. We proceed by induction. When $n = 1$, Cauchy's theorem gives us an element of order $2$, so we get a normal subgroup of order $m$ (index $2$) by the preceding results. Assume it to be true for $n = k-1$, we will show it is true for $n=k$. Let $t$ be the element of order $2^k$. Then by the above results it follows that there is a subgroup $H$ of $G$ of index $2$, i.e of order $ 2^{k-1} m $. By induction hypothesis there is a normal subgroup $J$ of $H$ which is of order $m$. We claim that $J$ is normal in $G$. 
Since $[G:H] = 2$, $H$ is normal in $G$. We will use a result which is easy to prove. If $ J\le H \le G$, $H$ is normal in $G$ and $J$ is a characteristic subgroup of $H$, then $J$ is normal in $G$. So now it suffices to show that $J$ is a characteristic subgroup of $G$. To show that it is sufficient to show that $J$ is the only subgroup of $H$ of order $m$. If not let $P$ be another subgroup of order $m$ of $H$. Note that we have $PJ= JP$ so $PJ \le H$. We have $|PJ| = \frac{|P||J|}{|P\cap J|} = \frac{m^2}{|P \cap J|}$. So $|PJ|$ is odd. If $|P \cap J| < m$, then $|PJ| >m$ and hence it can not divide $|H|$, contradicting Lagrange's theorem. So we must have $|P \cap J| = m$, implying $P= J$. Now the theorem is completely proved.
