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!
Best Answer
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 $|H|=|G|$.