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This question comes from M.A. Armstongs book Groups & Symmetry, Chapter 13, #5.

I came across a similar question here, but I am not familiar with the language yet and would rather solve using the tools I've amassed to this point in my learning.

The question tells us to use Cayley's and Cauchy's Theorem, as well as the fact that if a subgroup $H$ of $S_n$ is not contained in the alternating group $A_n$, then half the elements of $H$ are even.

At the moment a promising approach (to me) is to examine the two cases: $2n+1$ is prime or $2n+1$ is not prime. The former leads directly to the conclusion via Cauchy, but the more difficult one tells me nothing other than $2n+1$ is a product of primes.

Any suggestions or hints to pursue this course or another one is appreciated.

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Jacopo Stifani
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    To apply the hint from the question, consider the map to $S_{4n+2}$ given by Cayley's theorem. Now note that an element of order $2$ will not be sent to an even permutation by this map (show that it will be the product of an odd number of transpositions), and thus the image will not be contained in $A_{4n+2}$. Now show that the preimage of $A_{4n+2}$ is a subgroup of the desired order. – Tobias Kildetoft Mar 31 '16 at 08:52

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Lemma 1: If G is a subgroup of the symmetric group $S_{n}$ and G contains an odd permutation, then $|G: G \cap A_{n}| = 2$, that is, G contains a subgroup of index 2.

Proof: We have the following two maps: $i: G \mapsto S_{n}$ and $\pi: S_{n} \rightarrow S_{n}/A_{n}$. Let $\phi = \pi \circ i$. Then $\phi: G \rightarrow S_{n}/A_{n}$ with $ \ker \phi = G \cap A_{n}$ and since $|S_{n}: A_{n}| = 2$, we have $|G: G\cap A_{n}| = 2$ or $1$. But $|G: G\cap A_{n}| = 1$, implies that $G \subset A_{n}$, contradicting the assumption that G must contain an odd permutation.

Lemma 2: If G is a group of order $2m$, where m is odd and different from 1, then thinking of $G$ as a subgroup of $S_{2m}$, any element of order 2 corresponds to an odd permutation.

Proof: Let $g \in G$ have order 2, so that $g \neq 1$. Let $ \rho_{g}$ be the permutation corresponding to $g$ -[the idea comes from the proof of Cayley's Theorem]. Then for any $x \in G$, we have $\rho_{g}(x) \neq x$ for then the cancelation in the group will force $g = 1$. Hence there exists $y \in G$ such that $\rho_{g}(x) = y$ and as $g^{2} = 1$, $\rho_{g}(y) = x$ so that the cycle containing $x$ in $\rho_{g}$ is the transposition $(x,y)$ and so every cycle in the permutation is a transposition. Therefore $\rho_{g}$ is an element of $ S_{2m}$ whose cycle decomposition is a product of transpositions and no 1-cycles, and hence is a product of m transpositions and since m is odd, $\rho_{g}$ is odd.

Lemma 3: If g is a group of order $2m$, take $m = 2n +1$-an odd integer, then G contains a subgroup of index 2.

Proof: By Cayley's Theorem $G$ is a subgroup of $S_{2m}$. By Cauchy's Theorem, $G$ contains an element of order 2, which by Lemma 2 must be an odd permutation, and so by Lemma 1, G contains a subgroup of index 2 [what does it look like?]

Daniel Akech Thiong
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  • Thank you for the detailed response. Is there a way to formulate your first Lemma 1 without reference to indices? In Lemma 2 I understand the argument until the last sentence. Why must $\rho_g$ be a product of $m$ transpositions? I plan to mesh your ideas with @Tobias Kildetoft's comments. – Jacopo Stifani Apr 01 '16 at 08:32
  • This is really a question of whether a group of order 2m is simple or not for m odd. To show a group is not simple you search for nontrivial normal subgroups. Every subgroup of index p where p is the smallest prime dividing the order of the group is necessarily normal with p=2 being a popular case. Of course a normal subgroup may not be of index 2. So you can restate lemma 1 as any subgroup of the symmetric group that contains an odd permutation must be normal. – Daniel Akech Thiong Apr 01 '16 at 11:06
  • On Lemma 2, every permutation is a product of transpositions. The sign of a transposition is $-1$, making it odd. The product of transpositions is either odd or even depending on the number transpositions in the decomposition. I recommend you review this arithmetic of the permutation group. I believe that the finite group theory is really the study of $S_{n}$. – Daniel Akech Thiong Apr 01 '16 at 11:19