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:

  1. What is meant by left regular representation, is it an isomorphism or just an injection? and how would it be used here.
  2. 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?
  3. If you have a full, proof, I'd appreciate it, but good hints are just as good!

Thanks you!


3 Answers 3


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 $|H|=|G|$.


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.


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.

  • $\begingroup$ Note that $\phi$ is not usually an isomorphism onto $S_G$, but it is an isomorphism onto some subgroup of $S_G$ (which is good enough for your argument to go through). $\endgroup$ Oct 31, 2012 at 22:50
  • $\begingroup$ Yes, that one's ok, it's just that $\operatorname{Im}\phi\subsetneq S_G$ usually. It just wasn't clear to me if you were thinking "property of isomorphism" referred to $G\cong S_G$ (which is usually false) or $G\cong \operatorname{Im}\phi$ (which is true in this case). $\endgroup$ Oct 31, 2012 at 23:25
  • $\begingroup$ Ahhh ok! Thanks, I did not understand that difference yesterday. My professor made it clear today. Thanks for the clarification! $\endgroup$ Oct 31, 2012 at 23:49

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