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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.

darij grinberg
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Tuo
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  • Is $g$ this element of order $2$ ? In this case, yes, although you might want to be more precise here. (How do you define $x_1, x_2, ..., x_m$ ?) – darij grinberg Jul 03 '15 at 01:15
  • @ darij Yes g is the element of order 2. Be more precise how ? – Tuo Jul 03 '15 at 01:17
  • As I said, what are $x_1, x_2, ..., x_m$, and why is $\sigma_g$ equal to that product? The definition of $x_1, x_2, ..., x_m$ might involve a choice, but you still need to state what you are choosing from. – darij grinberg Jul 03 '15 at 01:19
  • @ darij the $x_{i}$'s are representatives of their (distinct) orbits. My thought was that each orbit will have two elements since $g$ has order 2 and so it will only permute two elements so that product of transpositions is what it will look like as a permutation. – Tuo Jul 03 '15 at 01:22
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    Very good. Probably it is best if you also explain why each orbit has exactly two elements. And, as I said -- why is $\sigma_g$ that product? The rest of the proof is perfectly fine. – darij grinberg Jul 03 '15 at 01:24
  • Alright thanks happy to know it's correct. – Tuo Jul 03 '15 at 02:01
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    I haven't tried, but you could probably prove this by induction. By Cauchy's theorem there exists an element of order $2$, hence a subgroup of order $2$, say $H$. Take the quotient $G/H$ which has order $2^{m-1}$ and then use the inverse map $\pi : G/H \rightarrow G$. – GiantTortoise1729 Jul 03 '15 at 02:27
  • @ GiantTortise I think that you could probably do it that way also. – Tuo Jul 03 '15 at 16:47

1 Answers1

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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

absolute friend
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