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Let G be a group has order of $2n$ with $n$ be odd. Prove that exists a unique subgroup $H$ of $G$ has order of $n$.

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The explicit definition of $H$ would be $H=\{g\in G\mid g^n=1\}$. But proving that is closed under multiplication is not obvious to me. Basically, $H$ is the elements of odd order in $G$. –  Thomas Andrews Oct 5 '12 at 13:45
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2 Answers 2

up vote 3 down vote accepted

The group $G$ acts on itself by left multiplication. So $G$ is isomorphic to a subgroup of $S_{2n}$. Now consider the subgroup $A_{2n}$ of index $2$. Now just prove that there is an element in group not lying in $A_{2n}$ and we are done. It can be done by using the fact that $G$ has an element of order $2$ and $n$ is odd, hence all elements of order $2$ in $S_{2n}$ are odd.


Alternatively, first we prove the following Lemma:

Lemma: If group $G$ acts on a finite set $S$, and if there exists an element in $G$ which induces an odd permutation of $S$, then there exists a subgroup $H$ of $G$ such that $[G:H]=2$.

Proof: Let $|S|=n$. Then action gives a representation $\phi :G \to S_n$. Let $\omega :S_n \to \{\pm 1\}$ be the parity homomorphism. The homomorphism $\omega \circ \phi :G \to \{ \pm 1\}$ is onto. Let $H$ be its kernel. Then using Isomorphism theorem, we are done. //

Now for our original question, we use the Lemma with $G$ acting on itself by left multiplication. We want an element inducing odd permutation. Let $a$ be element of order $2$. (It exists by Cauchy's theorem.)

Kerenel of this action is trivial. So $G$ is isomorphic to a subgroup of $S_{2n}$.

Now as $a$ has order $2$, we have if $a=\sigma_1 \cdots \sigma_m$ where $\sigma_i$'s are nontrivial disjoit cycles of $S_{2n}$, LCM of length of these cycles is $2$, i.e.: all of them are transpositions. Also $a$ fixes no elements (check yourself), $\sigma_i$'s must involve all $2n$ elements, i.e.: $m=n$. As $n$ is odd, $a$ induces odd permutation.

Now use the Lemma.

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I don't know what $H$ is? –  Firmino Oct 5 '12 at 13:26
I also don't know explicitly what $H$ is,but we can see that $A_{2n}$ corresponds to a subgroup of original group. also,there is an alternative proof if you are not very comfortable with $A_{2n}$ that I can provide for the fact $G$ has a subgroup of index $2$. –  TheJoker Oct 5 '12 at 13:57
Yes, I want you show $G$ has a subgroup of index 2, and this subgroup be uniqueness –  Firmino Oct 5 '12 at 14:03

$|G| = 2n$ where n is odd. Then any subgroup of $N$ of order $n$ must be normal in $G$, since $|G/N|=2$ and $2$ is the smallest prime that divides the order of $G$. Now suppose that there be two subgroups $H$ and $N$ of order $n$ in $G$. Then we can get a homomorphism $f: H \rightarrow G/N$ where $f = g\circ i$, $i$ is the inclusion map from $H$ to $G$, and $g:G \longrightarrow G/N$. Then $\text{Ker}f=H \cap N$.

$H/(H \cap N)$ is isomorphic to $\text{Image}(f)$, which is a subgroup of $G/N$. Thus $|H/(H \cap N)|$ divides $|G/N|$ and obviously $|H \cap N|$ divides $|H| =|N|$.

In this question, $\text{gcd}(|N|,|G/N|)=1$, whence $|H/(H \cap N)|=1$. Thus we get $H = H \cap N$. Therefore $H \leq N$ and $|H|=|N|$ so $H=N$, whence $H$ is the unique subgroup of order $n$.

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