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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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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.

EDIT:

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
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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
    
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. If you don't do this, people are less likely to answer your later questions. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, How does accept rate work?. –  The Chaz 2.0 Oct 29 '12 at 13:31

$|G| = 2n$ 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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