|G|=2n. Half the elemets are of order 2

[abhishekkgp]

Let \(\displaystyle G\) be a group of order \(\displaystyle 2n\).Suppose that half the elements of \(\displaystyle G\) are of order \(\displaystyle 2\), the other half form a subgroup of order \(\displaystyle n\). Prove that \(\displaystyle H\) is of odd order and is an abelian subgroup of \(\displaystyle G\).

ATTEMPT: since index of \(\displaystyle H\) in \(\displaystyle G\) = \(\displaystyle 2\)= smallest prime dividing \(\displaystyle o(G)\), we have \(\displaystyle H\) is normal subgroup of \(\displaystyle G\). Another observation i was able to make was:
Take \(\displaystyle g \not \in G\). Thus \(\displaystyle g^2=e\). Then \(\displaystyle H, \, gH\) partition \(\displaystyle G\). Now \(\displaystyle gh \not \in G \, \forall h \in H\). Hence by hypothesis \(\displaystyle (gh)^2=e \Rightarrow ghg=h^{-1} \Rightarrow ghg^{-1}=h^{-1}\).
This equivalent to saying that the automorphism \(\displaystyle T_g:G \rightarrow G\) defined by \(\displaystyle T_g(x)=gxg^{-1}\) sends each element of \(\displaystyle H\) to its inverse.
If i could show that \(\displaystyle T_g\) carries each element of \(\displaystyle G\) to its inverse then i would be able to show that \(\displaystyle G\) is abelian.
I have no idea how to show that \(\displaystyle o(H)\) is odd.

Please help.

 

[abhishekkgp]

Let \(\displaystyle G\) be a group of order \(\displaystyle 2n\).Suppose that half the elements of \(\displaystyle G\) are of order \(\displaystyle 2\), the other half form a subgroup of order \(\displaystyle n\). Prove that \(\displaystyle H\) is of odd order and is an abelian subgroup of \(\displaystyle G\).

ATTEMPT: since index of \(\displaystyle H\) in \(\displaystyle G\) = \(\displaystyle 2\)= smallest prime dividing \(\displaystyle o(G)\), we have \(\displaystyle H\) is normal subgroup of \(\displaystyle G\). Another observation i was able to make was:
Take \(\displaystyle g \not \in G\). Thus \(\displaystyle g^2=e\). Then \(\displaystyle H, \, gH\) partition \(\displaystyle G\). Now \(\displaystyle gh \not \in G \, \forall h \in H\). Hence by hypothesis \(\displaystyle (gh)^2=e \Rightarrow ghg=h^{-1} \Rightarrow ghg^{-1}=h^{-1}\).
This equivalent to saying that the automorphism \(\displaystyle T_g:G \rightarrow G\) defined by \(\displaystyle T_g(x)=gxg^{-1}\) sends each element of \(\displaystyle H\) to its inverse.
If i could show that \(\displaystyle T_g\) carries each element of \(\displaystyle G\) to its inverse then i would be able to show that \(\displaystyle G\) is abelian.
I have no idea how to show that \(\displaystyle o(H)\) is odd.

Please help.

don't bother i got it.