Let G be a group of order 2n and H a subgroup of G of order n. where n is and odd number. Moreover for all g in G-H and for all h in H we have ghg^-1 = h^-1.
Now Show that H is abelian and every element of G-H has order 2.
Let G be a group of order 2n and H a subgroup of G of order n. where n is and odd number. Moreover for all g in G-H and for all h in H we have ghg^-1 = h^-1.
Now Show that H is abelian and every element of G-H has order 2.
Hi this is just same problem which herstein topics in algebra has in the section of automorphism. I restate that the problem
Let $\displaystyle G$ be a finite group of order $\displaystyle 2n$. Suppose that half the elements of $\displaystyle G$ are of order 2 and the other half form a subgroup of order $\displaystyle n$. Then prove that $\displaystyle H$ is of odd order and is an abelian subgroup of $\displaystyle G$.
Soln: Suppose that $\displaystyle H$ is of even order. Then there exists an element $\displaystyle a \in H$ such that $\displaystyle a^{2}=e$, which implies that $\displaystyle a$ should belong to $\displaystyle S$(where $S$ is the set containing elements of even order). So $\displaystyle H$ is of odd order.
Now let $\displaystyle a \in S$ and $\displaystyle b \in H$ then $\displaystyle ab \in S$(because if $\displaystyle ab \in H$,then $\displaystyle a\in H$ which is a contradiction to the fact that $\displaystyle a\notin H$). Now $\displaystyle ab=a^{-1}b$ since $\displaystyle a \in S$ . Now $\displaystyle (ab)^{2}=e$ since $\displaystyle S$ is of order 2 and this will imply $\displaystyle (ab)=(ab)^{-1}=b^{-1}a^{-1}$.
Now since $H$ is a normal subgroup of $\displaystyle G$ therefore the mapping $\phi$ defined by $\displaystyle \phi:H \rightarrow H$ by $\displaystyle \phi(h)=ghg^{-1}$ is an automorphism(verify!).
Consider any element $\displaystyle k$ from the set S and $\displaystyle h \in H$,then $\displaystyle k^{2}=e$ and $\displaystyle (kh)^{2}=e$ which implies $\displaystyle khk=h^{-1}$;where $\displaystyle h \in H$. Consider the mapping $\displaystyle h\mapsto khk$ ie $\displaystyle h \mapsto k^{-1}.h.k$ is that same as $\displaystyle h \mapsto h^{-1}$ since $H$ is normal(therefore $ghg^{-1} \in H$ for all $\displaystyle g \in G$ and $\displaystyle h \in H$). But $\displaystyle h \mapsto h^{-1}$ is an automorphism iff $H$ is abelian. Hence H is of odd order and is an Abelian Subgroup of G.
The case $\displaystyle |G|=2$ is trivial, so we will assume that $\displaystyle n>1.$
Since $\displaystyle H$ has index 2, it is normal in $\displaystyle G,$ Therefore conjugation of the elements of $\displaystyle H$ is an automorphism of $\displaystyle H.$ As the mapping $\displaystyle x\mapsto x^{-1}$ is a group automorphism if and only if the group is Abelian, it follows that $\displaystyle H$ is Abelian.
Now, $\displaystyle H$ has odd order, so if $\displaystyle h\in H$ and $\displaystyle h^2=e,$ it must follow that $\displaystyle h=e$ (otherwise 2 would divide $\displaystyle |H|$ by Lagrange). As $\displaystyle H$ has index 2, it is easily proved that $\displaystyle g\notin H\ \implies\ g^2\in H.$ (Proof: The element $\displaystyle gH$ in the factor group $\displaystyle G/H$ has order 2.) But then $\displaystyle g^2=gg^2g^{-1}=g^{-2}\ \implies\ g^4=e.$ Hence $\displaystyle (g^2)^2=e$ and $\displaystyle g^2\in H.$ This shows that $\displaystyle g^2=e$ for all $\displaystyle g\in G-H.$