# Thread: Abelilan groups and subgroups

1. ## Abelilan groups and subgroups

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.

2. Originally Posted by Juancd08
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 (everything in G but not in H).
Further 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.
the sentence in red has no meaning because it's not complete. maybe this is what you meant:

Moreover for all g in G-H and all h in H we have ghg^-1 = h^-1.

3. ## Answer to ur problem

Hi this is just same problem which herstein topics in algebra has in the section of automorphism. I restate that the problem

Let $G$ be a finite group of order $2n$. Suppose that half the elements of $G$ are of order 2 and the other half form a subgroup of order $n$. Then prove that $H$ is of odd order and is an abelian subgroup of $G$.

Soln: Suppose that $H$ is of even order. Then there exists an element $a \in H$ such that $a^{2}=e$, which implies that $a$ should belong to $S$(where $S$ is the set containing elements of even order). So $H$ is of odd order.

Now let $a \in S$ and $b \in H$ then $ab \in S$(because if $ab \in H$,then $a\in H$ which is a contradiction to the fact that $a\notin H$). Now $ab=a^{-1}b$ since $a \in S$ . Now $(ab)^{2}=e$ since $S$ is of order 2 and this will imply $(ab)=(ab)^{-1}=b^{-1}a^{-1}$.

Now since $H$ is a normal subgroup of $G$ therefore the mapping $\phi$ defined by $\phi:H \rightarrow H$ by $\phi(h)=ghg^{-1}$ is an automorphism(verify!).

Consider any element $k$ from the set S and $h \in H$,then $k^{2}=e$ and $(kh)^{2}=e$ which implies $khk=h^{-1}$;where $h \in H$. Consider the mapping $h\mapsto khk$ ie $h \mapsto k^{-1}.h.k$ is that same as $h \mapsto h^{-1}$ since $H$ is normal(therefore $ghg^{-1} \in H$ for all $g \in G$ and $h \in H$). But $h \mapsto h^{-1}$ is an automorphism iff $H$ is abelian. Hence H is of odd order and is an Abelian Subgroup of G.

4. Originally Posted by Juancd08
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.
The case $|G|=2$ is trivial, so we will assume that $n>1.$

Since $H$ has index 2, it is normal in $G,$ Therefore conjugation of the elements of $H$ is an automorphism of $H.$ As the mapping $x\mapsto x^{-1}$ is a group automorphism if and only if the group is Abelian, it follows that $H$ is Abelian.

Now, $H$ has odd order, so if $h\in H$ and $h^2=e,$ it must follow that $h=e$ (otherwise 2 would divide $|H|$ by Lagrange). As $H$ has index 2, it is easily proved that $g\notin H\ \implies\ g^2\in H.$ (Proof: The element $gH$ in the factor group $G/H$ has order 2.) But then $g^2=gg^2g^{-1}=g^{-2}\ \implies\ g^4=e.$ Hence $(g^2)^2=e$ and $g^2\in H.$ This shows that $g^2=e$ for all $g\in G-H.$

5. I’d just like to say that the proof I originally gave was incorrect. I’ve edited my post, so hopefully my proof is correct now.