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 be a finite group of order . Suppose that half the elements of are of order 2 and the other half form a subgroup of order . Then prove that is of odd order and is an abelian subgroup of .
Soln: Suppose that is of even order. Then there exists an element such that , which implies that should belong to (where $S$ is the set containing elements of even order). So is of odd order.
Now let and then (because if ,then which is a contradiction to the fact that ). Now since . Now since is of order 2 and this will imply .
Now since $H$ is a normal subgroup of therefore the mapping $\phi$ defined by by is an automorphism(verify!).
Consider any element from the set S and ,then and which implies ;where . Consider the mapping ie is that same as since $H$ is normal(therefore $ghg^{-1} \in H$ for all and ). But is an automorphism iff $H$ is abelian. Hence H is of odd order and is an Abelian Subgroup of G.
The case is trivial, so we will assume that
Since has index 2, it is normal in Therefore conjugation of the elements of is an automorphism of As the mapping is a group automorphism if and only if the group is Abelian, it follows that is Abelian.
Now, has odd order, so if and it must follow that (otherwise 2 would divide by Lagrange). As has index 2, it is easily proved that (Proof: The element in the factor group has order 2.) But then Hence and This shows that for all