Thread: abelian normal group

Let be a group of order . Prove that if is a normal subgroup of order in then , where represents the center of .
(Please do not use sylow's or cauchy's theorems.)

Originally Posted by abhishekkgp

Let be a group of order . Prove that if is a normal subgroup of order in then , where represents the center of .
(Please do not use sylow's or cauchy's theorems.)

If you don't want Sylow and Cauchy theorems, then I suppose you want even less semidirect

products and stuff, thus the question is: what do you want? What stuff is allowed or wanted by you?

Tonio

Is there a nice way to do this if the Sylow theorems are allowed? I can't see anything short of a group classification argument.

Originally Posted by roninpro

Is there a nice way to do this if the Sylow theorems are allowed? I can't see anything short of a group classification argument.

Indeed, I feel as though the simplest way to do this is to note that it is of the form with prime and and thus abelian...that said the only ways I know how to prove this are Sylow's theorems or rep. theory.

Just noting the arithmetic: .

Originally Posted by abhishekkgp

Let be a group of order . Prove that if is a normal subgroup of order in then , where represents the center of .
(Please do not use sylow's or cauchy's theorems.)

(*) If H is a subgroup of G, then the factor group is isomorphic to a subgroup of Aut H (Hungerford, "Algebra", p92).

By assumption, . Since H is a group of order 17, H is isomorphic to the cyclic group of order 17, i.e., . Thus, |Aut(H)|=16.

Now, divides both 16 and 3825 by Lagrange's theorem and (*), it follows that . Can you conclude from here?