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.)

Printable View

- March 23rd 2011, 10:29 AMabhishekkgpabelian 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.) - March 23rd 2011, 01:21 PMtonio
- March 23rd 2011, 09:09 PMroninpro
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.

- March 23rd 2011, 09:48 PMDrexel28
- March 23rd 2011, 10:15 PMroninpro
Just noting the arithmetic: .

- March 23rd 2011, 10:55 PMTheArtofSymmetry
(*) 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?