# does there exist a non-abelian group such that all its subgroups are normal?

• Mar 17th 2010, 08:37 PM
snick
does there exist a non-abelian group such that all its subgroups are normal?
(1 0) (i 0) (0 1) (0 i) (-1 0 ) (-i 0) (0 -1) (0 -i)
(0 1) , (0 -i), (-1 0) , (i 0), (0 -1), (0 i), (1 0), (-i 0)

(*they all suppose to be 2x2 matrix)

where i refers to the imaginary number square root -1

A) prove that the set S forms a group of order 8

B) show that this group is non-abelian

c) Prove that all subgroups of this group are normal ( Hints: What are the possible subgroup orders and indexes?
• Mar 18th 2010, 03:15 AM
Swlabr
Quote:

Originally Posted by snick
(1 0) (i 0) (0 1) (0 i) (-1 0 ) (-i 0) (0 -1) (0 -i)
(0 1) , (0 -i), (-1 0) , (i 0), (0 -1), (0 i), (1 0), (-i 0)

(*they all suppose to be 2x2 matrix)

where i refers to the imaginary number square root -1

A) prove that the set S forms a group of order 8

B) show that this group is non-abelian

c) Prove that all subgroups of this group are normal ( Hints: What are the possible subgroup orders and indexes?

Which bit are you stuck on? Proving that it forms a group is just tedious, but do-able.

To prove that it is non-abelian, just take two matrices and show they do not commute (Hint: take any two matrices such that neither are the identity or the negative of the identity, and such that one is not the negative of the other).

To prove the last bit, you should notice that only two elements do not have order 4, the identity and negative the identity. You should also notice that negative the identity is a very special element. Very special.