# Subgroup with prime index

• Mar 29th 2009, 01:46 PM
didact273
Subgroup with prime index
Prove that \$\displaystyle A_6\$ has no subgroup of prime index.
• Mar 29th 2009, 02:30 PM
NonCommAlg
Quote:

Originally Posted by didact273

Prove that \$\displaystyle A_6\$ has no subgroup of prime index.

you can prove a much stronger result: if \$\displaystyle n \geq 5\$ and \$\displaystyle H \lneq A_n,\$ then \$\displaystyle [A_n:H] \geq n.\$ this is a quick result of this well-known fact that for any group G and any subgroup H of G of finite index, there

exists a normal subgroup N of G contained in H such that \$\displaystyle [G:N] \mid [G:H]!.\$
• Mar 29th 2009, 03:04 PM
didact273
Quote:

Originally Posted by NonCommAlg
you can prove a much stronger result: if \$\displaystyle n \geq 5\$ and \$\displaystyle H \lneq A_n,\$ then \$\displaystyle [A_n:H] \geq n.\$ this is a quick result of this well-known fact that for any group G and any subgroup H of G of finite index, there

exists a normal subgroup N of G contained in H such that \$\displaystyle [G:N] \mid [G:H]!.\$

Can't I deduce from the fact that since \$\displaystyle A_6\$ is a simple group, the only possible subgroup is {(1)} and itself, there is no subgroup of index p?
• Mar 29th 2009, 03:19 PM
NonCommAlg
Quote:

Originally Posted by didact273

Can't I deduce from the fact that since \$\displaystyle A_6\$ is a simple group, the only possible subgroup is {(1)} and itself, there is no subgroup of index p?

the only possible (proper) normal subgroup is {(1)}. see the definition of a simple group again!