Thread: Subgroup with prime index

1. Subgroup with prime index

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

2. 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 $\displaystyle [G:N] \mid [G:H]!.$

3. 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?

4. 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!