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

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- Mar 29th 2009, 01:46 PMdidact273Subgroup with prime index
Prove that $\displaystyle A_6$ has no subgroup of prime index.

- Mar 29th 2009, 02:30 PMNonCommAlg
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 PMdidact273
- Mar 29th 2009, 03:19 PMNonCommAlg