Question

Does `S_5` have a subgroup of order `40`?

Answer 1

I don't have a complete answer for you, but here are a few thoughts...

Explanation:

`S_5` is of order `5! = 120`.

By Lagrange's Theorem, any subgroup of `S_5` must have order a which divides evenly into of the order of `S_5`, i.e. must be a factor of `120`.

The converse is not true. That is, if a group has order a factor of `120` then it is not necessarily a subgroup of `S_5`. For example, `C_8` has order `8`, which is a factor of `120`, but `S_5` contains no element of order `8`, so no subgroup isomorphic to `C_8`.

It may help to look at what the possible generators of a subgroup of order `40` might be. Such a subgroup would contain an element of order `5`. Hence up to isomorphism, one of its generators is the `5` cycle `(1, 2, 3, 4, 5)`. It must also contain an element of order `2`. Up to isomorphism this element of order `2` can be assumed to permute `1`, so it can be taken to be one of:

(a)`" "(1, 2)" "` i.e. one adjacent transposition

(b)`" "(1, 3)" "` i.e. one non-adjacent transposition

(c)`" "(1, 2)(3, 4)" "` i.e. two adjacent transpositions

(d)`" "(1, 3)(2,4)" "` i.e. two non-adjacent transpositions

(e)`" "(1, 2)(3, 5)" "` i.e. one adjacent, one non-adjacent transpositions

Combined with `(1, 2, 3, 4, 5)` these five possibilities generate subgroups isomorphic to:

(a) `" "S_5" "` order `120`

(b) `" "S_5" "` order `120`

(c) `" "A_5" "` order `60`

(d) `" "D_5" "` order `10`

(e) ?

Anyway, we can look through possible generators and equivalences, hence enumerating and excluding possibilities.