# Does S_5 have a subgroup of order 40?

Sep 15, 2016

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 $\left(1 , 2 , 3 , 4 , 5\right)$. 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)$\text{ "(1, 2)" }$ i.e. one adjacent transposition

(b)$\text{ "(1, 3)" }$ i.e. one non-adjacent transposition

(c)$\text{ "(1, 2)(3, 4)" }$ i.e. two adjacent transpositions

(d)$\text{ "(1, 3)(2,4)" }$ i.e. two non-adjacent transpositions

(e)$\text{ "(1, 2)(3, 5)" }$ i.e. one adjacent, one non-adjacent transpositions

Combined with $\left(1 , 2 , 3 , 4 , 5\right)$ these five possibilities generate subgroups isomorphic to:

(a) $\text{ "S_5" }$ order $120$

(b) $\text{ "S_5" }$ order $120$

(c) $\text{ "A_5" }$ order $60$

(d) $\text{ "D_5" }$ order $10$

(e) ?

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