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

1 Answer
Sep 15, 2016

Answer:

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.