Drawing a Cayley table and finding a subgroup of H

Question

$${H = \{e, (1243), (14)(23), (1342)\} }$$

$\bullet$By drawing a Cayley table show that H forms a group under composition of permutations.

$\bullet$Find a subgroup of H

I have this question and I don't know where to start. What would the be Cayley table look like?

Answer 1

Brookes student?

$(1243)(1243) = (14)(23)$

$(1243)((14)(23)) = (1342)$

$((14)(23))(1243) = (1342)$

$(1342)(1243) = (1)(2)(3)(4) = e $

$((14)(23))((14)(23)) = (1)(2)(3)(4) = e $

$(1342)((14)(23)) =(1243) $

$(1243)(1342) = (1)(2)(3)(4) = e$

Closure holds since no new entries in table. Association carries over. Each element contains an inverse because e is contained in each row/column.

To find a subgroup, look for a set of even permutations in H.