clearly we have where now consider these two elements of

then a simple calculation shows that let

be the elements of corresponding to and then:

let (since and have the same presentation.) so

now let be the set of left cosets of define by

where clearly and is a non-trivial homomorphism. since G is simple,

we must have i.e. is an embedding. so G is isomorphic to a subgroup of since

and is the only subgroup of which has order 360, we must have