case 1 : so for all and therefore any subgroup of is a (right) ideal of but has no non-trivial right ideal. thus has no non-trivial subgroup and hence it
has to be of prime order.
case 2 : we need a couple of observations first:
(1) let clearly is a (right) ideal of hence either or if then which is a contradiction. thus so for all
therefore for any we have
(2) now suppose and let clearly is a right ideal of if then which is nonsense. thus i.e. implies that or
now let by (1): so there exists such that thus which gives us by (2). the claim is that : let obviously
is a right ideal of and because thus i.e. if then obviously if then since we
will get by (2). therefore now if then by (1): and so for some similarly for some and thus so
this proves that is a division ring.