one of the most notorious problems in Herstein's book as far as i remember! although it's an easy exercise for me now, it gave me a very hard time when i was an innocent undergrad student!

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.