for any there exists such that
note that an obvious result of is that for any and , there exists such that .
now, for any we have and so . we also have for some , by . but then and hence . therefore and so is closed under multiplication. let . then for some and since , we have . thus and so . therefore and so is a subgroup. now let . then, by , there exists such that . but then , because , and thus . so we have proved that , i.e. is normal.
finally, let and choose such that . then for some , by . but then , because , and so . hence
let be such that . then , because , and so . hence and we are done.