Wenn Sie die Gruppen-ID kennen
select member.* from member m
inner join member_has_group mg on m.id = mg.id_member
where mg.id_group = [x]
Wenn Sie nur den Gruppennamen kennen
select member.* from member m
inner join member_has_group mg on m.id = mg.id_member
inner join group g on g.id = mg.id_group
where g.name = 'group name'
