At (2) a) it was argumented, as if Proposition 4 would state a if-and-only-if-statement. But described and shown was in Proposition 4 only one direction. The other direction is true, too, but not trivial from my point of view..
Месяц назад
You're right, I should have stressed the "iff" - but I thought the other direction to be trivial using my definition of a subgroup. Reading your answer below showed me why I'm wrong. Thanks!
That (H,*H) is a group with a binary operation *H inherited from (G,*) implies that every element h from H has an (H,*H)-inverse. But for the closure property S2 under Inverses we have to show that H is closed unter (G,*)-inverses. That can be shown by showing, that the (H,*H)-inverses coincide with the (G,*)-inverses. In order to show this, I have to show first the auxiliary remark, that the neutral elements of (H,*H) and (G,*) coincide. In order to see, that even the last remark is not trivial, perhaps compare to monoids: If I would define "a submonoid of a monoid (G,*) is a subset H of G closed under *, such that (H,*H) (with a*Hb:=a*b) is a monoid" (Caution: That is NOT equivalent to the usual definition of a submonoid.) neutral elements of (G,*) and (H,*H) would not coincide in general: For example take (G,*) to be the monoid {0,1} with 0 and 1 different elements and 0*0=1*0=0*1=0 and 1*1=1. Then according to the (non-standard) definition given above H={0} is a submonoid of (G,*), but the neutral element 0 of H with respect to *H is not the neutral element 1 of (G,*).
Месяц назад+1
@@tobit4517 Thank you - NOW I finally get what you mean and agree with your criticism. "My" definition is used in a number of textbooks, though, where this problem isn't adressed at all. And I never saw a problem there until reading your comments. Thanks.
At (2) a) it was argumented, as if Proposition 4 would state a if-and-only-if-statement. But described and shown was in Proposition 4 only one direction. The other direction is true, too, but not trivial from my point of view..
You're right, I should have stressed the "iff" - but I thought the other direction to be trivial using my definition of a subgroup. Reading your answer below showed me why I'm wrong. Thanks!
That (H,*H) is a group with a binary operation *H inherited from (G,*) implies that every element h from H has an (H,*H)-inverse. But for the closure property S2 under Inverses we have to show that H is closed unter (G,*)-inverses. That can be shown by showing, that the (H,*H)-inverses coincide with the (G,*)-inverses. In order to show this, I have to show first the auxiliary remark, that the neutral elements of (H,*H) and (G,*) coincide.
In order to see, that even the last remark is not trivial, perhaps compare to monoids: If I would define "a submonoid of a monoid (G,*) is a subset H of G closed under *, such that (H,*H) (with a*Hb:=a*b) is a monoid" (Caution: That is NOT equivalent to the usual definition of a submonoid.) neutral elements of (G,*) and (H,*H) would not coincide in general: For example take (G,*) to be the monoid {0,1} with 0 and 1 different elements and 0*0=1*0=0*1=0 and 1*1=1. Then according to the (non-standard) definition given above H={0} is a submonoid of (G,*), but the neutral element 0 of H with respect to *H is not the neutral element 1 of (G,*).
@@tobit4517 Thank you - NOW I finally get what you mean and agree with your criticism. "My" definition is used in a number of textbooks, though, where this problem isn't adressed at all. And I never saw a problem there until reading your comments. Thanks.
I totally agree that many textbooks miss that point, too. When I see problems in your lectures, often the same problems are in many textbooks.