while proving the two sided inverse in the 2nd part you operated σ over element g-¹ag which is not in A. So how can we operate the function σ:A------A over it.
Yes but he operated over it with respect to its DEFINITION. That is, he applied the defining procedure which showed that it came down to applying the element g to g^-1 on the left hand side, and g^-1 to g on the right hand side of "a" (this is an already known to be valid "group operation"). Thus with g and g^-1 united on both sides of "a", they resolved into the identity leaving only the element "little a", which we know to be an element of A (by assumption). So although before applying the procedure it is true we don't know if g^-1ag is a member of A, upon applying the procedure to g^-1ag we were then able to conclude that it must be a member of A, namely "little a bar", which is to say, the same thing as g^-1ag that was previously operated on resulting in "little a". This is valid by virture of being a bijective map.
Thanks very much. it really helps me a lot.
Do you have videos about normal subgroup?
while proving the two sided inverse in the 2nd part you operated σ over element g-¹ag which is not in A. So how can we operate the function σ:A------A over it.
Yes but he operated over it with respect to its DEFINITION. That is, he applied the defining procedure which showed that it came down to applying the element g to g^-1 on the left hand side, and g^-1 to g on the right hand side of "a" (this is an already known to be valid "group operation"). Thus with g and g^-1 united on both sides of "a", they resolved into the identity leaving only the element "little a", which we know to be an element of A (by assumption). So although before applying the procedure it is true we don't know if g^-1ag is a member of A, upon applying the procedure to g^-1ag we were then able to conclude that it must be a member of A, namely "little a bar", which is to say, the same thing as g^-1ag that was previously operated on resulting in "little a". This is valid by virture of being a bijective map.
I think that's what he was trying to prove. that
for gx(g- )= x̄ : x,x̄∈ A (this is the definition of Normalizers)
( g-)xg = x̄ : x, x̄ ∈A,
@@dvashunz7880 Can you please organize this reply a bit? Thanks.
Plz always do 1 single with each topic..
Also show in a same group that normalizer is greater then centralizer plz
I saw a problem with the proof: you did not show that __sigma(g^-1)__ has range A, which is the requirement to being the inverse of __sigma(g)__
In fact, the definition of normalizers conflict with Wikipedia: en.wikipedia.org/wiki/Centralizer_and_normalizer.
Yeap, it's not a bijective map since _sigma(g^-1) may sit outside of A and _sigma(g) has no definition on it at all.