Well as H has prime index it means it is cyclic. Also we have that p=ord(H) | ord(N) | ord(G). So we should have ord(N)=##k_1##p and ord(G)=##k_2##ord(N)=##k_1## ##k_2## p. So we have to show that either ##k_1##=1 or ##k_2##=1. This is what I got by now. But I don't know what to do nextfresh_42 said:What do you know about the index of a subgroup? And where is ##N_G(H)## settled with respect to ##H## and ##G##?
So p=ord(H) | ord(N) | ord(G) means ##H \leq N_G(H) \leq G##. The index of ##H## is also equal to ##|G/H|##. Do you know what happens, if you factorize ##H## across these inclusions?Silviu said:Well as H has prime index it means it is cyclic. Also we have that p=ord(H) | ord(N) | ord(G). So we should have ord(N)=##k_1##p and ord(G)=##k_2##ord(N)=##k_1## ##k_2## p. So we have to show that either ##k_1##=1 or ##k_2##=1. This is what I got by now. But I don't know what to do next
Sorry I am a little confused. You mean something like ##|H/H| \le |N(H)/H| \le |G/H| ##? This means ##1 \le |N(H)/H| \le |G/H|##. So what next?fresh_42 said:So p=ord(H) | ord(N) | ord(G) means ##H \leq N_G(H) \leq G##. The index of ##H## is also equal to ##|G/H|##. Do you know what happens, if you factorize ##H## across these inclusions?
Thank you so much! I kept reading a subgroup of prime order. It is index not order. Now it is easy. Sorry for this and thank you again!fresh_42 said:If ##|G/H|=p## prime, how many orders fit in between ##1## and ##p##?
The essential part to be confused of, is the fact, that these cosets are possible disregarding whether ##H## is normal or not. Do you know why?
The normalizer of a subgroup of prime index is the largest subgroup of the entire group that contains the given subgroup. It consists of all elements in the group that commute with every element in the subgroup.
The normalizer plays a crucial role in group theory as it helps in understanding the structure of a group. It also has applications in other areas of mathematics such as algebraic geometry and number theory.
The normalizer can be calculated by finding all the elements in the group that commute with each element in the subgroup. This can be done by using the conjugacy class equation or by directly checking the commutativity of each element.
The normalizer and centralizer are related in that the centralizer is a subgroup of the normalizer. The centralizer consists of all elements in the group that commute with a given element, while the normalizer consists of all elements that commute with a given subgroup.
Yes, it is possible for the normalizer of a subgroup of prime index to be equal to the subgroup itself. This occurs when the given subgroup is a normal subgroup of the larger group, meaning that it commutes with all elements in the group. In this case, the normalizer is equal to the entire group.