If H is a subgroup of prime index in a finite group G, it is established that either the normalizer N(H) equals G or N(H) equals H. The discussion highlights that since H has prime index, it is cyclic, and the orders of H, N(H), and G are related by the divisibility condition p=ord(H) | ord(N) | ord(G). The conclusion drawn is that either k1=1 or k2=1, where ord(N)=k1p and ord(G)=k2ord(N).
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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?