# Normalizer of a subgroup of prime index

• I
• Silviu

#### Silviu

Hello! Can anyone help me with this problem?

If H is a subgroup of prime index in a finite group G, show that either N(H)=G or N(H) = H.

Thank you!

What do you know about the index of a subgroup? And where is ##N_G(H)## settled with respect to ##H## and ##G##?

What do you know about the index of a subgroup? And where is ##N_G(H)## settled with respect to ##H## and ##G##?
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

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
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?

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?
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?

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?

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?
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!