- #1

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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!

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- #1

- 624

- 11

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!

- #2

- 18,245

- 20,179

- #3

- 624

- 11

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

- #4

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- 20,179

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?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

- #5

- 624

- 11

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?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?

- #6

- 18,245

- 20,179

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?

- #7

- 624

- 11

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!

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?

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