# I Normalizer of a subgroup of prime index

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1. Oct 30, 2016

### 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!

2. Oct 30, 2016

### Staff: Mentor

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

3. Oct 30, 2016

### Silviu

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. Oct 30, 2016

### Staff: Mentor

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?

5. Oct 30, 2016

### Silviu

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?

6. Oct 30, 2016

### Staff: Mentor

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?

7. Oct 30, 2016

### Silviu

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!