# Thread: Normal subgroups and cosets

1. ## Normal subgroups and cosets

Let G be a finite group and N a normal subgroup. Suppose that x is an element of G of prime order p with x not in N. Show that $\displaystyle <x> \cap N = 1$ and that the cosets $\displaystyle N, Nx,..., N x^{p-1}$ are distinct. Deduce that p divides the index [G:N].

Now I have the ideas for the first two parts although I'm having trouble writing this concisely and properly. Also, I'm not sure where the last deduction comes into this - what am I missing?

Finally (sketch thoughts for this bit will be fine!) - if simply x is not in N and m is the smallest integer for which $\displaystyle x^m$ is in N, generalise the above result. What does this say about the quotient group G/N?

2. Originally Posted by Amanda1990
Let G be a finite group and N a normal subgroup. Suppose that x is an element of G of prime order p with x not in N. Show that $\displaystyle <x> \cap N = 1$ and that the cosets $\displaystyle N, Nx,..., N x^{p-1}$ are distinct. Deduce that p divides the index [G:N].
Construct $\displaystyle G/N$, and consider $\displaystyle xN$. Since $\displaystyle x\not \in N$ it means $\displaystyle |xN|=k > 1$. However, $\displaystyle (xN)^p = x^pN = N$, therefore $\displaystyle k|p\implies k=p$. Thus, the order of $\displaystyle xN$ is $\displaystyle p$. This immediately implies that $\displaystyle N,xN,...,x^{p-1}N$ are distinct cosets. If $\displaystyle \left< x\right> \cap N \not = \{ 1\}$ it would mean that $\displaystyle x^j \in N$ for some $\displaystyle 1<j<p$ and so $\displaystyle (xN)^j = N$ but that is a problem because we established that the order of $\displaystyle xN$ must be $\displaystyle p$.

3. OK so this idea is very neat and elegant. In order to deduce that p divides the index [G:N] I'm guessing that we show p does not divide |N| (since we know p divides |G| and then we'll be done). I can't see how we can use what we already have to see this though.

If x is not in N and m is the smallest integer for which $\displaystyle x^m$ is in N, then it looks as if $\displaystyle <x> \cap N = <x^m>$ and that $\displaystyle N, Nx...Nx^{m-1}$ are distinct cosets. Is this correct? And what is this supposed to say about G/N?? Thanks!

4. Originally Posted by Amanda1990
OK so this idea is very neat and elegant. In order to deduce that p divides the index [G:N] I'm guessing that we show p does not divide |N| (since we know p divides |G| and then we'll be done). I can't see how we can use what we already have to see this though.

If x is not in N and m is the smallest integer for which $\displaystyle x^m$ is in N, then it looks as if $\displaystyle <x> \cap N = <x^m>$ and that $\displaystyle N, Nx...Nx^{m-1}$ are distinct cosets. Is this correct? And what is this supposed to say about G/N?? Thanks!
Notice that $\displaystyle \{N,xN,x^2N,...,x^{p-1}N\} = \left< xN\right>$ is a subgroup of $\displaystyle G/N$.
Therefore, by Lagrange, $\displaystyle |\left< xN\right>|$ divides $\displaystyle [G:N]$.
However, $\displaystyle p = |\left< xN\right>|$.