# Normal subgroups and cosets

• Feb 22nd 2009, 10:54 AM
Amanda1990
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 $ \cap N = 1$ and that the cosets $N, Nx,..., N x^{p-1}$ 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 $x^m$ is in N, generalise the above result. What does this say about the quotient group G/N?
• Feb 22nd 2009, 02:20 PM
ThePerfectHacker
Quote:

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 $ \cap N = 1$ and that the cosets $N, Nx,..., N x^{p-1}$ are distinct. Deduce that p divides the index [G:N].

Construct $G/N$, and consider $xN$. Since $x\not \in N$ it means $|xN|=k > 1$. However, $(xN)^p = x^pN = N$, therefore $k|p\implies k=p$. Thus, the order of $xN$ is $p$. This immediately implies that $N,xN,...,x^{p-1}N$ are distinct cosets. If $\left< x\right> \cap N \not = \{ 1\}$ it would mean that $x^j \in N$ for some $1 and so $(xN)^j = N$ but that is a problem because we established that the order of $xN$ must be $p$.
• Feb 23rd 2009, 12:37 AM
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 $x^m$ is in N, then it looks as if $ \cap N = $ and that $N, Nx...Nx^{m-1}$ are distinct cosets. Is this correct? And what is this supposed to say about G/N?? Thanks!
• Feb 23rd 2009, 03:29 PM
ThePerfectHacker
Quote:

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 $x^m$ is in N, then it looks as if $ \cap N = $ and that $N, Nx...Nx^{m-1}$ are distinct cosets. Is this correct? And what is this supposed to say about G/N?? Thanks!

Notice that $\{N,xN,x^2N,...,x^{p-1}N\} = \left< xN\right>$ is a subgroup of $G/N$.
Therefore, by Lagrange, $|\left< xN\right>|$ divides $[G:N]$.
However, $p = |\left< xN\right>|$.