# cosets

• Oct 5th 2008, 10:48 AM
ll2008
cosets
Let $\displaystyle G$ be a group and let $\displaystyle N$ be a subgroup of $\displaystyle G$. Using the definitions of the sets $\displaystyle gN$, $\displaystyle Ng$, and $\displaystyle gNg^{-1}$, prove that $\displaystyle Ng = gN$ for every $\displaystyle g$ in $\displaystyle G$ if and only if $\displaystyle G = N_G(N)$.

$\displaystyle N_G(N)$ = {$\displaystyle g$ in $\displaystyle G$ | $\displaystyle gNg^{-1} = N$}
• Oct 5th 2008, 07:33 PM
ThePerfectHacker
Quote:

Originally Posted by ll2008
Let $\displaystyle G$ be a group and let $\displaystyle N$ be a subgroup of $\displaystyle G$. Using the definitions of the sets $\displaystyle gN$, $\displaystyle Ng$, and $\displaystyle gNg^{-1}$, prove that $\displaystyle Ng = gN$ for every $\displaystyle g$ in $\displaystyle G$ if and only if $\displaystyle G = N_G(N)$.

$\displaystyle N_G(N)$ = {$\displaystyle g$ in $\displaystyle G$ | $\displaystyle gNg^{-1} = N$}

$\displaystyle gN = Ng$ if and only if $\displaystyle N$ is normal subgroup if and only if the normalizer of $\displaystyle N$ is $\displaystyle G$.
• Oct 5th 2008, 09:06 PM
ll2008
I thought about that, but I don't think I can use those theorems to prove it. I would have to prove it using the definitions.
• Oct 7th 2008, 11:23 PM
gosualite
Hacker's way is valid. It's using the definitions, not theorems. Here it is essentially broken down into baby steps.

If $\displaystyle G=N_G(G)$, then $\displaystyle g\in G$ implies $\displaystyle g\in N_G(G)$, whence $\displaystyle gN=Ng$.

If $\displaystyle gN=Ng$ for all $\displaystyle g\in G$, it follows that $\displaystyle G \subseteq N_G(G)$ (and so equality too of course).