# cosets

• Oct 5th 2008, 11:48 AM
ll2008
cosets
Let $G$ be a group and let $N$ be a subgroup of $G$. Using the definitions of the sets $gN$, $Ng$, and $gNg^{-1}$, prove that $Ng = gN$ for every $g$ in $G$ if and only if $G = N_G(N)$.

$N_G(N)$ = { $g$ in $G$ | $gNg^{-1} = N$}
• Oct 5th 2008, 08:33 PM
ThePerfectHacker
Quote:

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

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

$gN = Ng$ if and only if $N$ is normal subgroup if and only if the normalizer of $N$ is $G$.
• Oct 5th 2008, 10:06 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 8th 2008, 12:23 AM
gosualite
Hacker's way is valid. It's using the definitions, not theorems. Here it is essentially broken down into baby steps.

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

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