Order of Right coset Na of a normal subgroup, N, of a group G

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by iarnab_kundu, Nov 13, 2013, 8:57 AM

For brevity we shall assume $N$ to be a subgroup of a group $G$ , with the index of $N$ in $G$ to be finite. Let's call the index $I$ .

Property - $N$ is normal in $G$ iff for any two element $a$ and $b$ in $G$ we have $NaNb=Nab$ .

Fact - The order of $Na$ divides $G/N$ .

Result - For all $x\in G$ we have $x^I\in N$ .

Consider the set(or possibly multiset) $\{N, Na, Na^2,\ldots, Na^I\}$ . Easily $Na^I=N\iff a^I\in N$ .

This post has been edited 3 times. Last edited by iarnab_kundu, Nov 29, 2013, 6:08 PM

Cracking the Nut Open

Order of Right coset Na of a normal subgroup, N, of a group G

by iarnab_kundu, Nov 13, 2013, 8:57 AM

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