-2
$\begingroup$

This question already has an answer here:

I had this as a statement in my book, but I am unable to prove it using the four basic properties of a group:

  1. Closure
  2. Associativity
  3. Existence of identity
  4. Existence of inverses.
$\endgroup$

marked as duplicate by Dietrich Burde abstract-algebra Jun 15 at 10:59

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 3
    $\begingroup$ First you need 0. multiplication in $G/N$ is well-defined. $\endgroup$ – Lord Shark the Unknown Jun 15 at 9:06
  • 1
    $\begingroup$ Multiplication as in "group operation"?! @LordSharktheUnknown $\endgroup$ – Aakash Singh Bais Jun 15 at 9:11
  • $\begingroup$ Use the fact $G$ satisfies all these properties. $\endgroup$ – Sunny Jun 15 at 9:18
  • $\begingroup$ The only non trivial thing to check that this binary operation is well defined and rest of things are clear because elements of $G/N$ are representatives by the elements of $G$ which satisfies these properties. $\endgroup$ – Sunny Jun 15 at 9:30
0
$\begingroup$

Use that $G$ is group. For example, if $1 \in G$ denotes the identity element of $G$, we get

$(1N)(gN) = (1 \cdot g) N = gN = (g \cdot 1)N = (gN)(1N)$

for all $g \in G$. Thus $1N \in G/N$ is the identity element in the quotient group.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.