# Prove the following statement: “Let $G$ be a group, and $N ⊲ G$. Then $G/N$ forms a group under the operation $(gN)(hN) = ghN$ ” [duplicate]

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.

## marked as duplicate by Dietrich Burde abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 15 at 10:59

• First you need 0. multiplication in $G/N$ is well-defined. – Lord Shark the Unknown Jun 15 at 9:06
• Use the fact $G$ satisfies all these properties. – Sunny Jun 15 at 9:18
• 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. – Sunny Jun 15 at 9:30
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.