Quotient Group G/G = {identity}? I know this is a basic question, but I'm trying to convince myself of Wikipedia's statement. "The quotient group $G / G$ is isomorphic to the trivial group."
I write the definition for left multiplication because left cosets = right cosets. $ G/G = \{g \in G : gG\} $ But how is this isomorphic to the trivial group, $ \{id_G\} $? $gG$ can't be simplified to $id_G$ ?
Thank you.
 A: Consider the zero homomorphism $\phi:G\rightarrow G$. Then the $\ker\phi=G$, $\mathrm{Im}\phi =\left\{id_G\right\}$ and from the first theorem of Isomorphism $\frac{G}{\ker\phi}\cong \mathrm{Im}\phi$ which of course implies that $G/G\cong \left\{id_G\right\}$
A: $\operatorname{id}_G$ is not an element of $G/G$. The elements of $G/G$ are sets of the form $gG$ with $g\in G$. But $gG=G$ for all $g\in G$, so there is only one co-set.
Now you just have to convince yourself that all groups with only one element are isomorphic. That's pretty easy to do.
A: Maybe it will help you to first look at a different quotient group, and then look at $G/G$.
Suppose that you have a subgroup $H$ of $G$ which is exactly half the size of $G$.  The factor group $G/H$ is the set of cosets of H in G.  What are these cosets?
If you multiply everything in $H$ by an element that is in $H$, you get back $H$.  If you do the same with some fixed $g_0\in G$ such that $g_0\notin H$, the resulting coset $g_0H=\{g_0h : h \in H\}$ has no elements in common with $H$.  (If it did have some $g_0h\in H$, by closure $g_0hh^{-1}=g_0\in H$, a contradiction.)  Since $g_0H$ is the same size as $H$, which is half the size of $G$, $H$ and $g_0H$ together make up all the elements of $G$. This means that for any $g'\in G$ you pick, either $g'H=H$ or $g'H=g_0H$; in other words, these are the only two cosets.
So we have that $G/H=\{H, g_0H\}$.  The operation for a factor group is $(aH)(bH)=(ab)H$, for any two cosets $aH,bH\in G/H$.  So, for example, $(g_0H)(g_0H)=(g_0^2)H$.  (For the coset "$H$" it's easiest to think of it as having an invisible $\text{id}_G$, so $(g_0H)(H)=(g_0\text{id}_G)H=g_0H$.  Thus we have that $H=\text{id}_{G/H}$.)  In particular, $G/H$ is a group with $2$ elements, so we know that it is isomorphic to $\mathbb{Z}_2$ because that is the only group of order $2$.
Now let's try doing it with $G/G$.
Well, $gG=G$ for any $g\in G$.  So that is the only coset.  Thus $G/G=\{G\}$, and $\text{id}_{G/G}=G$.  It has only one element, so it must be isomorphic to the only group with one element, the trivial group.
A: $G/G$ has only one element, which is $G$. Therefore this group is trivial. 
A: Well, forgive me for getting all technical, but this isn't right: $G/G = \{ g \in G \colon gG\}$. It should be the other way around: $G/G = \{ gG |\, g \in G \}$.
Next, we get rid of $g$ by realizing that $gG = G$ for any $g \in G$. Therefore, $G/G = \{G\}$, i.e. $G/G$ is a set with exactly 1 element, and this element is $G$ itself.
Now, the next thing is to realize how the operation in $G/G = \{G\}$ works. It works like this: $G \cdot G = G$. Exactly like that of the trivial group $\{1\}$: $1 \cdot 1 = 1$. So, the map $G/G \to \{1\}$ that sends $G$ to $1$ is a group isomorphism.
I know, I know, this may be the worst way to explain these things. There are way too many trivial formulas there. But it can be good for convincing oneself of something.
A: If G is a group and N is normal in G, then G/N is the quotient group. G/N as a group consists of cosets of the normal subgroup N in G and these cosets themselves satisfy the group properties because of normality of N. Now G is clearly normal in G. Hence G/G consists of the coset that is all of G. Thus this group has only one element, thence it must be isomorphic to the identity group
A: $G/G=id_G$ is false. It should be $G/G=\{[id_G]\}$ And actually you could use any element, where the [] mean the equivalence class. Saying G/G=all elements in G is also false, it should be that G/G is a group containing only one element, that is a set that contains all elements in G.
If we go to the definition of quotient group $xy^{-1}\in G\forall x,y\in G$, then, as all elements are related to each other and there's only one equivalence class, therefore the group G/G has one element. As it has one element and the identity must be in the group, then that element is the identity, but it's the identity of the group G/G, not the identity of the group G. The isomorphism G/G->{1} is trivial.
A: The notation of quotient group such $G/N$ is beautiful, but often misleads us to forget that $G/N$ is no more than a group where its elements are cosets $gN$ defined as:

For $N \unlhd G$ and $g \in G$ then $gN = \{gn \: | \: n \in N\}$.

Identity of the group is $N$. For elements $aN, bN \in G/N$, where $a, b \in G$, are considered the same element $aN = bN$ if $ab^{-1} \in N$.
In the case of $N = G$, we have that the identity of the group is element $G$. For $a, b \in G$, we ALWAYS have that $ab^{-1} \in G$. That means no matter which elements we choose they are always the same element $G$. So this new group $G/G$ has only one element $G$ which is also its identity. Written explicitly $G/G = \{G\} \cong \{1\}$.
