Monoid times monoid equal to itself I am reading about monoids.  It says that if $G$ is a commutative monoid, then $GG = G$ because G contains a unit element.  If $G = \{1,2,3\}$ and the composition law is multiplication, then $9$ should be in $GG$, but is not in $G$.  What am I not understanding?  Not only do I not see how $GG = G$, I don't see how $GG = G$ because of the unit element in $G$ either.  Thanks.
 A: Note that here $G = \{1,2,3\}$ denote elements in $G$ but these are not necessarily integers. To define a monoid structure on this set is to give a binary associative operation $*$ such that an element $e \in G$ on the set is a neuter element, that is
$$
g*e = e*g = g \ (\forall g \in G)
$$
So here $3*3$ should give an element that is again on $G$. One possible monoid structure on $G$ is that of the integers modulo $3$, that is 
$$
a*b = r_3(ab)
$$
where $r_3$ is the remainder in the division by $3$. What we definitely don't have is $3*3 = 9$ since $9 \not \in G$. Maybe your misconception comes from treating $*$ as the usual product, which is not necessarily well defined for this particular set.
Now, as for the result you're quoting, let's see both inclusions. For what we've just noted, $GG \subseteq G$, because an element of $GG$ is of the form $a*b$ with $a,b \in G$, and this by definition lies on $G$. As for the other, since we have a unity element $e \in G$ on our monoid, if $g \in G$
$$
g = e*g \in GG
$$ 
which concludes the proof. Note that here this can be done regardless of the set $G$, we have only used the existence of an identity element for the monoid operation. 
If we don't have a identity element these need not be true. For example, if $G = \mathbb{Z}$ and $m*n := 0$ for all $n,m$ then this opeartion is associative but there is no identity element (hence $G$ is not a monoid), since for example $1*a = 0 \neq 1$ for any $a \in \mathbb{Z}$. Here, $GG = \{0\} \neq G$.
A: Hint:
Check whether $G$ is monoid or not first.
If it is monoid, then when the operation is commutative, the property $GG = G$ will be satisfied.
