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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.

qualcuno
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Jonathan Tooker
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    $G$ is not monoid sonce it is not closed under the operation you defined. – Seewoo Lee May 30 '18 at 03:40
  • If G is all real numbers with multiplication, then I see that GG=G, but what does that have to do with the unit element? – Jonathan Tooker May 30 '18 at 03:43
  • @JonathanTooker For example, take $G = \lbrace 0, 2, 4, 6 \rbrace$ with multiplication being addition modulo $8$. If you compute $GG$, you'll only get $\lbrace 0, 4 \rbrace \neq G$. – Theo Bendit May 30 '18 at 04:40

2 Answers2

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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, $GG = \{0\} \neq G$.

qualcuno
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  • So if I'm understanding the example of your composition law, then the elements of the monoid would be the integers 0,1,2? – Jonathan Tooker May 30 '18 at 03:50
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    You can think of them as integers, yes, my point was that other products can be defined on the monoid such that these do not behave like the integers. I have added some comments on why the identity element is important to say $GG = G$. I hope it helps. – qualcuno May 30 '18 at 03:54
  • My misconception that was that I didn't see that the monoidal set needs to be closed under its composition law. Lang left that out of section 1.1 of his algebra book. – Jonathan Tooker May 30 '18 at 03:55
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    Oh, I see. In general a binary operation on a set $G$ is a function $$
    • : G \times G \longrightarrow G \

    (a,b) \to a*b $$

     – qualcuno May 30 '18 at 03:57
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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.

Our
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