Since one can carry out an induction on the number of elements, it suffices to show that for any two positive words $V$ and $W$ which have a common multiple a least common multiple exists. We prove this simultaneously for all $W$ by induction on the length of $V\text{.}$

Starting the induction: Let $V\equiv a$ and $U$ a common multiple of $a$ and $W\text{.}$ Then from (3.3) there is a term $W_i$ in the $a\text{-series}$ of $W$ with $W_i\equiv W_{i+1}\text{.}$ This $W_i$ is then a least common multiple of $a$ and $W,$ since by (3.3) both $a$ and $W$ divide $W_i,$ and from the construction of the $a\text{-series}$ and (3.1) it follows that $W_j$ divides $U$ for $j=1,2,\dots ,i\text{.}$

[Aside: Recall

(3.1) If $C$ is a chain from $a$ to $b$ and $D$ a positive word such that $CD$ is divisible by $a,$ then $D$ is divisible by $b\text{.}$

So if $W|U,$ then either $W_{k+1}\equiv W_k$ so $W_{k+1}|U,$ or $W_k$ is an $a\text{-chain}$ from $a$ to $b$ say, and $W_{k}K⩦U$ for some word $K,$ and $a|W_{k}K,$ so $K$ must be divisible by $b,$ so $U⩦W_{k}bK\prime$ for some word $K\prime ,$ but $W_{k+1}\equiv T_a\left(Wb\right)\text{.}$ So $U⩦W_{k+1}K\prime ,$ and $W_{k+1}|U\text{.}$

So we have that $W_i$ is a least common multiple. ]

Completing the induction: Let $V\equiv aV\prime$ and $U$ be a common multiple of $V$ and $W$ with $U\equiv aU\prime \text{.}$ Since $U$ is a common multiple of $a$ and $W,$ by the first induction step there is a least common multiple $aW\prime$ of $a$ and $W\text{.}$ By the reduction lemma, $U\prime$ is a common multiple of $V\prime$ and $W\prime ,$ so by induction hypothesis there exists a least common multiple $[V\prime ,W\prime ]\text{.}$ Then $a[V\prime ,w\prime ]$ is the least common multiple of $V$ and $W\text{.}$

$\square$

Let $X\subseteq G_M^{+}$ and $W\in X\text{.}$ The set of common divisors of the elements of $X$ is a finite set $\{A_1,\dots ,A_k\},$ since each of these elements must divide $W,$ and there are only finitely many divisors of $W\text{.}$

[Aside: A divisor of $W$ cannot be longer than $W,$ and (given a finite indexing set / set of letters) there are only finitely many words of length less than or equal to $W$ in $F^{+}\text{.}$ ]

Since $W$ is a common multiple of all $A_1,\cdots ,A_k,$ by (4.1), (Existence of least common multiple), the least common multiple $[A_1,\cdots ,A_k]$ exists, and this is clearly the greatest common divisor of the elements of $X\text{.}$

[Aside: Let $N=[A_1,\cdots ,A_k]\text{.}$ For all $W\in X,$ $W$ is a common multiple of $\{A_1,\cdots ,A_k\},$ so since $N$ is the least common multiple, $N|W\text{.}$ So $N$ is a common divisor of $X\text{.}$ So $N\in \{A_1,\cdots ,A_k\},$ and since it is a common multiple of this set, it must be the greatest common divisor. ]

$\square$

For the greatest common divisor it is clear, because in any pair of positive equivalent words exactly the same letters occur. For the least common multiple, the proof is an exact analogue of the existence proof in (4.1).

[Aside: Recall how we found $[a,W]\text{:}$ $W_1\equiv T_a\left(W\right),$ and $W_{i+1}\equiv W_i$ if $W_i$ starts with $a,$ or $W_{i+1}\equiv T_a\left(W_{i}b\right)$ if $W_i$ is an $a\text{-chain}$ from $a$ to $b\text{.}$ But if $b\ne a,$ then the only way we can have an $a\text{-chain}$ from $a$ to $b$ is if there is an elementary sub-chain somewhere in the $a\text{-chain}$ containing $b\text{.}$ So $W_{i+1}$ only contains letters which are already in $W_i\text{.}$ ]

$\square$