V is a variety of commutative semi group satisfying the identity $x^2 = x^3$. I need to prove that: $|F_V(\{x_1\dots,x_n\})|$ = $3^n -1$. Any hints on this ?
$F_V$ is V-free algebra.
V is a variety of commutative semi group satisfying the identity $x^2 = x^3$. I need to prove that: $|F_V(\{x_1\dots,x_n\})|$ = $3^n -1$. Any hints on this ?
$F_V$ is V-free algebra.
This question appears to be off-topic. The users who voted to close gave this specific reason:
Let $S_n = F_V(\{x_1, ..., x_n\})$. Since $S_n$ is commutative, its elements can be written in the form $x_1^{r_1} \cdots x_n^{r_n}$ where $r_1 + \dots + r_n > 0$. Morevover, since the semigroup $S_n$ satisfies the identity $x^2 = x^3$, you may assume that each $r_i$ is equal to $0$, $1$ or $2$. It follows that $|S_n| \leqslant 3^n -1$. To prove that this inequality is in fact an equality, it suffices to verify that the set of size $3^n -1$ $$ \{x_1^{r_1} \cdots x_n^{r_n} \mid 0 \leqslant r_i \leqslant 2 \text{ and } r_1 + \dots + r_n > 0\} $$ equipped with the product defined by $$ (x_1^{r_1} \cdots x_n^{r_n})(x_1^{s_1} \cdots x_n^{s_n}) = x_1^{\min\{2, r_1 + s_1\}} \cdots x_n^{\min\{2, r_n + s_n\}} $$ is a semigroup of $V$.