TITLE: Laws characterizing the trivial group
QUESTION [13 upvotes]: What are laws characterizing the trivial group? I mean all group words $w$ such that if the identity $w=1$ holds in a group $G$, then $G=1$.
For example, it can be easily verified that if a group word $w=x_{i_1}^{\alpha_1}x_{i_2}^{\alpha_2}\cdots x_{i_t}^{\alpha_t}$ has the  following property, then it characterizes the trivial group:
$$\exists J\subseteq \{x_{i_1},\ldots, x_{i_t}\}:\ gcd(deg_{x_{i_j}}(w): x_{i_j}\in J)=1.$$
It seems that such a problem must be already studied. Anybody knows a reference? 
P.S. Note that a given non-trivial group cannot be characterized by a law, because the identities of any group $A$ and its powers are the same.

REPLY [3 votes]: Every word $w$ on free generators $x_1,\dots,x_n$ can be written as 
$$w=x_1^{\alpha_1} \cdots x_n^{\alpha_n} c(x_1,\dots,x_n),$$
where $c$ is a word in the commutator subgroup of $\langle x_1,\dots,x_n \rangle$  and $\alpha_1,\dots, \alpha_n$ are some integers.
Note that the integer $\alpha_i$ is the same as the sum of the exponents of letter $x_i$ in the word $w$. You have denoted $\alpha_i$ by $\deg_{x_i}(w)$.
Now we can state our characterization of such words:
A word $w$ is charaterizing the trivial group if and only if $\gcd (\deg_{x_1}(w),\dots,\deg_{x_n}(w) )=1$.
This is because if $d:=\gcd(\alpha_1,\dots,\alpha_n)\neq 1$, then the cyclic group of order $d\neq 1$ satisfies $w$. Now  consider $w(1,\dots,a,\dots,1)$, where $a$ is an arbitrary element of the group satisfying the law $w$ and $a$ is in the possition $i$. It follows that $a^{\alpha_i}=1$ for all $i=1,\dots,n$ (note that $c(1,\dots,a,\dots,1)=1$). Thus $a^d=1$ and so $a=1$.