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.

  • 3
    $\begingroup$ These can be characterized as those elements in the free group which generate the free group as a fully characteristic subgroup. $\endgroup$
    – YCor
    Oct 18, 2016 at 6:25
  • $\begingroup$ In your first paragraph, do you mean to define a "law" as a set of words? I find your question rather difficult to parse. $\endgroup$ Oct 18, 2016 at 6:39
  • $\begingroup$ Thank you, but I like to know what are these words exactly. Is it possible to look at a group word and say that this word generates the free group as a fully characteristic subgroup (verbal subgroup)? $\endgroup$
    – Sh.M1972
    Oct 18, 2016 at 6:41
  • 1
    $\begingroup$ I doubt all people who voted to closed have really understood the question (if familiar with group varieties, maybe... anyway I don't think it's off-topic here...). $\endgroup$
    – YCor
    Oct 18, 2016 at 6:45
  • 1
    $\begingroup$ @RyanBudney a law just means some element $w(x_1,\dots,x_k)$ of a free group on $k$ generators; then one says that a group $G$ satisfies the law (or identity) $w$ if $w(g_1,\dots,g_k)=1$ for all $g_1,\dots,g_k\in G$. $\endgroup$
    – YCor
    Oct 18, 2016 at 6:47

2 Answers 2



  1. the law $w$ characterizes the trivial group among all groups;

  2. the law $w$ characterizes the trivial group among all cyclic groups of prime order;

  3. the image of $w$ in the abelianization of the free group is a primitive element.

That 1 implies 2 is trivial. Suppose conversely that 1 fails. Then there exists a nontrivial group $G$ satisfying the law $w$. Being nontrivial, $G$ has a nontrivial cyclic subgroup, and the latter has a quotient of prime order. Since satisfying a law passes to subgroups and quotients, we deduce that this group of prime order satisfies the law $w$, showing the failure of 2.

If 3 fails, then $w\in [F,F]F^p$ for some prime $p$ and hence $w$ holds in the cyclic group of order $p$, showing failure of 2. Conversely, assume 3. A primitive element of a free abelian group can be mapped by an automorphism to the first basis vector, and hence by an homomorphism to the generator of the infinite cyclic subgroup, and hence after composition to any element of any group. Let now $G$ be any nontrivial group; then we have an homomorphism $F\to G$ mapping $w$ to a nontrivial element, so $G$ fails to satisfy $w$. Thus $w$ characterizes the trivial group among all groups.

  • $\begingroup$ (Essentially) Restatement of the 1st equivalence for people familiar with group varieties: a group variety is nontrivial iff its free groups on 1 generators are nontrivial. $\endgroup$
    – YCor
    Oct 18, 2016 at 6:50

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 $a=1$.

  • 1
    $\begingroup$ Yes, it's a restatement of condition 3 of my answer (a $n$-tuple of integers being primitive iff its gcd is 1). $\endgroup$
    – YCor
    Mar 1, 2017 at 6:37
  • $\begingroup$ Thank you Alireza. Now such a description is completed. But one can ask another question: which first order sentences characterize the trivial group? Clearly for infinite groups, there is no such sentences, because of Los Theorem of Model Theory. But for finite groups, there are such first order sentences, for example, the sentence which describes the information of the Cyley table. $\endgroup$
    – Sh.M1972
    Mar 1, 2017 at 7:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.