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It happens that, $G$ being a group, we want to check if a finite subset $\{a_{1}, \ldots , a_{n}\}$ of $G$, with $n \geq 1$, is a subgroup of $G$.

It is a subgroup if and only if for every $i, j$ in $\{1, \ldots , n\}$, $a_{i} a_{j}$ is in $\{a_{1}, \ldots , a_{n}\}$. (A nonempty finite submagma of a group is a subgroup.)

If I'm not wrong, it is sufficient to verify that for every $i, j$ in $\{1, \ldots , n\}$ such that $i \leq j$, $a_{i} a_{j}$ is in $\{a_{1}, \ldots , a_{n}\}$, where we can clearly assume that the $a_{i}$'s are pairwise distinct.

(Edit : Thus (if I'm not wrong), instead of verifying that, for every $i, j$ such that $i \leq j$, the elements $a_{i} a_{j}$ AND $ a_{j} a_{i}$ are in $\{a_{1}, \ldots , a_{n}\}$, we need only to show that $ a_{i} a_{j}$ is in $\{a_{1}, \ldots , a_{n}\}$. So, we avoid almost half of the calculations.)

I have a proof (I give it below), but it is perhaps too complicated. Thus, my questions are :

1° is the statement correct ?

2° if the statement is correct, is the proof accurate ?

3° if the proof is accurate, is it the simplest possible ?

4° if the statement is correct, is it useful ?

5° do you know a reference to the literature about this question ?

Definition. Let $G$ be a group. We define a semistable sequence of elements of $G$ as a nonempty finite sequence $(a_{1}, \ldots , a_{n})$ of pairwise distinct elements of $G$ such that for every $i, j$ in $\{1, \ldots , n\}$ such that $i \leq j$, $a_{i} a_{j}$ is in $\{a_{1}, \ldots , a_{n}\}$.

Our problem is to prove that if $(a_{1}, \ldots , a_{n})$ is a semistable sequence of elements of $G$, then $\{a_{1}, \ldots , a_{n}\}$ is a subgroup of $G$.

Step 1. Let $(a_{1}, \ldots , a_{n})$ be a semistable sequence of elements of a group $G$, let $i$ be an index in $\{1, \ldots , n\}$. Then $a_{i}$ is of finite order ($\leq n$) and for every $s$ in $\mathbb{Z}$, $a_{i}^{s}$ is in $\{a_{1}, \ldots , a_{n}\}$. The function $a \mapsto a^{-1}$ is a permutation of $\{a_{1}, \ldots , a_{n}\}$.

Proof. Let us prove that for every natural number $r \geq 1$, $a_{i}^{r}$ is in $\{a_{1}, \ldots , a_{n}\}$. It is true if $r = 1$. Assume that it is true for a natural number $r$. Then $a_{i}^{r} = a_{j}$ for some $j$. It implies $a_{i}^{r+1} = a_{i}a_{j}$ and also $a_{i}^{r+1} = a_{j}a_{i}$. The first of these two results gives $a_{i}^{r+1} \in \{a_{1}, \ldots , a_{n}\}$ if $i \leq j$ and the second result also gives $a_{i}^{r+1} \in \{a_{1}, \ldots , a_{n}\}$ if $j \leq i$. By induction on $r$, we conclude that

(1) for every natural number $r \geq 1$, $a_{i}^{r}$ is in $\{a_{1}, \ldots , a_{n}\}$.

In particular, the $n+1$ elements $a_{i}, a_{i}^{2}, \ldots , a_{i}^{n+1}$ are all in $\{a_{1}, \ldots , a_{n}\}$. Since $\{a_{1}, \ldots , a_{n}\}$ has only $n$ elements, there are at least two exponents $r$ and $t$ in $\{1, \ldots , n+1\}$ such that $a_{i}^{t} = a_{i}^{r}$ and this implies that $a_{i}$ is of finite order ($\leq n$). Thus, for every $s$ in $\mathbb{Z}$, $a_{i}^{s}$ is of the form $a_{i}^{r}$ with some $r \geq 1$,thus, in view of (1), $a_{i}^{s}$ is in $\{a_{1}, \ldots , a_{n}\}$ for every $s$ in $\mathbb{Z}$. It is true in particular for $s=-1$, thus the inverse of each element of $\{a_{1}, \ldots , a_{n}\}$ is in $\{a_{1}, \ldots , a_{n}\}$, so the function $a \mapsto a^{-1}$ is a permutation of $\{a_{1}, \ldots , a_{n}\}$.

Step 2. Let $(a_{1}, \ldots , a_{n})$ be a semistable sequence of elements of a group $G$, let $i$ be an index in $\{1, \ldots , n\}$. The two following conditions are equivalent :

(i) for every $j \in \{1, \ldots , n\}$, $a_{i} a_{j}$ is in $\{a_{1}, \ldots , a_{n}\}$;

(ii) for every $j \in \{1, \ldots , n\}$, $a_{j} a_{i}$ is in $\{a_{1}, \ldots , a_{n}\}$.

Proof. Let's define (temporarily) a left universal of $(a_{1}, \ldots , a_{n})$ as an $a_{i}$ such that condition (i) is satisfied, i.e. such that for every $j \in \{1, \ldots , n\}$, $a_{i} a_{j}$ is in $\{a_{1}, \ldots , a_{n}\}$ and let's define a right universal of $(a_{1}, \ldots , a_{n})$ as an $a_{i}$ such that condition (ii) is staisfied, i.e. such that for every $j \in \{1, \ldots , n\}$, $a_{j} a_{i}$ is in $\{a_{1}, \ldots , a_{n}\}$. Thus, the statement amounts to say that the left universals are exactly the right universals.

Let us prove that if $a_{i}$ is a left universal, then $a_{i}^{-1}$ is a right universal. Since $a_{i}$ is a left universal, we have $a_{i}a_{j} \in \{a_{1}, \ldots , a_{n}\}$ for every $j$. We saw at step 1 that the function $a \mapsto a^{-1}$ is a permutation of $\{a_{1}, \ldots , a_{n}\}$, thus $a_{j}^{-1} a_{i}^{-1} \in \{a_{1}, \ldots , a_{n}\}$ for every $j$. Still because $a \mapsto a^{-1}$ is a permutation of $\{a_{1}, \ldots , a_{n}\}$, $a_{j}^{-1}$ runs over $\{a_{1}, \ldots , a_{n}\}$ as does $a_{j}$, thus, for every $j \in \{1, \ldots , n\}$, $a_{j} a_{i}^{-1} \in \{a_{1}, \ldots , a_{n}\}$. Since (step 1) $a_{i}^{-1}$ is in $\{a_{1}, \ldots , a_{n}\}$, this proves that

(1) if $a_{i}$ is a left universal, $a_{i}^{-1}$ is a right universal.

Similarly,

(2) if $a_{i}$ is a right universal, $a_{i}^{-1}$ is a left universal.

Now let us prove that the inverse of a left universal is also a left universal. Let $a_{i}$ be a left universal, let $j$ an index in $\{1, \ldots , n\}$. If $r$ is a natural number such that $a_{i}^{r} a_{j}$ is in $\{a_{1}, \ldots , a_{n}\}$, we have $a_{i}^{r} a_{j} = a_{k}$ for some $k$, whence $a_{i}^{r+1} a_{j} = a_{i}a_{k}$. Since $a_{i}$ is a left universal, the right member is in $\{a_{1}, \ldots , a_{n}\}$, thus $a_{i}^{r+1} a_{j}$ is in $\{a_{1}, \ldots , a_{n}\}$. By induction on $r$, we conclude that for every natural number $r \geq 1$, $a_{i}^{r}$ is a left universal. Since (step 1) $a_{i}$ is of finite order, $a_{i}^{-1}$ is of the form $a_{i}^{r}$ with a natural number $r \geq 1$, thus $a_{i}^{-1}$ is a left universal. We thus proved that

(3) if $a_{i}$ is a left universal, $a_{i}^{-1}$ is also a left universal.

Similarly,

(4) if $a_{i}$ is a right universal, $a_{i}^{-1}$ is also a right universal.

Let $a_{i}$ be a left universal. In view of (3), $a_{i}^{-1}$ is also a left universal. Thus, in view of (1), $a_{i}$ is a right universal. Thus every left universal is a right universal. Similarly, every right universal is a left universal. As noted, this proves step 2.

Definition. Let $(a_{1}, \ldots , a_{n})$ be a semistable sequence of elements of a group $G$. We define a universal of $(a_{1}, \ldots , a_{n})$ as an $a_{i}$ such that the two equivalent conditions of step 2 are satisfied.

Theorem. Let $(a_{1}, \ldots , a_{n})$ be a semistable sequence of elements of a group $G$. The set $\{a_{1}, \ldots , a_{n}\}$ is a subgroup of $G$.

Proof. Since $(a_{1}, \ldots , a_{n})$ is a semistable sequence of elements of a group $G$, we have $a_{1} a_{j} \in \{a_{1}, \ldots , a_{n}\}$ for every $j$ in $\{1, \ldots , n\}$, thus $a_{1}$ satisfies condition (i) of step 2 on $a_{i}$, thus

(1) $a_{1}$ is a universal of $(a_{1}, \ldots , a_{n})$.

If $n \geq 2$, we have $a_{2} a_{j} \in \{a_{1}, \ldots , a_{n}\}$ for every $j \geq 2$, because $(a_{1}, \ldots , a_{n})$ is a semistable sequence, and we have also $a_{2} a_{1} \in \{a_{1}, \ldots , a_{n}\}$, because of (1). Thus, if $n \geq 2$,

(2) $a_{2}$ is a universal of $(a_{1}, \ldots , a_{n})$.

Similarly, we deduce from (1) and (2) that (if $n \geq 3$) $a_{3}$ is a universal, and so on. Thus $a_{1}, \ldots , a_{n}$ are all universals, thus $\{a_{1}, \ldots , a_{n}\}$ is a subgroup of $G$.

Edit. The proof of relation (3) in step 2 (if $a_{i}$ is a left universal, $a_{i}^{-1}$ is also a left universal) can be simplified. Since $a_{i}$ is a left universal, we have $a_{i} A = A$, where $A$ denotes the set $\{a_{1}, \ldots , a_{n}\}$. Left multiplication by $a_{i}^{-1}$ gives $A = a_{i}^{-1} A$. Since, from step 1, $ a_{i}^{-1}$ is in A, this shows that $a_{i}^{-1}$ is a left universal. Similarly, the inverse of a right universal is a right universal.

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    $\begingroup$ I have looked through your proof and I believe it is correct. It is a little long winded, but I don't know whether there is any significantly simpler proof. $\endgroup$ – Derek Holt Nov 10 '19 at 15:34
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    $\begingroup$ E.g. for (3), if $a_i$ is left universal, so are its powers, including $a_i^{o(a_i)-1}=a_i^{-1}$. $\endgroup$ – Berci Nov 10 '19 at 16:16
  • $\begingroup$ Thanks to Derek Holt and Berci for the comments. $\endgroup$ – Panurge Nov 10 '19 at 16:55
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    $\begingroup$ Seems to be true, and the proof seems overly long but I'm struggling to come up with an argument that doesn't involve reinventing your little chain of three lemmas. $\endgroup$ – Jack M Nov 18 '19 at 13:17

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