Group conjecture

Question

Conjecture:

Given a finite group $G$ and a subset $A\subset G$. Then $\{A,A^2,A^3,\dots\}$ is a group iff $\forall n\in \mathbb N: |A^n|=|A^{n+1}|$.

Given that the composition between the subsets $A,B\subset G$ is $A\cdot B=\{g\in G|\exists a\in A\exists b\in B:g=a\cdot b\}$.

Example: suppose that $N\subset G$ is a normal subgroup, then the cosets $\{Ng,Ng^2,...\}$ have the same cardinality and constitutes a group, while for random sets the cardinality seems to grow when multiplying:

{ 3412 2143 4321 1234 } { 2143 1234 } nswap pnormal . -1  ok
{ 3412 2143 4321 1234 } { 2143 1234 } pquotient set. {{3412,4321},{2143,1234}} ok
{ 4321 3412 } go  ok
gen. {4321,3412} ok
gen. {2143,1234} ok
gen. {4321,3412} ok
ndrop  ok
{ 2431 2341 } go  ok
gen. {2431,2341} ok
gen. {4132,3142,4312,3412} ok
gen. {1234,1243,3214,4213,1324,1423,3124,4123} ok
gen. {2413,2314,3421,4321,1423,1324,2341,2431,2143,2134,3241,4231,1243,1234} ok
gen. {4123,3124,4321,3421,2143,2134,4132,3142,4213,3214,4231,3241,3412,4312,1432,1342,2413,2314,2431,2341} ok
gen. {4231,3241,1234,1243,3214,4213,1432,1342,1324,1423,2134,2143,2314,2413,4123,3124,4321,3421,4132,3142,4312,3412} ok
gen. {3412,4312,2314,2413,2341,2431,2143,2134,4321,3421,3241,4231,1342,1432,3142,4132,1234,1243,3214,4213,1324,1423,3124,4123} ok
gen. {4123,3124,3142,4132,3412,4312,1432,1342,3214,4213,2413,2314,3421,4321,1423,1324,2341,2431,2143,2134,3241,4231,1243,1234} ok
ndrop  ok

Answer 1

The first part:

If $x_1,\dots , x_m$ and $y_1,\dots ,y_n$ are two series of distinct elements of the finite group $G$ and $m\le n$, then $|\{x_1,\dots x_m\}\cdot\{y_1,\dots ,y_n\}|\ge n$. That's because$|\{x_iy_1,\dots,x_iy_n\}|= n$ (cancellation law) and $\{x_iy_1,\dots,x_iy_n\}\subseteq\{x_1,\dots x_{m}\}\cdot\{y_1,\dots ,y_n\}$. Therefore, $|A|,|A^2|,|A^3|,\dots$ is non decreasing and since $G$ is finite this series will become constant. If $|A^k|<|A^{k+1}|$ there could be no inverse to $A^k$ of the form $A^i$ since $A^k$ then would be in a cycle including $A^{k+1}$, and therefore $\{A,A^2,A^3\dots\}$ is not a group if $|A^n|\neq|A^{n+1}|$ for some $n$.

The second part:

Suppose that $\forall i:|A^i|=|A^{i+1}|$. Since $G$ is finite there is a $n$ so that $e\in A^n$ and thus $A^i\subseteq A^i\cdot A^n= A^n\cdot A^i$ and since $|A^i|=|A^i\cdot A^n|$ it must hold that $A^i= A^i\cdot A^n$, why $A^n$ is an identity in $\{A,A^2,A^3,...\}$. Obviously $A^{n-i}$ is the inverse to $A^i$ and hence $\{A,A^2,A^3,...\}$ is a group.

Answer 2

It's true that $A^n$ for large $n$ is a group (now I realize I think I misunderstood your question):

Finiteness of $G$ implies that $A^n$ eventually contains $1$ (each element has some $n$ so that $a^n = 1$). So your assumption on the orders implies that $A^n = A^{n+1}$ after that point.

Then for $a \in A^n$, multiplication by $a$ gives an injective map to $A^n \to A^{n+1}$ ($ag = ah$ implies $g = h$ in a group). By pigeon hole principal, this is also surjective. This implies that there is some $x \in A^n$ so that $ax = 1$, hence $A^n$ contains the inverse of all of its elements.

Hence $A^n$ is a subset of a group that contains $1$ and is closed under multiplication and inversion. So it is a subgroup.

Answer 3

Suppose that $A$ verifies the condition. There exists $a$ such that $a^n=1\in A^n$. This implies that $A^i\subset A^{n+1}$ $i\leq n$, since $\mid A^i\mid = \mid A^{n+1}\mid$, we deduce that $A^i=A^{n+1}$, thus $1\in A^n=A$ and $A=A^2$. This implies that for $a,b\in A, ab\in A^2\subset A$. Let $a\in A$, since $G$ is finite, there exists $l$ such that $a^l=1$, so $1\in A$ and $a^{l-1}\in A^{l-1}= A$ is the inverse of $a$.

Answer 4

The set of nonempty subsets of a finite group $G$ forms a semigroup. It is known that the subgroups of this semigroup are precisely the ones of the form

$$H/K = \{hK : h \in H \}$$

where $K \trianglelefteq H \leq G$. This is not difficult to prove, see for example 3.57 in "A Course in Group Theory" by Rose. In any case this shows that your condition is necessary, since $|hK| = |h'K|$.

See also this question.