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edited Feb 4, 2014 at 2:12

Ricardo Andrade

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Are There Always Group Generators Which Give Unimodal Growth?

Suppose $G$ is a $k$-generated finite group. Is there always a set of $k$ elements which generate the group and have a unimodal counting function?

Background:Background: The counting function, $f(n)$, is a function whose value at $n$ is the number of elementselements of $G$ of length $n$.

The length of an element $x$, relative to a given generating set, is the length of thethe shortest word, made up onlyonly of elements of thethe generating set, which is equal to $x$. We don't use the inverses of elements of the generating set for the purpose of determining length in this discussion.

A unimodal sequence is a finite sequence that first increases, and then decreases. A sequence

{$s_1, s_2,...,s_n$} $\{s_1, s_2,\ldots,s_n\}$ is unimodal if there exists a $t$ such that $s_1<=s_2<=...<=s_t$$s_1\leq s_2\leq\cdots\leq s_t$,

and $s_t>=s_{t+1}>=...>=s_n$$s_t \geq s_{t+1} \geq \cdots\geq s_n$.

Weisstein (Weisstein, Eric W. "Unimodal Sequence." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/UnimodalSequence.html)

Discussion:

 As pointed out in one of the comments, taking k to be the number of elements in the group will trivially give unimodal growth.  The question is more interesting if one considers values of k which are less than the number of elements in the group.

InDiscussion: As pointed out in one of the comments, taking $k$ to be the number of elements in the group will trivially give unimodal growth. The question is more interesting if one considers values of k which are less than the number of elements in the group. In a search that I made of some finite groups there was always a set of generators for which the counting function was unimodal. In fact, nonunimodal growth was a rarity. This may just be consequence of the fact that I looked only at groups with a small number of elements. I'm convinced that for commutative groups any choice of generators will yield a unimodal counting function.

Here is an example of a group with non-unimodal growth.

The two generators are the permutations:

$a = \{7, 9, 2, 3, 6, 5, 1, 8, 4\}$

$b = \{6, 8, 7, 5, 2, 1, 3, 4, 9\}$

In this notation $a$ is the permutation which takes 1 into 7, 2 into 9, 3 into 2,...

The counting function for this pair of generators is:

1,2,4,8,13,21,33,44,55,75,83,80,85,65,39,27,11,2.

Are There Always Group Generators Which Give Unimodal Growth

Suppose $G$ is a $k$-generated finite group. Is there always a set of $k$ elements which generate the group and have a unimodal counting function?

Background: The counting function, $f(n)$, is a function whose value at $n$ is the number of elements of $G$ of length $n$.

The length of an element $x$, relative to a given generating set, is the length of the shortest word, made up only of elements of the generating set, which is equal to $x$. We don't use the inverses of elements of the generating set for the purpose of determining length in this discussion.

A unimodal sequence is a finite sequence that first increases, and then decreases. A sequence

{$s_1, s_2,...,s_n$} is unimodal if there exists a $t$ such that $s_1<=s_2<=...<=s_t$,

and $s_t>=s_{t+1}>=...>=s_n$.

Weisstein, Eric W. "Unimodal Sequence." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/UnimodalSequence.html

Discussion:

 As pointed out in one of the comments, taking k to be the number of elements in the group will trivially give unimodal growth.  The question is more interesting if one considers values of k which are less than the number of elements in the group.

In a search that I made of some finite groups there was always a set of generators for which the counting function was unimodal. In fact nonunimodal growth was a rarity. This may just be consequence of the fact that I looked only at groups with a small number of elements. I'm convinced that for commutative groups any choice of generators will yield a unimodal counting function.

Here is an example of a group with non-unimodal growth.

The two generators are the permutations:

$a = \{7, 9, 2, 3, 6, 5, 1, 8, 4\}$

$b = \{6, 8, 7, 5, 2, 1, 3, 4, 9\}$

In this notation $a$ is the permutation which takes 1 into 7, 2 into 9, 3 into 2,...

The counting function for this pair of generators is:

1,2,4,8,13,21,33,44,55,75,83,80,85,65,39,27,11,2.

Are There Always Group Generators Which Give Unimodal Growth?

Suppose $G$ is a $k$-generated finite group. Is there always a set of $k$ elements which generate the group and have a unimodal counting function?

Background: The counting function, $f(n)$, is a function whose value at $n$ is the number of elements of $G$ of length $n$.

The length of an element $x$, relative to a given generating set, is the length of the shortest word, made up only of elements of the generating set, which is equal to $x$. We don't use the inverses of elements of the generating set for the purpose of determining length in this discussion.

A unimodal sequence is a finite sequence that first increases, and then decreases. A sequence $\{s_1, s_2,\ldots,s_n\}$ is unimodal if there exists a $t$ such that $s_1\leq s_2\leq\cdots\leq s_t$, and $s_t \geq s_{t+1} \geq \cdots\geq s_n$. (Weisstein, Eric W. "Unimodal Sequence." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/UnimodalSequence.html)

Discussion: As pointed out in one of the comments, taking $k$ to be the number of elements in the group will trivially give unimodal growth. The question is more interesting if one considers values of k which are less than the number of elements in the group. In a search that I made of some finite groups there was always a set of generators for which the counting function was unimodal. In fact, nonunimodal growth was a rarity. This may just be consequence of the fact that I looked only at groups with a small number of elements. I'm convinced that for commutative groups any choice of generators will yield a unimodal counting function.

Here is an example of a group with non-unimodal growth. The two generators are the permutations:

$a = \{7, 9, 2, 3, 6, 5, 1, 8, 4\}$

$b = \{6, 8, 7, 5, 2, 1, 3, 4, 9\}$

In this notation $a$ is the permutation which takes 1 into 7, 2 into 9, 3 into 2,...

The counting function for this pair of generators is:

1,2,4,8,13,21,33,44,55,75,83,80,85,65,39,27,11,2.

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edit approved Feb 4, 2014 at 1:20

user5794

Suppose G$G$ is a k$k$-generated finite group. Is there always a set of k$k$ elements which generate the group and have a unimodal counting function?

Background: The counting function, f(n)$f(n)$, is a function whose value at n$n$ is the number of elements of G$G$ of length n$n$.

The length of an element x$x$, relative to a given generating set, is the length of the shortest word, made up only of elements of the generating set, which is equal to x$x$. We don't use the inverses of elements of the generating set for the purpose of determining length in this discussion.

A unimodal sequence is a finite sequence that first increases, and then decreases. A sequence

{$s_1, s_2,...,s_n$} is unimodal if there exists a $t$ such that $s_1<=s_2<=...<=s_t$,

and $s_t>=s_{t+1}>=...>=s_n$.

Weisstein, Eric W. "Unimodal Sequence." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/UnimodalSequence.html

Discussion:

 As pointed out in one of the comments, taking k to be the number of elements in the group will trivially give unimodal growth.  The question is more interesting if one considers values of k which are less than the number of elements in the group.

In a search that I made of some finite groups there was always a set of generators for which the counting function was unimodal. In fact nonunimodal growth was a rarity. This may just be consequence of the fact that I looked only at groups with a small number of elements. I'm convinced that for commutative groups any choice of generators will yield a unimodal counting function.

Here is an example of a group with non-unimodal growth.

The two generators are the permutations:

a = {7, 9, 2, 3, 6, 5, 1, 8, 4}$a = \{7, 9, 2, 3, 6, 5, 1, 8, 4\}$

b = {6, 8, 7, 5, 2, 1, 3, 4, 9}$b = \{6, 8, 7, 5, 2, 1, 3, 4, 9\}$

In this notation a$a$ is the permutation which takes 1 into 7, 2 into 9, 3 into 2,...

The counting function for this pair of generators is:

1,2,4,8,13,21,33,44,55,75,83,80,85,65,39,27,11,2.

Suppose G is a k-generated finite group. Is there always a set of k elements which generate the group and have a unimodal counting function?

Background: The counting function, f(n), is a function whose value at n is the number of elements of G of length n.

The length of an element x, relative to a given generating set, is the length of the shortest word, made up only of elements of the generating set, which is equal to x. We don't use the inverses of elements of the generating set for the purpose of determining length in this discussion.

A unimodal sequence is a finite sequence that first increases, and then decreases. A sequence

{$s_1, s_2,...,s_n$} is unimodal if there exists a $t$ such that $s_1<=s_2<=...<=s_t$,

and $s_t>=s_{t+1}>=...>=s_n$.

Weisstein, Eric W. "Unimodal Sequence." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/UnimodalSequence.html

Discussion:

 As pointed out in one of the comments, taking k to be the number of elements in the group will trivially give unimodal growth.  The question is more interesting if one considers values of k which are less than the number of elements in the group.

In a search that I made of some finite groups there was always a set of generators for which the counting function was unimodal. In fact nonunimodal growth was a rarity. This may just be consequence of the fact that I looked only at groups with a small number of elements. I'm convinced that for commutative groups any choice of generators will yield a unimodal counting function.

Here is an example of a group with non-unimodal growth.

The two generators are the permutations:

a = {7, 9, 2, 3, 6, 5, 1, 8, 4}

b = {6, 8, 7, 5, 2, 1, 3, 4, 9}

In this notation a is the permutation which takes 1 into 7, 2 into 9, 3 into 2,...

The counting function for this pair of generators is:

1,2,4,8,13,21,33,44,55,75,83,80,85,65,39,27,11,2.

Suppose $G$ is a $k$-generated finite group. Is there always a set of $k$ elements which generate the group and have a unimodal counting function?

Background: The counting function, $f(n)$, is a function whose value at $n$ is the number of elements of $G$ of length $n$.

The length of an element $x$, relative to a given generating set, is the length of the shortest word, made up only of elements of the generating set, which is equal to $x$. We don't use the inverses of elements of the generating set for the purpose of determining length in this discussion.

A unimodal sequence is a finite sequence that first increases, and then decreases. A sequence

{$s_1, s_2,...,s_n$} is unimodal if there exists a $t$ such that $s_1<=s_2<=...<=s_t$,

and $s_t>=s_{t+1}>=...>=s_n$.

Weisstein, Eric W. "Unimodal Sequence." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/UnimodalSequence.html

Discussion:

 As pointed out in one of the comments, taking k to be the number of elements in the group will trivially give unimodal growth.  The question is more interesting if one considers values of k which are less than the number of elements in the group.

In a search that I made of some finite groups there was always a set of generators for which the counting function was unimodal. In fact nonunimodal growth was a rarity. This may just be consequence of the fact that I looked only at groups with a small number of elements. I'm convinced that for commutative groups any choice of generators will yield a unimodal counting function.

Here is an example of a group with non-unimodal growth.

The two generators are the permutations:

$a = \{7, 9, 2, 3, 6, 5, 1, 8, 4\}$

$b = \{6, 8, 7, 5, 2, 1, 3, 4, 9\}$

In this notation $a$ is the permutation which takes 1 into 7, 2 into 9, 3 into 2,...

The counting function for this pair of generators is:

1,2,4,8,13,21,33,44,55,75,83,80,85,65,39,27,11,2.