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A finite group $G$ and fixed $kgeq 1$ where for every $ngeq 1$, the $n$-direct product $G^n=Gtimesdotstimes G$ is $k$-generated?

Do “finitely generated abelian group” and “finite abelian group” mean the same thing?Is there an infinite group that contains every finite group (and no infinite group) as a subgroup?Is any pair of finite 2-generated perfect groups the quotient of a third finite 2-gen perfect group?$H,K$ normal subgroups of a finite group $G$ , $G cong H times K$ , every element of $H$ commutes with every element of $K$ , then is $G=HK$?Prove that locally finite nilpotent group is direct product of its normal maximal $p$- subgroupsSmallest finite group larger than the Quaternion group that isn't an outer Semi Direct ProductWhen the subgroup generated by all elements of some fixed finite order equals the groupDoes there exist a group that is both a free product and a direct product of nontrivial groups?Does the specific condition on a normal subgroup of a finite group imply that it is a direct factor?Does the specific condition on a normal subgroup of a finite group imply that it is a direct factor? v2.0

3

2

$begingroup$

Does exist a finite group $G$ and fixed $k geq 1$ such that the $n$-direct product $G^n = G times dots times G$ is $k$-generated for every $n geq 1$?

I suspect the answer is no. Does exist a simple proof of this fact (if true)?

group-theory finite-groups direct-product combinatorial-group-theory

share|cite|improve this question

edited 2 days ago

Shaun

9,334113684

asked 2 days ago

Luca SabatiniLuca Sabatini

858

$endgroup$

add a comment | 

3

2

$begingroup$

Does exist a finite group $G$ and fixed $k geq 1$ such that the $n$-direct product $G^n = G times dots times G$ is $k$-generated for every $n geq 1$?

I suspect the answer is no. Does exist a simple proof of this fact (if true)?

group-theory finite-groups direct-product combinatorial-group-theory

share|cite|improve this question

edited 2 days ago

Shaun

9,334113684

asked 2 days ago

Luca SabatiniLuca Sabatini

858

$endgroup$

add a comment | 

$begingroup$

Does exist a finite group $G$ and fixed $k geq 1$ such that the $n$-direct product $G^n = G times dots times G$ is $k$-generated for every $n geq 1$?

I suspect the answer is no. Does exist a simple proof of this fact (if true)?

group-theory finite-groups direct-product combinatorial-group-theory

share|cite|improve this question

edited 2 days ago

Shaun

9,334113684

asked 2 days ago

Luca SabatiniLuca Sabatini

858

$endgroup$

share|cite|improve this question

edited 2 days ago

Shaun

9,334113684

asked 2 days ago

Luca SabatiniLuca Sabatini

858

share|cite|improve this question

edited 2 days ago

Shaun

9,334113684

asked 2 days ago

Luca SabatiniLuca Sabatini

858

edited 2 days ago

Shaun

9,334113684

edited 2 days ago

Shaun

9,334113684

asked 2 days ago

Luca SabatiniLuca Sabatini

858

asked 2 days ago

Luca SabatiniLuca Sabatini

858

1 Answer
1

active

oldest

votes

9

$begingroup$

No. For given $k$ and sufficiently large $n$, the projections of all $k$ putative generators would be equal on two different components, and so they could not generate the full direct product.

To be more precise, if $h(k)$ is the number of surjective homomorphisms from the free group $F_k$ of rank $k$ to $G$ up to equivalence under automorphisms of $G$, then $G^{h(k)}$ is $k$-generated, but $G^{h(k)+1}$ is not. A well-known example (due I think to Philip Hall) is $A_5^{19}$ is $2$-generated but $A_5^{20}$ is not.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Derek HoltDerek Holt

54.1k53571

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you (the OP, not Derek) are not familiar with free groups, $h(k)$ is the number of generating $k$-tuples in $G$, modulo the action of $mathrm{Aut}(G)$. Then proving Derek's assertion that $G^{h(k)+1}$ is not $k$-generated is an exercise.
  $endgroup$
  – YCor
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another exercise is to calculate or estimate the probability that two randomly chosen elements of $A_5^{19}$ generate it. I don't remember the answer, but it's very small!
  $endgroup$
  – Derek Holt
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If I'm correct there are 780 non-generating pairs (among 3600 pairs), so already the probability $(1-780/3600)^{19}sim 0.0097$ that each of 19 pairs is generating, is quite small.
  $endgroup$
  – YCor
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I make it $1320 = 60 + 15times 36 + 20 times 24 + 24 times 10$ non-generating pairs $(g,h)$, with $60, 36, 24$, and $10$ possible $h$ when $g$ has order $1,2,3,5$ .That gives $.00017$ as the probability that a random pair projects onto each component. Of course the probability that they generate the direct product is less than that.
  $endgroup$
  – Derek Holt
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It seems I indeed missed many generating pairs (probably those generating copies of $A_4$). This makes $2280=120times 19$ generating pairs. Assuming that the orbits under the automorphism group all have the same size, this makes, for $19$ random generating pairs, the probability $1/r$ to make a generating 19-tuple, where $r=2280^{19}/(120^{19}times 19!)sim 16263866$. And I guess that this value of $1/r$ is an upper bound in any case. And this is just among 19-tuples of generating pairs. Among all 19-tuples of pairs, one has to multiply by your $.00017$, which gives $le 1/(95times10^9)$.
  $endgroup$
  – YCor
  yesterday
  
  
  

  
  
  
  

add a comment | 

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9

$begingroup$

No. For given $k$ and sufficiently large $n$, the projections of all $k$ putative generators would be equal on two different components, and so they could not generate the full direct product.

To be more precise, if $h(k)$ is the number of surjective homomorphisms from the free group $F_k$ of rank $k$ to $G$ up to equivalence under automorphisms of $G$, then $G^{h(k)}$ is $k$-generated, but $G^{h(k)+1}$ is not. A well-known example (due I think to Philip Hall) is $A_5^{19}$ is $2$-generated but $A_5^{20}$ is not.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Derek HoltDerek Holt

54.1k53571

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you (the OP, not Derek) are not familiar with free groups, $h(k)$ is the number of generating $k$-tuples in $G$, modulo the action of $mathrm{Aut}(G)$. Then proving Derek's assertion that $G^{h(k)+1}$ is not $k$-generated is an exercise.
  $endgroup$
  – YCor
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another exercise is to calculate or estimate the probability that two randomly chosen elements of $A_5^{19}$ generate it. I don't remember the answer, but it's very small!
  $endgroup$
  – Derek Holt
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If I'm correct there are 780 non-generating pairs (among 3600 pairs), so already the probability $(1-780/3600)^{19}sim 0.0097$ that each of 19 pairs is generating, is quite small.
  $endgroup$
  – YCor
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I make it $1320 = 60 + 15times 36 + 20 times 24 + 24 times 10$ non-generating pairs $(g,h)$, with $60, 36, 24$, and $10$ possible $h$ when $g$ has order $1,2,3,5$ .That gives $.00017$ as the probability that a random pair projects onto each component. Of course the probability that they generate the direct product is less than that.
  $endgroup$
  – Derek Holt
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It seems I indeed missed many generating pairs (probably those generating copies of $A_4$). This makes $2280=120times 19$ generating pairs. Assuming that the orbits under the automorphism group all have the same size, this makes, for $19$ random generating pairs, the probability $1/r$ to make a generating 19-tuple, where $r=2280^{19}/(120^{19}times 19!)sim 16263866$. And I guess that this value of $1/r$ is an upper bound in any case. And this is just among 19-tuples of generating pairs. Among all 19-tuples of pairs, one has to multiply by your $.00017$, which gives $le 1/(95times10^9)$.
  $endgroup$
  – YCor
  yesterday
  
  
  

  
  
  
  

add a comment | 

9

$begingroup$

No. For given $k$ and sufficiently large $n$, the projections of all $k$ putative generators would be equal on two different components, and so they could not generate the full direct product.

To be more precise, if $h(k)$ is the number of surjective homomorphisms from the free group $F_k$ of rank $k$ to $G$ up to equivalence under automorphisms of $G$, then $G^{h(k)}$ is $k$-generated, but $G^{h(k)+1}$ is not. A well-known example (due I think to Philip Hall) is $A_5^{19}$ is $2$-generated but $A_5^{20}$ is not.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Derek HoltDerek Holt

54.1k53571

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you (the OP, not Derek) are not familiar with free groups, $h(k)$ is the number of generating $k$-tuples in $G$, modulo the action of $mathrm{Aut}(G)$. Then proving Derek's assertion that $G^{h(k)+1}$ is not $k$-generated is an exercise.
  $endgroup$
  – YCor
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Another exercise is to calculate or estimate the probability that two randomly chosen elements of $A_5^{19}$ generate it. I don't remember the answer, but it's very small!
  $endgroup$
  – Derek Holt
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If I'm correct there are 780 non-generating pairs (among 3600 pairs), so already the probability $(1-780/3600)^{19}sim 0.0097$ that each of 19 pairs is generating, is quite small.
  $endgroup$
  – YCor
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I make it $1320 = 60 + 15times 36 + 20 times 24 + 24 times 10$ non-generating pairs $(g,h)$, with $60, 36, 24$, and $10$ possible $h$ when $g$ has order $1,2,3,5$ .That gives $.00017$ as the probability that a random pair projects onto each component. Of course the probability that they generate the direct product is less than that.
  $endgroup$
  – Derek Holt
  yesterday
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It seems I indeed missed many generating pairs (probably those generating copies of $A_4$). This makes $2280=120times 19$ generating pairs. Assuming that the orbits under the automorphism group all have the same size, this makes, for $19$ random generating pairs, the probability $1/r$ to make a generating 19-tuple, where $r=2280^{19}/(120^{19}times 19!)sim 16263866$. And I guess that this value of $1/r$ is an upper bound in any case. And this is just among 19-tuples of generating pairs. Among all 19-tuples of pairs, one has to multiply by your $.00017$, which gives $le 1/(95times10^9)$.
  $endgroup$
  – YCor
  yesterday
  
  
  

  
  
  
  

add a comment | 

$begingroup$

No. For given $k$ and sufficiently large $n$, the projections of all $k$ putative generators would be equal on two different components, and so they could not generate the full direct product.

To be more precise, if $h(k)$ is the number of surjective homomorphisms from the free group $F_k$ of rank $k$ to $G$ up to equivalence under automorphisms of $G$, then $G^{h(k)}$ is $k$-generated, but $G^{h(k)+1}$ is not. A well-known example (due I think to Philip Hall) is $A_5^{19}$ is $2$-generated but $A_5^{20}$ is not.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Derek HoltDerek Holt

54.1k53571

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

Derek HoltDerek Holt

54.1k53571

answered 2 days ago

Derek HoltDerek Holt

54.1k53571

answered 2 days ago

Derek HoltDerek Holt

54.1k53571