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Does there exist a verbally simple group, which is not characteristically simple?

Does there exist some sort of classification of finite verbally simple groups?Does there exist an Artinian verbally simple group, which is not characteristically simple?Does there exist a group without automorphisms such that…Infinite group not isomorphic to proper subgroup$G$ is characteristically simple $iff$ there is simple $T$ such that $G cong Ttimes T times cdots times T$Question about an equivalent definition of a simple groupDoes there exist an infinite non-abelian group such that all of its proper subgroups become cyclic?Does there exist some sort of classification of finite “characteristically simple” groups?Is it true, that for any two non-isomorphic finite groups $G$ and $H$ there exists such a group word $w$, that $|V_w(G)| neq |V_w(H)|$?Does there exist some sort of classification of finite verbally simple groups?Large counterexamples to “Non-isomorphic finite groups have verbal subgroups of different order”Does there exist some sort of classification of finite marginally simple groups?

1

2

$begingroup$

Does there exist a verbally simple group, which is not characteristically simple? A characteristically simple group is a group without non-trivial proper characteristic subgroups, a verbally simple group is a group without non-trivial proper verbal subgroups.

If such group exists, it has to be infinite, as every finite group is characteristically simple iff it is verbally simple (the proof of this fact can be found here: Does there exist some sort of classification of finite verbally simple groups?). However that proof, relies strongly on mathematical induction by the group order and is thus valid only for finite groups. And I do not know whether such infinite group exists.

abstract-algebra group-theory infinite-groups verbal-subgroups characteristic-subgroups

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edited Mar 22 at 10:04

Yanior Weg

asked Mar 22 at 9:39

Yanior WegYanior Weg

2,69711446

$endgroup$

add a comment | 

1

2

$begingroup$

Does there exist a verbally simple group, which is not characteristically simple? A characteristically simple group is a group without non-trivial proper characteristic subgroups, a verbally simple group is a group without non-trivial proper verbal subgroups.

If such group exists, it has to be infinite, as every finite group is characteristically simple iff it is verbally simple (the proof of this fact can be found here: Does there exist some sort of classification of finite verbally simple groups?). However that proof, relies strongly on mathematical induction by the group order and is thus valid only for finite groups. And I do not know whether such infinite group exists.

abstract-algebra group-theory infinite-groups verbal-subgroups characteristic-subgroups

share|cite|improve this question

edited Mar 22 at 10:04

Yanior Weg

asked Mar 22 at 9:39

Yanior WegYanior Weg

2,69711446

$endgroup$

add a comment | 

$begingroup$

Does there exist a verbally simple group, which is not characteristically simple? A characteristically simple group is a group without non-trivial proper characteristic subgroups, a verbally simple group is a group without non-trivial proper verbal subgroups.

If such group exists, it has to be infinite, as every finite group is characteristically simple iff it is verbally simple (the proof of this fact can be found here: Does there exist some sort of classification of finite verbally simple groups?). However that proof, relies strongly on mathematical induction by the group order and is thus valid only for finite groups. And I do not know whether such infinite group exists.

abstract-algebra group-theory infinite-groups verbal-subgroups characteristic-subgroups

share|cite|improve this question

edited Mar 22 at 10:04

Yanior Weg

asked Mar 22 at 9:39

Yanior WegYanior Weg

2,69711446

$endgroup$

share|cite|improve this question

edited Mar 22 at 10:04

Yanior Weg

asked Mar 22 at 9:39

Yanior WegYanior Weg

2,69711446

share|cite|improve this question

edited Mar 22 at 10:04

Yanior Weg

asked Mar 22 at 9:39

Yanior WegYanior Weg

2,69711446

asked Mar 22 at 9:39

Yanior WegYanior Weg

2,69711446

asked Mar 22 at 9:39

Yanior WegYanior Weg

2,69711446

1 Answer
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oldest

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7

$begingroup$

Let $S$ be a nonabelian finite simple group, and let $G=prod_i=0^inftyS$ be the direct product of infinitely many copies of $S$.

Then $G$ is verbally simple, since every element $(g_i)$ involves only finitely many different $g_i$, and so is in a subgroup $H<G$ isomorphic to $S^n$ for some finite $n$. But $S^n$ is verbally simple, so for any word $w$, either

• 
  $w$ vanishes on $S$, in which case it vanishes on $G$, or


• the values of $w$ on $S^n$ generate $S^ncong H$, in which case the verbal subgroup of $G$ determined by $w$ is the whole of $G$.

But $G$ is not characteristically simple, since the restricted direct product $bigoplus_i=0^inftyS$ is a characteristic subgroup, characterized by the fact that its elements are precisely the elements of $G$ whose centralizers have finite index in $G$.

share|cite|improve this answer

answered Mar 22 at 10:48

Jeremy RickardJeremy Rickard

16.9k11746

$endgroup$

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7

$begingroup$

Let $S$ be a nonabelian finite simple group, and let $G=prod_i=0^inftyS$ be the direct product of infinitely many copies of $S$.

Then $G$ is verbally simple, since every element $(g_i)$ involves only finitely many different $g_i$, and so is in a subgroup $H<G$ isomorphic to $S^n$ for some finite $n$. But $S^n$ is verbally simple, so for any word $w$, either

• 
  $w$ vanishes on $S$, in which case it vanishes on $G$, or


• the values of $w$ on $S^n$ generate $S^ncong H$, in which case the verbal subgroup of $G$ determined by $w$ is the whole of $G$.

But $G$ is not characteristically simple, since the restricted direct product $bigoplus_i=0^inftyS$ is a characteristic subgroup, characterized by the fact that its elements are precisely the elements of $G$ whose centralizers have finite index in $G$.

share|cite|improve this answer

answered Mar 22 at 10:48

Jeremy RickardJeremy Rickard

16.9k11746

$endgroup$

add a comment | 

7

$begingroup$

Let $S$ be a nonabelian finite simple group, and let $G=prod_i=0^inftyS$ be the direct product of infinitely many copies of $S$.

Then $G$ is verbally simple, since every element $(g_i)$ involves only finitely many different $g_i$, and so is in a subgroup $H<G$ isomorphic to $S^n$ for some finite $n$. But $S^n$ is verbally simple, so for any word $w$, either

• 
  $w$ vanishes on $S$, in which case it vanishes on $G$, or


• the values of $w$ on $S^n$ generate $S^ncong H$, in which case the verbal subgroup of $G$ determined by $w$ is the whole of $G$.

But $G$ is not characteristically simple, since the restricted direct product $bigoplus_i=0^inftyS$ is a characteristic subgroup, characterized by the fact that its elements are precisely the elements of $G$ whose centralizers have finite index in $G$.

share|cite|improve this answer

answered Mar 22 at 10:48

Jeremy RickardJeremy Rickard

16.9k11746

$endgroup$

add a comment | 

$begingroup$

Let $S$ be a nonabelian finite simple group, and let $G=prod_i=0^inftyS$ be the direct product of infinitely many copies of $S$.

Then $G$ is verbally simple, since every element $(g_i)$ involves only finitely many different $g_i$, and so is in a subgroup $H<G$ isomorphic to $S^n$ for some finite $n$. But $S^n$ is verbally simple, so for any word $w$, either

• 
  $w$ vanishes on $S$, in which case it vanishes on $G$, or


• the values of $w$ on $S^n$ generate $S^ncong H$, in which case the verbal subgroup of $G$ determined by $w$ is the whole of $G$.

But $G$ is not characteristically simple, since the restricted direct product $bigoplus_i=0^inftyS$ is a characteristic subgroup, characterized by the fact that its elements are precisely the elements of $G$ whose centralizers have finite index in $G$.

share|cite|improve this answer

answered Mar 22 at 10:48

Jeremy RickardJeremy Rickard

16.9k11746

$endgroup$

share|cite|improve this answer

answered Mar 22 at 10:48

Jeremy RickardJeremy Rickard

16.9k11746

answered Mar 22 at 10:48

Jeremy RickardJeremy Rickard

16.9k11746

answered Mar 22 at 10:48

Jeremy RickardJeremy Rickard

16.9k11746