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Is it always true that $g$ is bounded left-Engel element iff it belongs to an abelian subnormal subgroup?

1

$begingroup$

Suppose $G$ is a group. Let’s call $g in G$ a bounded left-Engel element iff $exists n in mathbbN forall h in G [h, g]_n = e$. Here $[h, g]_n$ is defined by recurrence:

$$[h, g]_n = begincases [[h, g]_n-1, g] & quad g > 0 \ h & quad g = 0 endcases$$

Is it always true that $g$ is bounded left-Engel element iff it belongs to an abelian subnormal subgroup?

Suppose $g$ belongs to an abelian subgroup $H$, which is subnormal in $G$ of length $n$ (We call $H < G$ subnormal of length $n$, iff $exists H_k_k = 0^n$ such that $H_0 = H$, $H_n = G$ and $forall 0 < k < n-1 H_n triangleleft H_n+1$)

Now we will prove, that $g$ is a bounded left-Engel element by induction:

Base: If $n = 0$, then $H = G$ is abelian and the statement is trivially true for any element.

Step: Suppose, it is true for $n-1$. Suppose $H$ is an abelian subgroup, which is subnormal in $G$ of length $n$, and $g in H$. Then there exists $K triangleleft G$, such that $H$ is subnormal in $K$ of length $n - 1$. So, by the supposition of induction $exists k in mathbbN forall h in K [h, g]_k = e$. Now it is sufficient to prove, that $forall h in G [h, g] = (hgh^-1)g in K$, which is rather obvious.

And so we have proved, that every element of an abelian subnormal subgroup is bounded left-Engel.

However, the inverse statement seems to be more difficult, and I do not know how to prove it. I have tried to construct the corresponding subnormal series here: $C_G^n(g) triangleleft C_G^n + 1$?, but, according to what was said in the comments, the series constructed that way are not always subnormal.

abstract-algebra group-theory normal-subgroups

share|cite|improve this question

asked Mar 27 at 16:37

Yanior WegYanior Weg

2,97821549

$endgroup$

add a comment | 

1

$begingroup$

Suppose $G$ is a group. Let’s call $g in G$ a bounded left-Engel element iff $exists n in mathbbN forall h in G [h, g]_n = e$. Here $[h, g]_n$ is defined by recurrence:

$$[h, g]_n = begincases [[h, g]_n-1, g] & quad g > 0 \ h & quad g = 0 endcases$$

Is it always true that $g$ is bounded left-Engel element iff it belongs to an abelian subnormal subgroup?

Suppose $g$ belongs to an abelian subgroup $H$, which is subnormal in $G$ of length $n$ (We call $H < G$ subnormal of length $n$, iff $exists H_k_k = 0^n$ such that $H_0 = H$, $H_n = G$ and $forall 0 < k < n-1 H_n triangleleft H_n+1$)

Now we will prove, that $g$ is a bounded left-Engel element by induction:

Base: If $n = 0$, then $H = G$ is abelian and the statement is trivially true for any element.

Step: Suppose, it is true for $n-1$. Suppose $H$ is an abelian subgroup, which is subnormal in $G$ of length $n$, and $g in H$. Then there exists $K triangleleft G$, such that $H$ is subnormal in $K$ of length $n - 1$. So, by the supposition of induction $exists k in mathbbN forall h in K [h, g]_k = e$. Now it is sufficient to prove, that $forall h in G [h, g] = (hgh^-1)g in K$, which is rather obvious.

And so we have proved, that every element of an abelian subnormal subgroup is bounded left-Engel.

However, the inverse statement seems to be more difficult, and I do not know how to prove it. I have tried to construct the corresponding subnormal series here: $C_G^n(g) triangleleft C_G^n + 1$?, but, according to what was said in the comments, the series constructed that way are not always subnormal.

abstract-algebra group-theory normal-subgroups

share|cite|improve this question

asked Mar 27 at 16:37

Yanior WegYanior Weg

2,97821549

$endgroup$

add a comment | 

$begingroup$

Suppose $G$ is a group. Let’s call $g in G$ a bounded left-Engel element iff $exists n in mathbbN forall h in G [h, g]_n = e$. Here $[h, g]_n$ is defined by recurrence:

$$[h, g]_n = begincases [[h, g]_n-1, g] & quad g > 0 \ h & quad g = 0 endcases$$

Is it always true that $g$ is bounded left-Engel element iff it belongs to an abelian subnormal subgroup?

Suppose $g$ belongs to an abelian subgroup $H$, which is subnormal in $G$ of length $n$ (We call $H < G$ subnormal of length $n$, iff $exists H_k_k = 0^n$ such that $H_0 = H$, $H_n = G$ and $forall 0 < k < n-1 H_n triangleleft H_n+1$)

Now we will prove, that $g$ is a bounded left-Engel element by induction:

Base: If $n = 0$, then $H = G$ is abelian and the statement is trivially true for any element.

Step: Suppose, it is true for $n-1$. Suppose $H$ is an abelian subgroup, which is subnormal in $G$ of length $n$, and $g in H$. Then there exists $K triangleleft G$, such that $H$ is subnormal in $K$ of length $n - 1$. So, by the supposition of induction $exists k in mathbbN forall h in K [h, g]_k = e$. Now it is sufficient to prove, that $forall h in G [h, g] = (hgh^-1)g in K$, which is rather obvious.

And so we have proved, that every element of an abelian subnormal subgroup is bounded left-Engel.

However, the inverse statement seems to be more difficult, and I do not know how to prove it. I have tried to construct the corresponding subnormal series here: $C_G^n(g) triangleleft C_G^n + 1$?, but, according to what was said in the comments, the series constructed that way are not always subnormal.

abstract-algebra group-theory normal-subgroups

share|cite|improve this question

asked Mar 27 at 16:37

Yanior WegYanior Weg

2,97821549

$endgroup$

share|cite|improve this question

asked Mar 27 at 16:37

Yanior WegYanior Weg

2,97821549

share|cite|improve this question

asked Mar 27 at 16:37

Yanior WegYanior Weg

2,97821549

asked Mar 27 at 16:37

Yanior WegYanior Weg

2,97821549

asked Mar 27 at 16:37

Yanior WegYanior Weg

2,97821549