Skip to main content
minor latex edit
Source Link
Myshkin
  • 17.6k
  • 5
  • 71
  • 137

I am thinking the following question: for a group that splits as a free product $G = G_1 \ast F_n $, where $G_1$ is any infinite group (preferebly freely indecomposable) and we denote by $F_n$ the free group on $n$ generators. Can we find an (outer) automorhism $\phi$ of $G$ with $\phi(G_1) = G_1$ which also satisfies the properties:

  1. there is no $\phi$-invariant (up to cojugacy) proper free factor $H$ of $G$ which strictly contains $G_1$ and

  2. there is no any free product decomposition of $G$, $G = G_1 \ast H$, where $ H = <w, v>$ for some elements $w,v$ of $G$ with $\phi(H) = H$.

For example for the automorphism of $G = G_1 \ast <a> \ast <b>$ (where $<a> \ast <b>$ is isomorphic to F_2$F_2$) which induces the identity on $G_1$ and sends $a$ to $ab$ and to $b$ to $ag$, where $g \in G_1$, is it possible to prove that it is satisfies these properties?

In general, is there any such automorphism? Is there anything like this in the literature?

I am thinking the following question: for a group that splits as a free product $G = G_1 \ast F_n $, where $G_1$ is any infinite group (preferebly freely indecomposable) and we denote by $F_n$ the free group on $n$ generators. Can we find an (outer) automorhism $\phi$ of $G$ with $\phi(G_1) = G_1$ which also satisfies the properties:

  1. there is no $\phi$-invariant (up to cojugacy) proper free factor $H$ of $G$ which strictly contains $G_1$ and

  2. there is no any free product decomposition of $G$, $G = G_1 \ast H$, where $ H = <w, v>$ for some elements $w,v$ of $G$ with $\phi(H) = H$.

For example for the automorphism of $G = G_1 \ast <a> \ast <b>$ (where $<a> \ast <b>$ is isomorphic to F_2) which induces the identity on $G_1$ and sends $a$ to $ab$ and to $b$ to $ag$, where $g \in G_1$, is it possible to prove that it is satisfies these properties?

In general, is there any such automorphism? Is there anything like this in the literature?

I am thinking the following question: for a group that splits as a free product $G = G_1 \ast F_n $, where $G_1$ is any infinite group (preferebly freely indecomposable) and we denote by $F_n$ the free group on $n$ generators. Can we find an (outer) automorhism $\phi$ of $G$ with $\phi(G_1) = G_1$ which also satisfies the properties:

  1. there is no $\phi$-invariant (up to cojugacy) proper free factor $H$ of $G$ which strictly contains $G_1$ and

  2. there is no any free product decomposition of $G$, $G = G_1 \ast H$, where $ H = <w, v>$ for some elements $w,v$ of $G$ with $\phi(H) = H$.

For example for the automorphism of $G = G_1 \ast <a> \ast <b>$ (where $<a> \ast <b>$ is isomorphic to $F_2$) which induces the identity on $G_1$ and sends $a$ to $ab$ and to $b$ to $ag$, where $g \in G_1$, is it possible to prove that it is satisfies these properties?

In general, is there any such automorphism? Is there anything like this in the literature?

deleted 2 characters in body; edited title
Source Link
user75691
  • 339
  • 1
  • 7

Invariant subgroupsfree factors for automorphisms of free products

Invariant subgroupsfree factors for automorphisms of free products

I am thinking the following question: for a group that splits as a free product $G = G_1 \ast F_n $, where $G_1$ is any infinite group (preferebly freely indecomposable) and we denote by $F_n$ the free group on $n$ generators. Can we find an (outer) automorhism $\phi$ of $G$ with $\phi(G_1) = G_1$ which also satisfies the properties:

  1. there is no $\phi$-invariant (up to cojugacy) proper free factor $H$ of $G$ which strictly contains $G_1$ and

  2. there is no any free product decomposition of $G$, $G = G_1 \ast H$, where $ H = <w, v>$ for some elements $w,v$ of $G$ with $\phi(H) = H$.

For example for the automorphism of $G = G_1 \ast <a> \ast <b>$ (where $<a> \ast <b>$ is isomorphic to F_2) which induces the identity on $G_1$ and sends $a$ to $ab$ and to $b$ to $ag$, where $g \in G_1$, is it possible to prove that it is satisfies these properties?

In general, is there any such automorphism? Is there anything like this in the literature?

Invariant subgroups for automorphisms of free products

I am thinking the following question: for a group that splits as a free product $G = G_1 \ast F_n $, where $G_1$ is any infinite group (preferebly freely indecomposable) and we denote by $F_n$ the free group on $n$ generators. Can we find an (outer) automorhism $\phi$ of $G$ with $\phi(G_1) = G_1$ which also satisfies the properties:

  1. there is no $\phi$-invariant (up to cojugacy) proper free factor $H$ of $G$ which strictly contains $G_1$ and

  2. there is no any free product decomposition of $G$, $G = G_1 \ast H$, where $ H = <w, v>$ for some elements $w,v$ of $G$ with $\phi(H) = H$.

For example for the automorphism of $G = G_1 \ast <a> \ast <b>$ (where $<a> \ast <b>$ is isomorphic to F_2) which induces the identity on $G_1$ and sends $a$ to $ab$ and to $b$ to $ag$, where $g \in G_1$, is it possible to prove that it is satisfies these properties?

In general, is there any such automorphism? Is there anything like this in the literature?

Invariant free factors for automorphisms of free products

I am thinking the following question: for a group that splits as a free product $G = G_1 \ast F_n $, where $G_1$ is any infinite group (preferebly freely indecomposable) and we denote by $F_n$ the free group on $n$ generators. Can we find an (outer) automorhism $\phi$ of $G$ with $\phi(G_1) = G_1$ which also satisfies the properties:

  1. there is no $\phi$-invariant (up to cojugacy) proper free factor $H$ of $G$ which strictly contains $G_1$ and

  2. there is no any free product decomposition of $G$, $G = G_1 \ast H$, where $ H = <w, v>$ for some elements $w,v$ of $G$ with $\phi(H) = H$.

For example for the automorphism of $G = G_1 \ast <a> \ast <b>$ (where $<a> \ast <b>$ is isomorphic to F_2) which induces the identity on $G_1$ and sends $a$ to $ab$ and to $b$ to $ag$, where $g \in G_1$, is it possible to prove that it is satisfies these properties?

In general, is there any such automorphism? Is there anything like this in the literature?

deleted 2 characters in body; edited title
Source Link
user75691
  • 339
  • 1
  • 7

Invariant subgroups for automorphisms of directfree products

Invariant subgroups for automorphisms of directfree products

I am thinking the following question: for a group that splits as a free product $G = G_1 \ast F_n $, where $G_1$ is any infinite group (preferebly freely indecomposable) and we denote by $F_n$ the free group on $n$ generators. Can we find an (outer) automorhism $\phi$ of $G$ with $\phi(G_1) = G_1$ which also satisfies the properties:

  1. there is no $\phi$-invariant (up to cojugacy) proper free factor $H$ of $G$ which strictly contains $G_1$ and

  2. there is no any free product decomposition of $G$, $G = G_1 \ast H$, where $ H = <w, v>$ for some elements $w,v$ of $G$ with $\phi(H) = H$.

For example for the automorphism of $G = G_1 \ast <a> \ast <b>$ (where $<a> \ast <b>$ is isomorphic to F_2) which induces the identity on $G_1$ and sends $a$ to $ab$ and to $b$ to $ag$, where $g \in G_1$, is it possible to prove that it is satisfies these properties?

In general, is there any such automorphism? Is there anything like this in the literature?

Invariant subgroups for automorphisms of direct products

I am thinking the following question: for a group that splits as a free product $G = G_1 \ast F_n $, where $G_1$ is any infinite group (preferebly freely indecomposable) and we denote by $F_n$ the free group on $n$ generators. Can we find an (outer) automorhism $\phi$ of $G$ with $\phi(G_1) = G_1$ which also satisfies the properties:

  1. there is no $\phi$-invariant (up to cojugacy) proper free factor $H$ of $G$ which strictly contains $G_1$ and

  2. there is no any free product decomposition of $G$, $G = G_1 \ast H$, where $ H = <w, v>$ for some elements $w,v$ of $G$ with $\phi(H) = H$.

For example for the automorphism of $G = G_1 \ast <a> \ast <b>$ (where $<a> \ast <b>$ is isomorphic to F_2) which induces the identity on $G_1$ and sends $a$ to $ab$ and to $b$ to $ag$, where $g \in G_1$, is it possible to prove that it is satisfies these properties?

In general, is there any such automorphism? Is there anything like this in the literature?

Invariant subgroups for automorphisms of free products

I am thinking the following question: for a group that splits as a free product $G = G_1 \ast F_n $, where $G_1$ is any infinite group (preferebly freely indecomposable) and we denote by $F_n$ the free group on $n$ generators. Can we find an (outer) automorhism $\phi$ of $G$ with $\phi(G_1) = G_1$ which also satisfies the properties:

  1. there is no $\phi$-invariant (up to cojugacy) proper free factor $H$ of $G$ which strictly contains $G_1$ and

  2. there is no any free product decomposition of $G$, $G = G_1 \ast H$, where $ H = <w, v>$ for some elements $w,v$ of $G$ with $\phi(H) = H$.

For example for the automorphism of $G = G_1 \ast <a> \ast <b>$ (where $<a> \ast <b>$ is isomorphic to F_2) which induces the identity on $G_1$ and sends $a$ to $ab$ and to $b$ to $ag$, where $g \in G_1$, is it possible to prove that it is satisfies these properties?

In general, is there any such automorphism? Is there anything like this in the literature?

Source Link
user75691
  • 339
  • 1
  • 7
Loading