On automorphisms of groups which extend as automorphisms to every larger group
For a group $G$, let $operatorname{Aut}(G)$ denote the group of all automorphisms of $G$ and $operatorname{Inn}(G)$ denote the subgroup of all autmorphisms which is of the form $f_h(g)=hgh^{-1}, forall gin G$, where $hin G$ . Now if $G_1$ is a group containing $G$ as a subgroup then every $f_h in operatorname{Inn}(G)$ extends to an inner automorphism of $G'$ as $f_h(x)=hxh^{-1},forall xin G_1$, so in other words, for every $fin operatorname{Inn} (G)$ and every group $G_1$ containing $G$ as a subgroup, $exists bar fin operatorname{Inn}(G_1) subseteq operatorname{Aut} (G_1)$ such that $bar f|_G =f$.
Now my question is : Let $f in operatorname{Aut} (G)$ be such that for every group $G_1$ containing $G$ as a subgroup, $exists bar fin operatorname{Aut}(G_1)$ such that $bar f|_G =f$. Then is it necessarily true that $f in operatorname{Inn}(G)$ ? If this is not true in general, then does some extra condition on $G$ makes it true (like $G$ being finite, or simple)?
group-theory finite-groups simple-groups solvable-groups automorphism-group
edited 21 hours ago
the_fox
2,44411431
asked yesterday
user521337
9781315
add a comment |
For a group $G$, let $operatorname{Aut}(G)$ denote the group of all automorphisms of $G$ and $operatorname{Inn}(G)$ denote the subgroup of all autmorphisms which is of the form $f_h(g)=hgh^{-1}, forall gin G$, where $hin G$ . Now if $G_1$ is a group containing $G$ as a subgroup then every $f_h in operatorname{Inn}(G)$ extends to an inner automorphism of $G'$ as $f_h(x)=hxh^{-1},forall xin G_1$, so in other words, for every $fin operatorname{Inn} (G)$ and every group $G_1$ containing $G$ as a subgroup, $exists bar fin operatorname{Inn}(G_1) subseteq operatorname{Aut} (G_1)$ such that $bar f|_G =f$.
Now my question is : Let $f in operatorname{Aut} (G)$ be such that for every group $G_1$ containing $G$ as a subgroup, $exists bar fin operatorname{Aut}(G_1)$ such that $bar f|_G =f$. Then is it necessarily true that $f in operatorname{Inn}(G)$ ? If this is not true in general, then does some extra condition on $G$ makes it true (like $G$ being finite, or simple)?
group-theory finite-groups simple-groups solvable-groups automorphism-group
edited 21 hours ago
the_fox
2,44411431
asked yesterday
user521337
9781315
1
Take a look at George Bergman’s paper An inner automorphism is only an inner automorphism but an inner endomorphism can be something strange; inner automorphisms can be characterized as precisely the automorphisms $f$ of $G$ that can be extended functiorally to an automorphism of $H$ for any morphism $hcolon Gto H$. This seems related to your condition, so you may be able to use this characterization to show that $f$ is inner.
– Arturo Magidin
yesterday
@ArturoMagidin: thanks ... that looks interesting ... will take a look
– user521337
yesterday
Bergman's paper is very interesting, but I think the condition given there is much stronger than the condition you've given here, so the statement seems likely to either be false or hard to prove. Particularly since he says explicitly that he doesn't know whether the stronger condition that an automorphism be "extensible" (my words not his) along all homomorphisms implies that it is inner, which is why the coherence/functorial condition is necessary. (I'm admittedly quite tired, and the paper is certainly worth reading, but this is intended as a brief summary for future readers)
– jgon
yesterday
add a comment |
For a group $G$, let $operatorname{Aut}(G)$ denote the group of all automorphisms of $G$ and $operatorname{Inn}(G)$ denote the subgroup of all autmorphisms which is of the form $f_h(g)=hgh^{-1}, forall gin G$, where $hin G$ . Now if $G_1$ is a group containing $G$ as a subgroup then every $f_h in operatorname{Inn}(G)$ extends to an inner automorphism of $G'$ as $f_h(x)=hxh^{-1},forall xin G_1$, so in other words, for every $fin operatorname{Inn} (G)$ and every group $G_1$ containing $G$ as a subgroup, $exists bar fin operatorname{Inn}(G_1) subseteq operatorname{Aut} (G_1)$ such that $bar f|_G =f$.
Now my question is : Let $f in operatorname{Aut} (G)$ be such that for every group $G_1$ containing $G$ as a subgroup, $exists bar fin operatorname{Aut}(G_1)$ such that $bar f|_G =f$. Then is it necessarily true that $f in operatorname{Inn}(G)$ ? If this is not true in general, then does some extra condition on $G$ makes it true (like $G$ being finite, or simple)?
group-theory finite-groups simple-groups solvable-groups automorphism-group
edited 21 hours ago
the_fox
2,44411431
asked yesterday
user521337
9781315
For a group $G$, let $operatorname{Aut}(G)$ denote the group of all automorphisms of $G$ and $operatorname{Inn}(G)$ denote the subgroup of all autmorphisms which is of the form $f_h(g)=hgh^{-1}, forall gin G$, where $hin G$ . Now if $G_1$ is a group containing $G$ as a subgroup then every $f_h in operatorname{Inn}(G)$ extends to an inner automorphism of $G'$ as $f_h(x)=hxh^{-1},forall xin G_1$, so in other words, for every $fin operatorname{Inn} (G)$ and every group $G_1$ containing $G$ as a subgroup, $exists bar fin operatorname{Inn}(G_1) subseteq operatorname{Aut} (G_1)$ such that $bar f|_G =f$.
Now my question is : Let $f in operatorname{Aut} (G)$ be such that for every group $G_1$ containing $G$ as a subgroup, $exists bar fin operatorname{Aut}(G_1)$ such that $bar f|_G =f$. Then is it necessarily true that $f in operatorname{Inn}(G)$ ? If this is not true in general, then does some extra condition on $G$ makes it true (like $G$ being finite, or simple)?
group-theory finite-groups simple-groups solvable-groups automorphism-group
group-theory finite-groups simple-groups solvable-groups automorphism-group
edited 21 hours ago
the_fox
2,44411431
asked yesterday
user521337
9781315
edited 21 hours ago
the_fox
2,44411431
asked yesterday
user521337
9781315
edited 21 hours ago
the_fox
2,44411431
edited 21 hours ago
the_fox
2,44411431
edited 21 hours ago
the_fox
2,44411431
2,44411431
asked yesterday
user521337
9781315
asked yesterday
user521337
9781315
asked yesterday
user521337
9781315
9781315
1
Take a look at George Bergman’s paper An inner automorphism is only an inner automorphism but an inner endomorphism can be something strange; inner automorphisms can be characterized as precisely the automorphisms $f$ of $G$ that can be extended functiorally to an automorphism of $H$ for any morphism $hcolon Gto H$. This seems related to your condition, so you may be able to use this characterization to show that $f$ is inner.
– Arturo Magidin
yesterday
@ArturoMagidin: thanks ... that looks interesting ... will take a look
– user521337
yesterday
Bergman's paper is very interesting, but I think the condition given there is much stronger than the condition you've given here, so the statement seems likely to either be false or hard to prove. Particularly since he says explicitly that he doesn't know whether the stronger condition that an automorphism be "extensible" (my words not his) along all homomorphisms implies that it is inner, which is why the coherence/functorial condition is necessary. (I'm admittedly quite tired, and the paper is certainly worth reading, but this is intended as a brief summary for future readers)
– jgon
yesterday
add a comment |
1
Take a look at George Bergman’s paper An inner automorphism is only an inner automorphism but an inner endomorphism can be something strange; inner automorphisms can be characterized as precisely the automorphisms $f$ of $G$ that can be extended functiorally to an automorphism of $H$ for any morphism $hcolon Gto H$. This seems related to your condition, so you may be able to use this characterization to show that $f$ is inner.
– Arturo Magidin
yesterday
@ArturoMagidin: thanks ... that looks interesting ... will take a look
– user521337
yesterday
Bergman's paper is very interesting, but I think the condition given there is much stronger than the condition you've given here, so the statement seems likely to either be false or hard to prove. Particularly since he says explicitly that he doesn't know whether the stronger condition that an automorphism be "extensible" (my words not his) along all homomorphisms implies that it is inner, which is why the coherence/functorial condition is necessary. (I'm admittedly quite tired, and the paper is certainly worth reading, but this is intended as a brief summary for future readers)
– jgon
yesterday
1
1
Take a look at George Bergman’s paper An inner automorphism is only an inner automorphism but an inner endomorphism can be something strange; inner automorphisms can be characterized as precisely the automorphisms $f$ of $G$ that can be extended functiorally to an automorphism of $H$ for any morphism $hcolon Gto H$. This seems related to your condition, so you may be able to use this characterization to show that $f$ is inner.
– Arturo Magidin
yesterday
Take a look at George Bergman’s paper An inner automorphism is only an inner automorphism but an inner endomorphism can be something strange; inner automorphisms can be characterized as precisely the automorphisms $f$ of $G$ that can be extended functiorally to an automorphism of $H$ for any morphism $hcolon Gto H$. This seems related to your condition, so you may be able to use this characterization to show that $f$ is inner.
– Arturo Magidin
yesterday
@ArturoMagidin: thanks ... that looks interesting ... will take a look
– user521337
yesterday
@ArturoMagidin: thanks ... that looks interesting ... will take a look
– user521337
yesterday
Bergman's paper is very interesting, but I think the condition given there is much stronger than the condition you've given here, so the statement seems likely to either be false or hard to prove. Particularly since he says explicitly that he doesn't know whether the stronger condition that an automorphism be "extensible" (my words not his) along all homomorphisms implies that it is inner, which is why the coherence/functorial condition is necessary. (I'm admittedly quite tired, and the paper is certainly worth reading, but this is intended as a brief summary for future readers)
– jgon
yesterday
Bergman's paper is very interesting, but I think the condition given there is much stronger than the condition you've given here, so the statement seems likely to either be false or hard to prove. Particularly since he says explicitly that he doesn't know whether the stronger condition that an automorphism be "extensible" (my words not his) along all homomorphisms implies that it is inner, which is why the coherence/functorial condition is necessary. (I'm admittedly quite tired, and the paper is certainly worth reading, but this is intended as a brief summary for future readers)
– jgon
yesterday
add a comment |
1 Answer
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How about this:
A Characterization of Inner Automorphisms
Paul E. Schupp
Proceedings of the American Mathematical Society
Vol. 101, No. 2 (Oct., 1987), pp. 226-228
https://www.jstor.org/stable/2045986?seq=1#page_scan_tab_contents
Abstract: It turns out that one can characterize inner automorphisms without mentioning either conjugation or specific elements. We prove the following
THEOREM Let $G$ be a group and let $alpha$ an automorphism of $G$. The automorphism $alpha$ is an inner automorphism of $G$ if and only if $alpha$ has the property that whenever $G$ is embedded in a group $H$, then $alpha$ extends to some automorphism of $H$.
edited yesterday
j.p.
9011118
answered yesterday
verret
2,9541818
Where can I find a proof (specifically for finite groups) that every (finite) group embeds as a malnormal subgroup of a complete group?
– the_fox
23 hours ago
Well, the paper in the answer has the proof in the general case. Apparently, this was proved earlier in the finite case in "Miller, Charles F., III; Schupp, Paul E. Embeddings into Hopfian groups. J. Algebra 17 1971 171–176."
– verret
23 hours ago
I've looked at both, but neither contains a satisfactory answer.
– the_fox
23 hours ago
Why are the proofs in those papers unsatisfactory?
– verret
22 hours ago
They are not satisfactory to me because I am only interested in finite groups. The methods and techniques involved are often very different from the infinite counterpart.
– the_fox
22 hours ago
|
show 4 more comments
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1 Answer
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How about this:
A Characterization of Inner Automorphisms
Paul E. Schupp
Proceedings of the American Mathematical Society
Vol. 101, No. 2 (Oct., 1987), pp. 226-228
https://www.jstor.org/stable/2045986?seq=1#page_scan_tab_contents
Abstract: It turns out that one can characterize inner automorphisms without mentioning either conjugation or specific elements. We prove the following
THEOREM Let $G$ be a group and let $alpha$ an automorphism of $G$. The automorphism $alpha$ is an inner automorphism of $G$ if and only if $alpha$ has the property that whenever $G$ is embedded in a group $H$, then $alpha$ extends to some automorphism of $H$.
edited yesterday
j.p.
9011118
answered yesterday
verret
2,9541818
Where can I find a proof (specifically for finite groups) that every (finite) group embeds as a malnormal subgroup of a complete group?
– the_fox
23 hours ago
Well, the paper in the answer has the proof in the general case. Apparently, this was proved earlier in the finite case in "Miller, Charles F., III; Schupp, Paul E. Embeddings into Hopfian groups. J. Algebra 17 1971 171–176."
– verret
23 hours ago
I've looked at both, but neither contains a satisfactory answer.
– the_fox
23 hours ago
Why are the proofs in those papers unsatisfactory?
– verret
22 hours ago
They are not satisfactory to me because I am only interested in finite groups. The methods and techniques involved are often very different from the infinite counterpart.
– the_fox
22 hours ago
|
show 4 more comments
How about this:
A Characterization of Inner Automorphisms
Paul E. Schupp
Proceedings of the American Mathematical Society
Vol. 101, No. 2 (Oct., 1987), pp. 226-228
https://www.jstor.org/stable/2045986?seq=1#page_scan_tab_contents
Abstract: It turns out that one can characterize inner automorphisms without mentioning either conjugation or specific elements. We prove the following
THEOREM Let $G$ be a group and let $alpha$ an automorphism of $G$. The automorphism $alpha$ is an inner automorphism of $G$ if and only if $alpha$ has the property that whenever $G$ is embedded in a group $H$, then $alpha$ extends to some automorphism of $H$.
edited yesterday
j.p.
9011118
answered yesterday
verret
2,9541818
Where can I find a proof (specifically for finite groups) that every (finite) group embeds as a malnormal subgroup of a complete group?
– the_fox
23 hours ago
Well, the paper in the answer has the proof in the general case. Apparently, this was proved earlier in the finite case in "Miller, Charles F., III; Schupp, Paul E. Embeddings into Hopfian groups. J. Algebra 17 1971 171–176."
– verret
23 hours ago
I've looked at both, but neither contains a satisfactory answer.
– the_fox
23 hours ago
Why are the proofs in those papers unsatisfactory?
– verret
22 hours ago
They are not satisfactory to me because I am only interested in finite groups. The methods and techniques involved are often very different from the infinite counterpart.
– the_fox
22 hours ago
|
show 4 more comments
How about this:
A Characterization of Inner Automorphisms
Paul E. Schupp
Proceedings of the American Mathematical Society
Vol. 101, No. 2 (Oct., 1987), pp. 226-228
https://www.jstor.org/stable/2045986?seq=1#page_scan_tab_contents
Abstract: It turns out that one can characterize inner automorphisms without mentioning either conjugation or specific elements. We prove the following
THEOREM Let $G$ be a group and let $alpha$ an automorphism of $G$. The automorphism $alpha$ is an inner automorphism of $G$ if and only if $alpha$ has the property that whenever $G$ is embedded in a group $H$, then $alpha$ extends to some automorphism of $H$.
edited yesterday
j.p.
9011118
answered yesterday
verret
2,9541818
How about this:
A Characterization of Inner Automorphisms
Paul E. Schupp
Proceedings of the American Mathematical Society
Vol. 101, No. 2 (Oct., 1987), pp. 226-228
https://www.jstor.org/stable/2045986?seq=1#page_scan_tab_contents
Abstract: It turns out that one can characterize inner automorphisms without mentioning either conjugation or specific elements. We prove the following
THEOREM Let $G$ be a group and let $alpha$ an automorphism of $G$. The automorphism $alpha$ is an inner automorphism of $G$ if and only if $alpha$ has the property that whenever $G$ is embedded in a group $H$, then $alpha$ extends to some automorphism of $H$.
edited yesterday
j.p.
9011118
answered yesterday
verret
2,9541818
edited yesterday
j.p.
9011118
edited yesterday
j.p.
9011118
edited yesterday
j.p.
9011118
9011118
answered yesterday
verret
2,9541818
answered yesterday
verret
2,9541818
answered yesterday
verret
2,9541818
2,9541818
Where can I find a proof (specifically for finite groups) that every (finite) group embeds as a malnormal subgroup of a complete group?
– the_fox
23 hours ago
Well, the paper in the answer has the proof in the general case. Apparently, this was proved earlier in the finite case in "Miller, Charles F., III; Schupp, Paul E. Embeddings into Hopfian groups. J. Algebra 17 1971 171–176."
– verret
23 hours ago
I've looked at both, but neither contains a satisfactory answer.
– the_fox
23 hours ago
Why are the proofs in those papers unsatisfactory?
– verret
22 hours ago
They are not satisfactory to me because I am only interested in finite groups. The methods and techniques involved are often very different from the infinite counterpart.
– the_fox
22 hours ago
|
show 4 more comments
Where can I find a proof (specifically for finite groups) that every (finite) group embeds as a malnormal subgroup of a complete group?
– the_fox
23 hours ago
Well, the paper in the answer has the proof in the general case. Apparently, this was proved earlier in the finite case in "Miller, Charles F., III; Schupp, Paul E. Embeddings into Hopfian groups. J. Algebra 17 1971 171–176."
– verret
23 hours ago
I've looked at both, but neither contains a satisfactory answer.
– the_fox
23 hours ago
Why are the proofs in those papers unsatisfactory?
– verret
22 hours ago
They are not satisfactory to me because I am only interested in finite groups. The methods and techniques involved are often very different from the infinite counterpart.
– the_fox
22 hours ago
Where can I find a proof (specifically for finite groups) that every (finite) group embeds as a malnormal subgroup of a complete group?
– the_fox
23 hours ago
Where can I find a proof (specifically for finite groups) that every (finite) group embeds as a malnormal subgroup of a complete group?
– the_fox
23 hours ago
Well, the paper in the answer has the proof in the general case. Apparently, this was proved earlier in the finite case in "Miller, Charles F., III; Schupp, Paul E. Embeddings into Hopfian groups. J. Algebra 17 1971 171–176."
– verret
23 hours ago
Well, the paper in the answer has the proof in the general case. Apparently, this was proved earlier in the finite case in "Miller, Charles F., III; Schupp, Paul E. Embeddings into Hopfian groups. J. Algebra 17 1971 171–176."
– verret
23 hours ago
I've looked at both, but neither contains a satisfactory answer.
– the_fox
23 hours ago
I've looked at both, but neither contains a satisfactory answer.
– the_fox
23 hours ago
Why are the proofs in those papers unsatisfactory?
– verret
22 hours ago
Why are the proofs in those papers unsatisfactory?
– verret
22 hours ago
They are not satisfactory to me because I am only interested in finite groups. The methods and techniques involved are often very different from the infinite counterpart.
– the_fox
22 hours ago
They are not satisfactory to me because I am only interested in finite groups. The methods and techniques involved are often very different from the infinite counterpart.
– the_fox
22 hours ago
|
show 4 more comments
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1
Take a look at George Bergman’s paper An inner automorphism is only an inner automorphism but an inner endomorphism can be something strange; inner automorphisms can be characterized as precisely the automorphisms $f$ of $G$ that can be extended functiorally to an automorphism of $H$ for any morphism $hcolon Gto H$. This seems related to your condition, so you may be able to use this characterization to show that $f$ is inner.
– Arturo Magidin
yesterday
@ArturoMagidin: thanks ... that looks interesting ... will take a look
– user521337
yesterday
Bergman's paper is very interesting, but I think the condition given there is much stronger than the condition you've given here, so the statement seems likely to either be false or hard to prove. Particularly since he says explicitly that he doesn't know whether the stronger condition that an automorphism be "extensible" (my words not his) along all homomorphisms implies that it is inner, which is why the coherence/functorial condition is necessary. (I'm admittedly quite tired, and the paper is certainly worth reading, but this is intended as a brief summary for future readers)
– jgon
yesterday