Mixing
On automorphisms of groups which extend as automorphisms to every larger group
$begingroup$
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 Jan 4 at 1:20
the_fox
2,58711533
asked Jan 3 at 2:10
user521337user521337
9891415
$endgroup$
add a comment |
$begingroup$
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 Jan 4 at 1:20
the_fox
2,58711533
asked Jan 3 at 2:10
user521337user521337
9891415
$endgroup$
1
$begingroup$
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.
$endgroup$
– Arturo Magidin
Jan 3 at 2:20
$begingroup$
@ArturoMagidin: thanks ... that looks interesting ... will take a look
$endgroup$
– user521337
Jan 3 at 2:23
$begingroup$
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)
$endgroup$
– jgon
Jan 3 at 2:51
add a comment |
$begingroup$
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 Jan 4 at 1:20
the_fox
2,58711533
asked Jan 3 at 2:10
user521337user521337
9891415
$endgroup$
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 Jan 4 at 1:20
the_fox
2,58711533
asked Jan 3 at 2:10
user521337user521337
9891415
edited Jan 4 at 1:20
the_fox
2,58711533
asked Jan 3 at 2:10
user521337user521337
9891415
edited Jan 4 at 1:20
the_fox
2,58711533
edited Jan 4 at 1:20
the_fox
2,58711533
edited Jan 4 at 1:20
the_fox
2,58711533
2,58711533
asked Jan 3 at 2:10
user521337user521337
9891415
asked Jan 3 at 2:10
user521337user521337
9891415
asked Jan 3 at 2:10
user521337user521337
9891415
9891415
1
$begingroup$
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.
$endgroup$
– Arturo Magidin
Jan 3 at 2:20
$begingroup$
@ArturoMagidin: thanks ... that looks interesting ... will take a look
$endgroup$
– user521337
Jan 3 at 2:23
$begingroup$
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)
$endgroup$
– jgon
Jan 3 at 2:51
add a comment |
1
$begingroup$
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.
$endgroup$
– Arturo Magidin
Jan 3 at 2:20
$begingroup$
@ArturoMagidin: thanks ... that looks interesting ... will take a look
$endgroup$
– user521337
Jan 3 at 2:23
$begingroup$
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)
$endgroup$
– jgon
Jan 3 at 2:51
1
1
$begingroup$
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.
$endgroup$
– Arturo Magidin
Jan 3 at 2:20
$begingroup$
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.
$endgroup$
– Arturo Magidin
Jan 3 at 2:20
$begingroup$
@ArturoMagidin: thanks ... that looks interesting ... will take a look
$endgroup$
– user521337
Jan 3 at 2:23
$begingroup$
@ArturoMagidin: thanks ... that looks interesting ... will take a look
$endgroup$
– user521337
Jan 3 at 2:23
$begingroup$
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)
$endgroup$
– jgon
Jan 3 at 2:51
$begingroup$
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)
$endgroup$
– jgon
Jan 3 at 2:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 Jan 3 at 22:27
j.p.
9031118
answered Jan 3 at 17:31
verretverret
3,1041921
$endgroup$
$begingroup$
Where can I find a proof (specifically for finite groups) that every (finite) group embeds as a malnormal subgroup of a complete group?
$endgroup$
– the_fox
Jan 3 at 23:42
$begingroup$
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."
$endgroup$
– verret
Jan 3 at 23:55
$begingroup$
I've looked at both, but neither contains a satisfactory answer.
$endgroup$
– the_fox
Jan 4 at 0:09
$begingroup$
Why are the proofs in those papers unsatisfactory?
$endgroup$
– verret
Jan 4 at 0:57
$begingroup$
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.
$endgroup$
– the_fox
Jan 4 at 1:09
|
show 5 more comments
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1 Answer
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1 Answer
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$begingroup$
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 Jan 3 at 22:27
j.p.
9031118
answered Jan 3 at 17:31
verretverret
3,1041921
$endgroup$
$begingroup$
Where can I find a proof (specifically for finite groups) that every (finite) group embeds as a malnormal subgroup of a complete group?
$endgroup$
– the_fox
Jan 3 at 23:42
$begingroup$
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."
$endgroup$
– verret
Jan 3 at 23:55
$begingroup$
I've looked at both, but neither contains a satisfactory answer.
$endgroup$
– the_fox
Jan 4 at 0:09
$begingroup$
Why are the proofs in those papers unsatisfactory?
$endgroup$
– verret
Jan 4 at 0:57
$begingroup$
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.
$endgroup$
– the_fox
Jan 4 at 1:09
|
show 5 more comments
$begingroup$
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 Jan 3 at 22:27
j.p.
9031118
answered Jan 3 at 17:31
verretverret
3,1041921
$endgroup$
$begingroup$
Where can I find a proof (specifically for finite groups) that every (finite) group embeds as a malnormal subgroup of a complete group?
$endgroup$
– the_fox
Jan 3 at 23:42
$begingroup$
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."
$endgroup$
– verret
Jan 3 at 23:55
$begingroup$
I've looked at both, but neither contains a satisfactory answer.
$endgroup$
– the_fox
Jan 4 at 0:09
$begingroup$
Why are the proofs in those papers unsatisfactory?
$endgroup$
– verret
Jan 4 at 0:57
$begingroup$
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.
$endgroup$
– the_fox
Jan 4 at 1:09
|
show 5 more comments
$begingroup$
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 Jan 3 at 22:27
j.p.
9031118
answered Jan 3 at 17:31
verretverret
3,1041921
$endgroup$
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 Jan 3 at 22:27
j.p.
9031118
answered Jan 3 at 17:31
verretverret
3,1041921
edited Jan 3 at 22:27
j.p.
9031118
edited Jan 3 at 22:27
j.p.
9031118
edited Jan 3 at 22:27
j.p.
9031118
9031118
answered Jan 3 at 17:31
verretverret
3,1041921
answered Jan 3 at 17:31
verretverret
3,1041921
answered Jan 3 at 17:31
verretverret
3,1041921
3,1041921
$begingroup$
Where can I find a proof (specifically for finite groups) that every (finite) group embeds as a malnormal subgroup of a complete group?
$endgroup$
– the_fox
Jan 3 at 23:42
$begingroup$
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."
$endgroup$
– verret
Jan 3 at 23:55
$begingroup$
I've looked at both, but neither contains a satisfactory answer.
$endgroup$
– the_fox
Jan 4 at 0:09
$begingroup$
Why are the proofs in those papers unsatisfactory?
$endgroup$
– verret
Jan 4 at 0:57
$begingroup$
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.
$endgroup$
– the_fox
Jan 4 at 1:09
|
show 5 more comments
$begingroup$
Where can I find a proof (specifically for finite groups) that every (finite) group embeds as a malnormal subgroup of a complete group?
$endgroup$
– the_fox
Jan 3 at 23:42
$begingroup$
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."
$endgroup$
– verret
Jan 3 at 23:55
$begingroup$
I've looked at both, but neither contains a satisfactory answer.
$endgroup$
– the_fox
Jan 4 at 0:09
$begingroup$
Why are the proofs in those papers unsatisfactory?
$endgroup$
– verret
Jan 4 at 0:57
$begingroup$
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.
$endgroup$
– the_fox
Jan 4 at 1:09
$begingroup$
Where can I find a proof (specifically for finite groups) that every (finite) group embeds as a malnormal subgroup of a complete group?
$endgroup$
– the_fox
Jan 3 at 23:42
$begingroup$
Where can I find a proof (specifically for finite groups) that every (finite) group embeds as a malnormal subgroup of a complete group?
$endgroup$
– the_fox
Jan 3 at 23:42
$begingroup$
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."
$endgroup$
– verret
Jan 3 at 23:55
$begingroup$
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."
$endgroup$
– verret
Jan 3 at 23:55
$begingroup$
I've looked at both, but neither contains a satisfactory answer.
$endgroup$
– the_fox
Jan 4 at 0:09
$begingroup$
I've looked at both, but neither contains a satisfactory answer.
$endgroup$
– the_fox
Jan 4 at 0:09
$begingroup$
Why are the proofs in those papers unsatisfactory?
$endgroup$
– verret
Jan 4 at 0:57
$begingroup$
Why are the proofs in those papers unsatisfactory?
$endgroup$
– verret
Jan 4 at 0:57
$begingroup$
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.
$endgroup$
– the_fox
Jan 4 at 1:09
$begingroup$
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.
$endgroup$
– the_fox
Jan 4 at 1:09
|
show 5 more comments
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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.
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– Arturo Magidin
Jan 3 at 2:20
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@ArturoMagidin: thanks ... that looks interesting ... will take a look
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– user521337
Jan 3 at 2:23
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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)
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– jgon
Jan 3 at 2:51