Mixing
Hölder's theorem (Group Theory). Proof collection and applications
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
In (finite) group theory, the number $6$ seems to be special in the sense that we have the result of Hölder that this is the only natural number $n=6$ such that there exists an outer automorphism of $S_n$. My question is twofolds
I would like a collection of possible proofs of this result. For instance, I would like to also know if there is a proof that is purely number theoretic or a non-constructive proof of this result. I don't know if Hölder originally proved this constructively, but most of the proof that I see are constructive (and more or less the same). One starts with a transitive subgroup of $S_6$ of order 120 (this will be isomorphic to $S_5$)
I would like to know of any (easy?) application of this result leading to some result in group theory where the number $6$ stands out just because of this result.
I would appreciate any ideas.
group-theory reference-request finite-groups permutations automorphism-group
asked Jul 18 at 12:14
quantum
40429
add a comment |Â
up vote
4
down vote
favorite
In (finite) group theory, the number $6$ seems to be special in the sense that we have the result of Hölder that this is the only natural number $n=6$ such that there exists an outer automorphism of $S_n$. My question is twofolds
I would like a collection of possible proofs of this result. For instance, I would like to also know if there is a proof that is purely number theoretic or a non-constructive proof of this result. I don't know if Hölder originally proved this constructively, but most of the proof that I see are constructive (and more or less the same). One starts with a transitive subgroup of $S_6$ of order 120 (this will be isomorphic to $S_5$)
I would like to know of any (easy?) application of this result leading to some result in group theory where the number $6$ stands out just because of this result.
I would appreciate any ideas.
group-theory reference-request finite-groups permutations automorphism-group
asked Jul 18 at 12:14
quantum
40429
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
In (finite) group theory, the number $6$ seems to be special in the sense that we have the result of Hölder that this is the only natural number $n=6$ such that there exists an outer automorphism of $S_n$. My question is twofolds
I would like a collection of possible proofs of this result. For instance, I would like to also know if there is a proof that is purely number theoretic or a non-constructive proof of this result. I don't know if Hölder originally proved this constructively, but most of the proof that I see are constructive (and more or less the same). One starts with a transitive subgroup of $S_6$ of order 120 (this will be isomorphic to $S_5$)
I would like to know of any (easy?) application of this result leading to some result in group theory where the number $6$ stands out just because of this result.
I would appreciate any ideas.
group-theory reference-request finite-groups permutations automorphism-group
asked Jul 18 at 12:14
quantum
40429
In (finite) group theory, the number $6$ seems to be special in the sense that we have the result of Hölder that this is the only natural number $n=6$ such that there exists an outer automorphism of $S_n$. My question is twofolds
I would like a collection of possible proofs of this result. For instance, I would like to also know if there is a proof that is purely number theoretic or a non-constructive proof of this result. I don't know if Hölder originally proved this constructively, but most of the proof that I see are constructive (and more or less the same). One starts with a transitive subgroup of $S_6$ of order 120 (this will be isomorphic to $S_5$)
I would like to know of any (easy?) application of this result leading to some result in group theory where the number $6$ stands out just because of this result.
I would appreciate any ideas.
group-theory reference-request finite-groups permutations automorphism-group
asked Jul 18 at 12:14
quantum
40429
edited Jul 18 at 13:10
asked Jul 18 at 12:14
quantum
40429
asked Jul 18 at 12:14
quantum
40429
asked Jul 18 at 12:14
quantum
40429
40429
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Here is a nice direct application: subgroups of small index in $S_n$. Assume that $ngeq 5$, and let $Hleq S_n$ be a subgroup of index at most $n$. Then either $H$ is $A_n, S_n$, or a stabilizer of some element $1leq i leq n$ in $S_n$, except if $n=6$.
If $n=6$, then $S_6$ can be viewed as the group of all permutation of the 6 elements of the projective line over $mathbbF_5$. But the group of projective transformations $PGL(2,5)$ form an index 6 subgroup.
The connection is: if $Hleq S_n$ with index $k$, then the action of $S_n$ on the left cosets of $H$ by left multiplication defines a homomorphism $varphi: S_nrightarrow S_k$. If $k< n$, then $varphi$ cannot be injective. But as $ngeq 5$, the only possible kernels are $A_n, S_n$, yielding these subgroups of small indices.
If $k=n$, then $varphi$ can be injective. However, in that case it has to be surjective. So $varphiin Aut(S_n)$. Inner automorphims correspond to the stabilizers. Outer automorphisms correspond to different index $n$ subgroups. (Think about it.)
This line of thought can be extended to obtain simple proofs to results such as $PGL(2,5)cong S_5$ and $PSL(2,5)cong A_5$.
answered Jul 18 at 12:34
A. Pongrácz
2,574321
add a comment |Â
up vote
3
down vote
There are at least two fundamentally different ways to get the exceptional outer automorphism of $S_6$. One is to find the permutation representation of $S_5$ on six letters, as already alluded to. The other is to notice that $A_6cong PSL(2,9)$ is an index four normal subgroup of $PGamma L(2,9)$. The quotient is the Klein four group, and the intermediate index two subgroups are of distinct isomorphism types: One is $S_6$, one is $PGL(2,9)$, and one is $M_10$ (a point stabilizer in $M_11$). Since they are pairwise non-isomorphic, they are characteristic subgroups, so since $PGamma L(2,9)$ has no order $2$ normal subgroup (and since $S_6$ has trivial center) it acts faithfully by conjugation on $S_6$, and thus embeds monomorphically into $Aut(S_6)$.
So it boils down to whether you like $BbbF_5$ or $BbbF_9$ better.
If you are unfamiliar with the notation, $Gamma L(n,k)$ is the group of semilinear automorphisms of $k^n$, where a semilinear map $f$ satisfies $f(u+v)=f(u)+f(v)$ and $f(av)=sigma_f(a)f(v)$ for some $sigma_fin Aut(k)$. $PGamma L(n,k)$ is just that group mod the scalar matrices.
answered Jul 19 at 2:50
C Monsour
4,511221
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Here is a nice direct application: subgroups of small index in $S_n$. Assume that $ngeq 5$, and let $Hleq S_n$ be a subgroup of index at most $n$. Then either $H$ is $A_n, S_n$, or a stabilizer of some element $1leq i leq n$ in $S_n$, except if $n=6$.
If $n=6$, then $S_6$ can be viewed as the group of all permutation of the 6 elements of the projective line over $mathbbF_5$. But the group of projective transformations $PGL(2,5)$ form an index 6 subgroup.
The connection is: if $Hleq S_n$ with index $k$, then the action of $S_n$ on the left cosets of $H$ by left multiplication defines a homomorphism $varphi: S_nrightarrow S_k$. If $k< n$, then $varphi$ cannot be injective. But as $ngeq 5$, the only possible kernels are $A_n, S_n$, yielding these subgroups of small indices.
If $k=n$, then $varphi$ can be injective. However, in that case it has to be surjective. So $varphiin Aut(S_n)$. Inner automorphims correspond to the stabilizers. Outer automorphisms correspond to different index $n$ subgroups. (Think about it.)
This line of thought can be extended to obtain simple proofs to results such as $PGL(2,5)cong S_5$ and $PSL(2,5)cong A_5$.
answered Jul 18 at 12:34
A. Pongrácz
2,574321
add a comment |Â
up vote
4
down vote
accepted
Here is a nice direct application: subgroups of small index in $S_n$. Assume that $ngeq 5$, and let $Hleq S_n$ be a subgroup of index at most $n$. Then either $H$ is $A_n, S_n$, or a stabilizer of some element $1leq i leq n$ in $S_n$, except if $n=6$.
If $n=6$, then $S_6$ can be viewed as the group of all permutation of the 6 elements of the projective line over $mathbbF_5$. But the group of projective transformations $PGL(2,5)$ form an index 6 subgroup.
The connection is: if $Hleq S_n$ with index $k$, then the action of $S_n$ on the left cosets of $H$ by left multiplication defines a homomorphism $varphi: S_nrightarrow S_k$. If $k< n$, then $varphi$ cannot be injective. But as $ngeq 5$, the only possible kernels are $A_n, S_n$, yielding these subgroups of small indices.
If $k=n$, then $varphi$ can be injective. However, in that case it has to be surjective. So $varphiin Aut(S_n)$. Inner automorphims correspond to the stabilizers. Outer automorphisms correspond to different index $n$ subgroups. (Think about it.)
This line of thought can be extended to obtain simple proofs to results such as $PGL(2,5)cong S_5$ and $PSL(2,5)cong A_5$.
answered Jul 18 at 12:34
A. Pongrácz
2,574321
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Here is a nice direct application: subgroups of small index in $S_n$. Assume that $ngeq 5$, and let $Hleq S_n$ be a subgroup of index at most $n$. Then either $H$ is $A_n, S_n$, or a stabilizer of some element $1leq i leq n$ in $S_n$, except if $n=6$.
If $n=6$, then $S_6$ can be viewed as the group of all permutation of the 6 elements of the projective line over $mathbbF_5$. But the group of projective transformations $PGL(2,5)$ form an index 6 subgroup.
The connection is: if $Hleq S_n$ with index $k$, then the action of $S_n$ on the left cosets of $H$ by left multiplication defines a homomorphism $varphi: S_nrightarrow S_k$. If $k< n$, then $varphi$ cannot be injective. But as $ngeq 5$, the only possible kernels are $A_n, S_n$, yielding these subgroups of small indices.
If $k=n$, then $varphi$ can be injective. However, in that case it has to be surjective. So $varphiin Aut(S_n)$. Inner automorphims correspond to the stabilizers. Outer automorphisms correspond to different index $n$ subgroups. (Think about it.)
This line of thought can be extended to obtain simple proofs to results such as $PGL(2,5)cong S_5$ and $PSL(2,5)cong A_5$.
answered Jul 18 at 12:34
A. Pongrácz
2,574321
Here is a nice direct application: subgroups of small index in $S_n$. Assume that $ngeq 5$, and let $Hleq S_n$ be a subgroup of index at most $n$. Then either $H$ is $A_n, S_n$, or a stabilizer of some element $1leq i leq n$ in $S_n$, except if $n=6$.
If $n=6$, then $S_6$ can be viewed as the group of all permutation of the 6 elements of the projective line over $mathbbF_5$. But the group of projective transformations $PGL(2,5)$ form an index 6 subgroup.
The connection is: if $Hleq S_n$ with index $k$, then the action of $S_n$ on the left cosets of $H$ by left multiplication defines a homomorphism $varphi: S_nrightarrow S_k$. If $k< n$, then $varphi$ cannot be injective. But as $ngeq 5$, the only possible kernels are $A_n, S_n$, yielding these subgroups of small indices.
If $k=n$, then $varphi$ can be injective. However, in that case it has to be surjective. So $varphiin Aut(S_n)$. Inner automorphims correspond to the stabilizers. Outer automorphisms correspond to different index $n$ subgroups. (Think about it.)
This line of thought can be extended to obtain simple proofs to results such as $PGL(2,5)cong S_5$ and $PSL(2,5)cong A_5$.
answered Jul 18 at 12:34
A. Pongrácz
2,574321
answered Jul 18 at 12:34
A. Pongrácz
2,574321
answered Jul 18 at 12:34
A. Pongrácz
2,574321
answered Jul 18 at 12:34
A. Pongrácz
2,574321
2,574321
add a comment |Â
add a comment |Â
up vote
3
down vote
There are at least two fundamentally different ways to get the exceptional outer automorphism of $S_6$. One is to find the permutation representation of $S_5$ on six letters, as already alluded to. The other is to notice that $A_6cong PSL(2,9)$ is an index four normal subgroup of $PGamma L(2,9)$. The quotient is the Klein four group, and the intermediate index two subgroups are of distinct isomorphism types: One is $S_6$, one is $PGL(2,9)$, and one is $M_10$ (a point stabilizer in $M_11$). Since they are pairwise non-isomorphic, they are characteristic subgroups, so since $PGamma L(2,9)$ has no order $2$ normal subgroup (and since $S_6$ has trivial center) it acts faithfully by conjugation on $S_6$, and thus embeds monomorphically into $Aut(S_6)$.
So it boils down to whether you like $BbbF_5$ or $BbbF_9$ better.
If you are unfamiliar with the notation, $Gamma L(n,k)$ is the group of semilinear automorphisms of $k^n$, where a semilinear map $f$ satisfies $f(u+v)=f(u)+f(v)$ and $f(av)=sigma_f(a)f(v)$ for some $sigma_fin Aut(k)$. $PGamma L(n,k)$ is just that group mod the scalar matrices.
answered Jul 19 at 2:50
C Monsour
4,511221
add a comment |Â
up vote
3
down vote
There are at least two fundamentally different ways to get the exceptional outer automorphism of $S_6$. One is to find the permutation representation of $S_5$ on six letters, as already alluded to. The other is to notice that $A_6cong PSL(2,9)$ is an index four normal subgroup of $PGamma L(2,9)$. The quotient is the Klein four group, and the intermediate index two subgroups are of distinct isomorphism types: One is $S_6$, one is $PGL(2,9)$, and one is $M_10$ (a point stabilizer in $M_11$). Since they are pairwise non-isomorphic, they are characteristic subgroups, so since $PGamma L(2,9)$ has no order $2$ normal subgroup (and since $S_6$ has trivial center) it acts faithfully by conjugation on $S_6$, and thus embeds monomorphically into $Aut(S_6)$.
So it boils down to whether you like $BbbF_5$ or $BbbF_9$ better.
If you are unfamiliar with the notation, $Gamma L(n,k)$ is the group of semilinear automorphisms of $k^n$, where a semilinear map $f$ satisfies $f(u+v)=f(u)+f(v)$ and $f(av)=sigma_f(a)f(v)$ for some $sigma_fin Aut(k)$. $PGamma L(n,k)$ is just that group mod the scalar matrices.
answered Jul 19 at 2:50
C Monsour
4,511221
add a comment |Â
up vote
3
down vote
up vote
3
down vote
There are at least two fundamentally different ways to get the exceptional outer automorphism of $S_6$. One is to find the permutation representation of $S_5$ on six letters, as already alluded to. The other is to notice that $A_6cong PSL(2,9)$ is an index four normal subgroup of $PGamma L(2,9)$. The quotient is the Klein four group, and the intermediate index two subgroups are of distinct isomorphism types: One is $S_6$, one is $PGL(2,9)$, and one is $M_10$ (a point stabilizer in $M_11$). Since they are pairwise non-isomorphic, they are characteristic subgroups, so since $PGamma L(2,9)$ has no order $2$ normal subgroup (and since $S_6$ has trivial center) it acts faithfully by conjugation on $S_6$, and thus embeds monomorphically into $Aut(S_6)$.
So it boils down to whether you like $BbbF_5$ or $BbbF_9$ better.
If you are unfamiliar with the notation, $Gamma L(n,k)$ is the group of semilinear automorphisms of $k^n$, where a semilinear map $f$ satisfies $f(u+v)=f(u)+f(v)$ and $f(av)=sigma_f(a)f(v)$ for some $sigma_fin Aut(k)$. $PGamma L(n,k)$ is just that group mod the scalar matrices.
answered Jul 19 at 2:50
C Monsour
4,511221
There are at least two fundamentally different ways to get the exceptional outer automorphism of $S_6$. One is to find the permutation representation of $S_5$ on six letters, as already alluded to. The other is to notice that $A_6cong PSL(2,9)$ is an index four normal subgroup of $PGamma L(2,9)$. The quotient is the Klein four group, and the intermediate index two subgroups are of distinct isomorphism types: One is $S_6$, one is $PGL(2,9)$, and one is $M_10$ (a point stabilizer in $M_11$). Since they are pairwise non-isomorphic, they are characteristic subgroups, so since $PGamma L(2,9)$ has no order $2$ normal subgroup (and since $S_6$ has trivial center) it acts faithfully by conjugation on $S_6$, and thus embeds monomorphically into $Aut(S_6)$.
So it boils down to whether you like $BbbF_5$ or $BbbF_9$ better.
If you are unfamiliar with the notation, $Gamma L(n,k)$ is the group of semilinear automorphisms of $k^n$, where a semilinear map $f$ satisfies $f(u+v)=f(u)+f(v)$ and $f(av)=sigma_f(a)f(v)$ for some $sigma_fin Aut(k)$. $PGamma L(n,k)$ is just that group mod the scalar matrices.
answered Jul 19 at 2:50
C Monsour
4,511221
edited Jul 19 at 4:40
answered Jul 19 at 2:50
C Monsour
4,511221
answered Jul 19 at 2:50
C Monsour
4,511221
answered Jul 19 at 2:50
C Monsour
4,511221
4,511221
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2855524%2fh%25c3%25b6lders-theorem-group-theory-proof-collection-and-applications%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password