Connection between sum of Graphs and their automorphism groupsConnection Between Automorphism Groups of a Graph and its Line GraphCombinatorial vs. geometric symmetries of graphs and their drawingsAutomorphisms groups of direct product of graphsVertex-transitive graphs and deletion of verticesautomorphism group of direct product of groupsExpository articles on Cayley Graphs?Groups which can not occur as automorphism group of a groupNon-trivial graph automorphism groups with $D_n$ as subgroupGraph Automorphism GroupsRealizable automorphism groups of graphs
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Connection between sum of Graphs and their automorphism groups
Connection Between Automorphism Groups of a Graph and its Line GraphCombinatorial vs. geometric symmetries of graphs and their drawingsAutomorphisms groups of direct product of graphsVertex-transitive graphs and deletion of verticesautomorphism group of direct product of groupsExpository articles on Cayley Graphs?Groups which can not occur as automorphism group of a groupNon-trivial graph automorphism groups with $D_n$ as subgroupGraph Automorphism GroupsRealizable automorphism groups of graphs
$begingroup$
can we say something about the automorphism group of a graph $G$ that has the property: $ G cong A + B $ , if we know the automorphism groups of $A$ and $B$ respectively. The $+$ is the union $ cup$ of the $A$ and $B$ with the only addition that $ V(A) cap V(B)= emptyset$.
Thank you in advance, any view on this would be helpful!
abstract-algebra graph-theory algebraic-graph-theory automorphism-group
asked Mar 14 at 7:54
Someone86Someone86
176
$endgroup$
add a comment |
$begingroup$
can we say something about the automorphism group of a graph $G$ that has the property: $ G cong A + B $ , if we know the automorphism groups of $A$ and $B$ respectively. The $+$ is the union $ cup$ of the $A$ and $B$ with the only addition that $ V(A) cap V(B)= emptyset$.
Thank you in advance, any view on this would be helpful!
abstract-algebra graph-theory algebraic-graph-theory automorphism-group
asked Mar 14 at 7:54
Someone86Someone86
176
$endgroup$
add a comment |
$begingroup$
can we say something about the automorphism group of a graph $G$ that has the property: $ G cong A + B $ , if we know the automorphism groups of $A$ and $B$ respectively. The $+$ is the union $ cup$ of the $A$ and $B$ with the only addition that $ V(A) cap V(B)= emptyset$.
Thank you in advance, any view on this would be helpful!
abstract-algebra graph-theory algebraic-graph-theory automorphism-group
asked Mar 14 at 7:54
Someone86Someone86
176
$endgroup$
can we say something about the automorphism group of a graph $G$ that has the property: $ G cong A + B $ , if we know the automorphism groups of $A$ and $B$ respectively. The $+$ is the union $ cup$ of the $A$ and $B$ with the only addition that $ V(A) cap V(B)= emptyset$.
Thank you in advance, any view on this would be helpful!
abstract-algebra graph-theory algebraic-graph-theory automorphism-group
abstract-algebra graph-theory algebraic-graph-theory automorphism-group
asked Mar 14 at 7:54
Someone86Someone86
176
asked Mar 14 at 7:54
Someone86Someone86
176
asked Mar 14 at 7:54
Someone86Someone86
176
asked Mar 14 at 7:54
Someone86Someone86
176
asked Mar 14 at 7:54
Someone86Someone86
176
176
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Assuming A is not isomorphic to B, the automorphism group is the direct product of the groups of A and B. The direct product can be thought of as the set of elements (a,b) for a in the group for A and b in the group for B. Then (a,c) times (b,d) is (ab,cd), because the automorphism of B does not affect that of A and vice versa.
answered Mar 14 at 12:56
Michael GintzMichael Gintz
865
$endgroup$
$begingroup$
Thank you for the answer my actual problem is that i have the situation: $ xoverlineK_p,q cong K_p + K_q $ where the graphs , are the complete graphs of p and q vertices respectively. Is it still the product? I mean i know that $ Aut(K_n)=S_n$ ..soo can I conclude to the fact that $Aut(xoverlineK_p,q=S_p times Sq $?
$endgroup$
– Someone86
Mar 14 at 13:22
$begingroup$
Absolutely! Any pair of permutations of $p,q$ still corresponds to an element of the automorphism group. Don't forget that this is only if $pneq q$, because otherwise we get things where the points can be mapped to the other subgraph.
$endgroup$
– Michael Gintz
Mar 14 at 16:06
$begingroup$
Thank you mate!
$endgroup$
– Someone86
Mar 14 at 17:40
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Assuming A is not isomorphic to B, the automorphism group is the direct product of the groups of A and B. The direct product can be thought of as the set of elements (a,b) for a in the group for A and b in the group for B. Then (a,c) times (b,d) is (ab,cd), because the automorphism of B does not affect that of A and vice versa.
answered Mar 14 at 12:56
Michael GintzMichael Gintz
865
$endgroup$
$begingroup$
Thank you for the answer my actual problem is that i have the situation: $ xoverlineK_p,q cong K_p + K_q $ where the graphs , are the complete graphs of p and q vertices respectively. Is it still the product? I mean i know that $ Aut(K_n)=S_n$ ..soo can I conclude to the fact that $Aut(xoverlineK_p,q=S_p times Sq $?
$endgroup$
– Someone86
Mar 14 at 13:22
$begingroup$
Absolutely! Any pair of permutations of $p,q$ still corresponds to an element of the automorphism group. Don't forget that this is only if $pneq q$, because otherwise we get things where the points can be mapped to the other subgraph.
$endgroup$
– Michael Gintz
Mar 14 at 16:06
$begingroup$
Thank you mate!
$endgroup$
– Someone86
Mar 14 at 17:40
add a comment |
$begingroup$
Assuming A is not isomorphic to B, the automorphism group is the direct product of the groups of A and B. The direct product can be thought of as the set of elements (a,b) for a in the group for A and b in the group for B. Then (a,c) times (b,d) is (ab,cd), because the automorphism of B does not affect that of A and vice versa.
answered Mar 14 at 12:56
Michael GintzMichael Gintz
865
$endgroup$
$begingroup$
Thank you for the answer my actual problem is that i have the situation: $ xoverlineK_p,q cong K_p + K_q $ where the graphs , are the complete graphs of p and q vertices respectively. Is it still the product? I mean i know that $ Aut(K_n)=S_n$ ..soo can I conclude to the fact that $Aut(xoverlineK_p,q=S_p times Sq $?
$endgroup$
– Someone86
Mar 14 at 13:22
$begingroup$
Absolutely! Any pair of permutations of $p,q$ still corresponds to an element of the automorphism group. Don't forget that this is only if $pneq q$, because otherwise we get things where the points can be mapped to the other subgraph.
$endgroup$
– Michael Gintz
Mar 14 at 16:06
$begingroup$
Thank you mate!
$endgroup$
– Someone86
Mar 14 at 17:40
add a comment |
$begingroup$
Assuming A is not isomorphic to B, the automorphism group is the direct product of the groups of A and B. The direct product can be thought of as the set of elements (a,b) for a in the group for A and b in the group for B. Then (a,c) times (b,d) is (ab,cd), because the automorphism of B does not affect that of A and vice versa.
answered Mar 14 at 12:56
Michael GintzMichael Gintz
865
$endgroup$
Assuming A is not isomorphic to B, the automorphism group is the direct product of the groups of A and B. The direct product can be thought of as the set of elements (a,b) for a in the group for A and b in the group for B. Then (a,c) times (b,d) is (ab,cd), because the automorphism of B does not affect that of A and vice versa.
answered Mar 14 at 12:56
Michael GintzMichael Gintz
865
answered Mar 14 at 12:56
Michael GintzMichael Gintz
865
answered Mar 14 at 12:56
Michael GintzMichael Gintz
865
answered Mar 14 at 12:56
Michael GintzMichael Gintz
865
865
$begingroup$
Thank you for the answer my actual problem is that i have the situation: $ xoverlineK_p,q cong K_p + K_q $ where the graphs , are the complete graphs of p and q vertices respectively. Is it still the product? I mean i know that $ Aut(K_n)=S_n$ ..soo can I conclude to the fact that $Aut(xoverlineK_p,q=S_p times Sq $?
$endgroup$
– Someone86
Mar 14 at 13:22
$begingroup$
Absolutely! Any pair of permutations of $p,q$ still corresponds to an element of the automorphism group. Don't forget that this is only if $pneq q$, because otherwise we get things where the points can be mapped to the other subgraph.
$endgroup$
– Michael Gintz
Mar 14 at 16:06
$begingroup$
Thank you mate!
$endgroup$
– Someone86
Mar 14 at 17:40
add a comment |
$begingroup$
Thank you for the answer my actual problem is that i have the situation: $ xoverlineK_p,q cong K_p + K_q $ where the graphs , are the complete graphs of p and q vertices respectively. Is it still the product? I mean i know that $ Aut(K_n)=S_n$ ..soo can I conclude to the fact that $Aut(xoverlineK_p,q=S_p times Sq $?
$endgroup$
– Someone86
Mar 14 at 13:22
$begingroup$
Absolutely! Any pair of permutations of $p,q$ still corresponds to an element of the automorphism group. Don't forget that this is only if $pneq q$, because otherwise we get things where the points can be mapped to the other subgraph.
$endgroup$
– Michael Gintz
Mar 14 at 16:06
$begingroup$
Thank you mate!
$endgroup$
– Someone86
Mar 14 at 17:40
$begingroup$
Thank you for the answer my actual problem is that i have the situation: $ xoverlineK_p,q cong K_p + K_q $ where the graphs , are the complete graphs of p and q vertices respectively. Is it still the product? I mean i know that $ Aut(K_n)=S_n$ ..soo can I conclude to the fact that $Aut(xoverlineK_p,q=S_p times Sq $?
$endgroup$
– Someone86
Mar 14 at 13:22
$begingroup$
Thank you for the answer my actual problem is that i have the situation: $ xoverlineK_p,q cong K_p + K_q $ where the graphs , are the complete graphs of p and q vertices respectively. Is it still the product? I mean i know that $ Aut(K_n)=S_n$ ..soo can I conclude to the fact that $Aut(xoverlineK_p,q=S_p times Sq $?
$endgroup$
– Someone86
Mar 14 at 13:22
$begingroup$
Absolutely! Any pair of permutations of $p,q$ still corresponds to an element of the automorphism group. Don't forget that this is only if $pneq q$, because otherwise we get things where the points can be mapped to the other subgraph.
$endgroup$
– Michael Gintz
Mar 14 at 16:06
$begingroup$
Absolutely! Any pair of permutations of $p,q$ still corresponds to an element of the automorphism group. Don't forget that this is only if $pneq q$, because otherwise we get things where the points can be mapped to the other subgraph.
$endgroup$
– Michael Gintz
Mar 14 at 16:06
$begingroup$
Thank you mate!
$endgroup$
– Someone86
Mar 14 at 17:40
$begingroup$
Thank you mate!
$endgroup$
– Someone86
Mar 14 at 17:40
add a comment |
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