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!


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.

  • $\begingroup$ Thank you for the answer my actual problem is that i have the situation: $ \xoverline{K_{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(\xoverline{K_{p,q}}=S_{p} \times S{q} $? $\endgroup$ – 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 $p\neq q$, 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

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