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verret
verret
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The answer to Q2 is no. Let $H$ be an elementary abelian $2$-group, in its regular action, so $n=2^a$ for some $a$. Clearly, to generate $H$, we need $a$ elements with full support, so the total support is $an= n \log_2 n$. Moreover, it is known that, for $a\geq 5$, $H$ is the full automorphism group of a$H=\mathrm{Aut}(G)$ for some graph $G$. (In other words, an elementary abelian $2$-group of order at least $32$ admits a graphical regular representation.)

The answer to Q2 is no. Let $H$ be an elementary abelian $2$-group, in its regular action, so $n=2^a$ for some $a$. Clearly, to generate $H$, we need $a$ elements with full support, so the total support is $an= n \log_2 n$. Moreover, it is known that, for $a\geq 5$, $H$ is the full automorphism group of a graph. (In other words, an elementary abelian $2$-group of order at least $32$ admits a graphical regular representation.)

The answer to Q2 is no. Let $H$ be an elementary abelian $2$-group, in its regular action, so $n=2^a$ for some $a$. Clearly, to generate $H$, we need $a$ elements with full support, so the total support is $an= n \log_2 n$. Moreover, it is known that, for $a\geq 5$, $H=\mathrm{Aut}(G)$ for some graph $G$. (In other words, an elementary abelian $2$-group of order at least $32$ admits a graphical regular representation.)

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verret
verret
  • 3.3k
  • 1
  • 22
  • 30

The answer to Q2 is no. Let $H$ be an elementary abelian $2$-group, in its regular action, so $n=2^a$ for some $a$. Clearly, to generate $H$, we need $a$ elements with full support, so the total support is $an= n \log_2 n$. Moreover, it is known that, for $a\geq 5$, $H$ is the full automorphism group of a graph. (In other words, an elementary abelian $2$-group of order at least $32$ admits a graphical regular representation.)