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This question answers which the smallest symmetric group to embeds $D_n$ of order $2 n$.

I would like to know what is the smallest symmetric group which embeds $D_n \rtimes Aut(D_n)$.

I understand that to know the answer I need to know the order of the elements in the product. From here, I can say that the order of a semidirect product is the product of the orders of the operand groups. We know that $$Aut(D_n) = C_n \rtimes_\phi Aut(C_n)$$. So, $$|D_n \rtimes Aut(D_n)| = |D_n \rtimes (C_n \rtimes_\phi Aut(C_n))| $$.

Then I have shown here that:

$$ |D_n \rtimes Aut(D_n)| = 2 n^3 \prod_{p|n} \left(1 - \frac{1}{p}\right) $$

I am not sure how to proceed from here.

PS. I initially thought that $|Aut (D_n)|$ is also $2 n$ which is incorrect as Dietrich pointed out. So, I fixed it.

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    $\begingroup$ Your last sentence is not true, since $\operatorname{Aut}(\mathbb {D}_n) \cong \mathbb {Z}_n \rtimes \operatorname{Aut}(\mathbb {Z}_n)$, see here. $\endgroup$ – Dietrich Burde Jan 7 '17 at 14:55
  • $\begingroup$ @DietrichBurde, are you suggesting that the order of the automorphism group may not be the same as the order of the original group? $\endgroup$ – Omar Shehab Jan 8 '17 at 21:48
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    $\begingroup$ Yes, I do. Do you see this differently? Take for example $Q_8$ and its automorphism group $S_4$, see here. $\endgroup$ – Dietrich Burde Jan 8 '17 at 23:08
  • $\begingroup$ @DietrichBurde, I just checked, there is no reason that automorphism preserves order. $\endgroup$ – Omar Shehab Jan 9 '17 at 8:23
  • $\begingroup$ @DietrichBurde, I have updated the question. $\endgroup$ – Omar Shehab Jan 9 '17 at 9:12

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