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Let $\text{Aut}(G)$ denote the group of automorphisms of $G$ and let $A\subseteq B$ denote $A$ is a subgroup of $B$. Does the following hold:

$$A\subseteq B\implies \text{Aut}(A)\subseteq \text{Aut}(B)$$

If not, is there a necessary and sufficient condition for this to hold?

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  • $\begingroup$ Welcome to Math.SE! Please read this post and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you edit your question to include some motivation, and an explanation of your own attempts. $\endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Nov 30, 2020 at 9:34
  • $\begingroup$ Could you let me know how I can improve this quesiton? $\endgroup$
    – Mathew
    Nov 30, 2020 at 9:36
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    $\begingroup$ @HallaSurvivor, thanks, this is exactly the kind of thing I was looking for. $\endgroup$
    – Mathew
    Nov 30, 2020 at 9:39
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    $\begingroup$ Yes, I will accept that, thank you $\endgroup$
    – Mathew
    Nov 30, 2020 at 9:40
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    $\begingroup$ Note that you are trying to deduce something about the symmetries of a larger object from the symmetries of a smaller one. It is then possible that the smaller, simpler, object will have more symmetry than the larger more complicated one in which it is embedded (the reverse also could be true - symmetries could be added). You will know from the answers that there are counterexamples: I'm just adding an intuition as to why they might exist. $\endgroup$
    – Mark Bennet
    Nov 30, 2020 at 13:37

2 Answers 2

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It is certainly not true in general.

For a counterexample take $A:=S_6$ whose automorphism group is well-known to have order $2\cdot 6!$; and take $B:=S_7$ whose automorphism group is well-known to have order $7!=7\cdot 6!$.

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This question is hard, and there's not a simple answer. In general, there's no reason a function preserving only the $A$ structure should extend to one preserving all the extra $B$ structure. You can find a discussion about which automorphisms can be extended here, but the answer may not satisfy you.

So I suppose the answer is "no", and the answer to the follow-up question about a necessary and sufficient condition is "not a simple one".

There is also some good discussion in the comments of this question

I hope this helps ^_^

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  • $\begingroup$ Can you give an example of $A$ a subgroup of $B$, and the automorphism group of $A$ not a subgroup of the sutomorphism group of $B$? $\endgroup$
    – Gerry Myerson
    Nov 30, 2020 at 9:43
  • $\begingroup$ GerryMyerson, good question. @HallaSurvivor, I would also like to know about that before accepting the answer. Thanks $\endgroup$
    – Mathew
    Nov 30, 2020 at 9:45
  • $\begingroup$ Not off the top of my head. I'll ask sage about it tomorrow. It's easy to see that for simple cases $\text{Aut}(A)$ is not naturally a subgroup of $\text{Aut}(B)$. But the few examples I tried all worked (though I suspect this is an accident of small groups) $\endgroup$
    – HallaSurvivor
    Nov 30, 2020 at 9:51
  • $\begingroup$ thanks for this, I look forward to hearing more tomorrow. $\endgroup$
    – Mathew
    Nov 30, 2020 at 9:53
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    $\begingroup$ A smaller example is $A=C_2 \times C_2$ as a subgroup of $B=D_8$ (dihedral of order $8$). $A$ has an automorphism of order $3$ that does not extend to $D_8$. $\endgroup$
    – Derek Holt
    Nov 30, 2020 at 13:25

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