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Couple of days ago, i actually asked this question : Existence of a normal subgroup with $|\operatorname{Aut}{(H)}|>|\operatorname{Aut}{(G)}|$.

I was thinking about the converse of this statement. Suppose $G$ is a finite group, with subgroups $H$ and satisfies $$|Aut(G)|=|Aut(H)|,$$ then can anything be said about, $H$ or $G$ in terms of their structure. Will $H$ be normal or ....? Or one can even ask, that find all finite groups $G$ such that $|Aut(H)|=|Aut(G)|$ for every subgroup $H$ of $G$! I am not interested that much in this question as there seems one can have lot of Groups of this type, but i am more curious to know the behaviour of $H$.

Also, anyone interested can very well read this article in MO:

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I'd be amazed if there was anything general that could be said about this situation. But here's an example where it holds. For most values of $n$, $S_n$ and $A_n$ have the same order of automorphism group. – Robin Chapman Aug 16 '10 at 21:05
In the group $C_2\times C_4 \times S_3$, the subgroup $H=C_2\times C_4\times C_2$ satisfies this property, but is not normal. – user641 Aug 16 '10 at 21:20
@Robin Chapman: Please provide me with proof of that result, between $S_{n}$ and $A_{n}$. – anonymous Aug 17 '10 at 4:05
@Chandru1: See pages 299ff in Suzuki's "Group Theory", vol. 1. – user641 Aug 17 '10 at 7:04
@Steve D: For my problem or the one which Robin Chapman had stated. – anonymous Aug 17 '10 at 7:08

1 Answer 1

In response to:

...find all finite groups $G$ such that $|\mathrm{Aut}(H)|=|\mathrm{Aut}(G)|$ for every subgroup $H$ of $G$

the only groups (finite or infinite) for which this is true is the trivial group and the group of order $2$.

Proof: All groups have the trivial group $\{1\}$ as a subgroup, which has a trivial automorphism group, i.e., $|\mathrm{Aut}(\{1\})|=1$. By the assumptions on $G$, we have $|\mathrm{Aut}(\{1\})|=|\mathrm{Aut}(G)|=1$. The only groups with $|\mathrm{Aut}(G)|=1$ are the trivial group and the group of order $2$ (ref.). The property is true for these two groups.

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Aren't statements like this usually read with an implicit 'nontrivial' just to avoid this sort of pedantry? E.g., talking about simple groups as 'those with no normal subgroups'... – Steven Stadnicki Jan 26 '13 at 0:03
I have spent like 10 minutes on this, but suppose $G$ is finite with $ \#G \geq 2$. Consider the prime factorisation of $ \#G$ and note that for every prime $p \mid \#G$ there is a subgroup $H_p$ of order $p$ by Cauchy's theorem. In particular, each $H_p$ is isomorphic to $\mathbb{Z}/p\mathbb{Z}$ and, hence, $|\text{Aut}(H_p)| = |U(\mathbb{Z}/p\mathbb{Z})| = p-1$. This leaves only the groups of prime order? – J.H. Feb 19 '13 at 19:30
Even without including the trivial subgroup, any cyclic subgroup of order $p$ has automorphism group of size $p-1$. So if $|G|$ is divisible by two primes, you're out of luck. If $G$ is a p-group, looking at $G/\Phi(G)$ shows it would have to be cyclic. A quick count then shows $G$ would have to be cyclic of prime order. – user641 May 1 '13 at 22:05

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