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This fact seems rather obvious but for some reason I can’t wrap my head around a proof.

Also, just generally, if we have two groups $G_1$ and $G_2$ equal up to isomorphism, then is the amount of automorphisms between the two equal to $|{\rm Aut}(G_1)|= |{\rm Aut}(G_2)|$?

Is there a proof for this? I have been trying but since I’m new to all this it feels like Im just getting more and more confused on what an isomorphism even is.

Thanks!

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  • $\begingroup$ I think you should talk about the number of isomorphisms, not about amounts. $\endgroup$
    – Carsten S
    Jul 5 at 9:54

2 Answers 2

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Hint

There is only one cyclic group of given order. Sometimes denoted $C_n$. It is also $\Bbb Z_n$.

As to the other question, isomorphic groups have isomorphic automorphism groups. Let $\phi: G\to H$ be an isomorphism. Then $\hat\phi:\rm {Aut} G\to\rm{Aut}H$ given by $\hat\phi(\psi)=\phi\circ\psi\circ \phi^{-1}$ is an isomorphism.

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Here is more detailed answer. The statement doesn't depend on the groups being cyclic.

Let $\psi \in \text{Iso}(G_1, G_2)$. Then for each $\phi \in \text{Aut}(G_2)$ we can construct an isomorphism as follows:

$$ g \mapsto \phi(\psi(g)).$$

Thus, we have an injective map $\text{Aut}(G_2) \to \text{Iso}(G_1,G_2)$.

Let $\hat{\psi}$ be an isomorphism from $G_1$ to $G_2$. Now the map $$ g \mapsto \hat{\psi}(\psi^{-1}(g)) $$ is an automorphism on $G_2$. Thus, choosing $\phi$ to be this automorphism, we see that the map is also surjective.

Thus the two have the same cardinality.

Note this had nothing to do with them being cyclic groups. Philosophically, two groups being isomorphic mean that they are equal. Thus an isomorphism is in some an automorphism, (since the two groups are "equal" and vice verse).

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    $\begingroup$ You can say more: the automorphism groups are actually isomorphic. $\endgroup$
    – Noobie
    Jul 4 at 23:14
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    $\begingroup$ I see how the injective-ness of the second mapping follows from the one to one-ness of $\phi$, but how can you prove the first mapping is injective? $\endgroup$
    – DownstairsPanda
    Jul 5 at 4:18
  • $\begingroup$ They're all isomorphisms. $\endgroup$
    – Noobie
    Jul 5 at 4:49
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    $\begingroup$ @Cpc wait, sorry, could you elaborate? I’m asking about how we know the mapping (say, $f$) from Iso$(G_1, G_2)$ to Aut($G_2)$ is injective - that is $f(\psi_1) = f(\psi_2)$ implies $\psi_1 = \psi_2$. In other words, showing $\psi_1(\phi(\psi_1^{-1})))= \psi_2(\phi\psi_2^{-1})))$ implies $\psi_1 = \psi_2$. How does the fact that they are all isomorphisms remedy this? $\endgroup$
    – DownstairsPanda
    Jul 5 at 5:12
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    $\begingroup$ You are right. I modified my argument. $\endgroup$
    – Rishi Sonthalia
    Jul 5 at 8:23

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