Is the number of homomorphisms between two isomorphic groups equals to the number of Automorphisms for each group?

Question

Let us first divide to cases:

finite groups
infinite groups

Let $G,H$ be some groups such that: $G \cong H$.

Is the number of homomorphisms between $G\to H$ equals to the number of elements in Aut$(G)$?

My intuition is that the answer is true, at least for one of the cases, but I don't know how to prove that.

The number of isomorphisms $G\to H$ equals the number of elements of $\mathrm{Aut}(G)$, but not the number of homomorphisms — Mathmo123, Dec 08 '22 at 12:15
Possibly helpful general principle: https://math.stackexchange.com/questions/2039702/what-is-an-homomorphism-isomorphism-saying/2039715#2039715 — Ethan Bolker, Dec 08 '22 at 15:32

Yos · Accepted Answer · 2022-12-08T12:35:43.037

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It is false.

When $G=H=\mathbb{Z}$, $\#$Hom$(G,H)=\#$Hom$(\mathbb{Z}, \mathbb{Z})=\# \mathbb{Z}=\infty.$

But $\#$Aut$(G)=\#$Aut$(\mathbb{Z})=\# \{\pm1\}=2$.

edited Dec 08 '22 at 12:35

answered Dec 08 '22 at 11:05

Yos

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They are clearly asking for group homomorphisms, not ring homomorphisms – alvoi Dec 08 '22 at 12:09
Certainly. But they are not isomorphic as a group. – Yos Dec 08 '22 at 12:31
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Yeah in your original answer you spoke about ring homomorphism which was wrong – alvoi Dec 08 '22 at 15:17
is this always false or only with specific conditions? $\mathbb{Z}$ is a unique group – Noa Even Dec 11 '22 at 21:25
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Always false if not $G＝0$. In fact 0-map is homomorphism and not isomorphism. – Yos Dec 12 '22 at 00:15

alvoi · Answer 2 · 2022-12-08T15:29:27.523

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Fix an isomorphism $\varphi:G→ H$. Every isomorphism $\psi:G→ H$ can be seen as the composition of $\varphi$ with an automorphism of $H$. So there is a bijection between the set of isomorphism from $G$ to $H$, and the set of automorphisms of $H$.

EDIT: I read in a comment that you wrote homomorphisms. If you meant isomorphism, then my other answer is right. If you really meant homomorphism, the result you are talking about is clearly false, even in the case that $G=H$! In that case, every automorphism of $G$ is also a morphism from $G$ to $H$, but there are some morphisms from $G$ to $H$ (for example, the trivial morphism if $|G|\ge2$) which aren't automorphisms

edited Dec 08 '22 at 15:29

answered Dec 08 '22 at 11:03

alvoi

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The user wrote homomorphisms. – Pedro Dec 08 '22 at 11:07
Oh lol, then it's clearly wrong – alvoi Dec 08 '22 at 12:02
number of homomorphisms between two isomorphic groups. – Noa Even Dec 11 '22 at 21:23

score 0 · Answer 3 · answered Dec 08 '22 at 11:23

To be explicit: Let $I(G,H)$ be the set of Isomorphisms between $G$ and $H$. Take an element $\phi \in I(G,H)$:
Now $\phi^{-1} I(G,H) \subseteq Aut(G)$. Now $\phi Aut(G) \subseteq I(G,H)$.
So composing $|I(G,H)| = |Aut(G)|$.

So the number of Isomorphisms is equal to the number of Automorphisms.

score 0 · Answer 4 · answered Dec 16 '22 at 00:38

A finite group example (inspired by answer by Nakayama):

Let $G$ be a cyclic group of order $n$ generated by $g$.

Then $Hom(G,G) = \{f : f(g) = g^i, i \in [n-1] \}$.
Then $Aut(G,G) = \{f : f(g) = g^i, i \in [n-1], gcd(i,n) = 1 \}$.

Clearly $|Hom(G,G)| > |Aut(G)|$ for $n$ not a prime number.

Is the number of homomorphisms between two isomorphic groups equals to the number of Automorphisms for each group?

4 Answers4