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How shall I establish that the number of homomorphisms from the group $G$ to $G_1\oplus G_2\oplus \cdots G_n$ is same as $h_1h_2\cdots h_n$ where $h_i$ is the number of homomorphisms from $G$ to $G_i$ ? Here $G, G_1, G_2, \cdots, G_n$ are all groups.

Matthew Towers
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KON3
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  • I can only say that it's not the same, for example consider homomorphisms from $\mathbb{Z_2} $ to $\mathbb{Z_2}\times \mathbb{Z_2}$ – Belgi Aug 28 '14 at 11:45
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    @Belgi: There are four morphisms $\mathbb{Z}/2 \to \mathbb{Z}/2 \times \mathbb{Z}/2$, and $4 = 2 \times 2$ where $2$ is the number of morphisms $\mathbb{Z}/2 \to \mathbb{Z}/2$... – Najib Idrissi Aug 28 '14 at 11:47
  • why aren't they the same? – Quang Hoang Aug 28 '14 at 11:47
  • The cross product have three elements of order two giving us one more homomorphisms, which is also an isomorphism – Belgi Aug 28 '14 at 13:44
  • @Belgi: I'm sorry but don't understand what you are trying to say. $\mathbb{Z}/2 \times \mathbb{Z}/2$ has three elements of order two, yes, but what is the "one more homomorphism" you are talking about? Isomorphism between what and what? – Najib Idrissi Aug 28 '14 at 13:53
  • @Idrissi I'm sorry that I was unclear, I'm writing from a mobile phone so it's difficult to write math. I hope this will be more clear:define the following isomorphism $h$:$h(0)=(0,0),h(1)=(1,1)$ – Belgi Aug 28 '14 at 13:57
  • @Belgi: This is not an isomorphism... $(1,0)$ is not in the image! To make everything clear (I think there's a miscommunication somewhere), the four homomorphisms I'm talking about map $1$ respectively to $(0,0)$, $(1,0)$, $(0,1)$ and $(1,1)$, and they all map $0$ to $(0,0)$. None of these are isomorphisms. – Najib Idrissi Aug 28 '14 at 14:00
  • Yes, I'm sorry, I meant that it's isomorphic to its image. You're also right about the four homomorphisms – Belgi Aug 28 '14 at 14:03

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A finite direct sum is a product. By the universal property of products, a homomorphism $G \to G_1 \times \dots \times G_n$ is uniquely determined by its projections $G \to G_i$ and vice-versa. In other words, there is a bijection $$\hom(G, G_1 \times \dots \times G_n) \cong \hom(G, G_1) \times \dots \times \hom(G, G_n)$$ and the cardinality of this last set is $h_1 \dots h_n$.

Najib Idrissi
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  • Please see my comment to the post – Belgi Aug 28 '14 at 13:51
  • @Najib Idrissi Sorry, how does a bijection between $\hom(G, G_1 \times \dots \times G_n)$ and $\hom(G, G_1) \times \dots \times \hom(G, G_n)$ suggest that $G \to G_1 \times \dots \times G_n$ is uniquely determined by its projections? – Brofessor Aug 18 '20 at 17:01
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    @Brofessor Because that's how the bijection is defined. It sends a map $f : G \to G_1 \times \dots \times G_n$ to $(p_1 \circ f, \dots, p_n \circ f)$ where $p_i : G_1 \times \dots \times G_n \to G_i$ is the projection. – Najib Idrissi Aug 19 '20 at 09:04
  • @NajibIdrissi since im pretty new to this, is there somewhere where i could read about how the bijection is defined in this context? or perhaps, something i could search up? – Brofessor Aug 19 '20 at 13:50
  • @Brofessor I defined the bijection in my comment, what else did you want to know? – Najib Idrissi Aug 21 '20 at 07:46