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This is directly related to this question, but should be easier: Suppose for finite groups $G_1$ and $G_2,$ we know that for any group $H,$ $$|\rm{Hom}(G_1, H)| = |\rm{Hom}(G_2, H)|$$

Does it follow that $G_1 \simeq G_2?$ I would think that requiring that $H$ is finite should not make it less true, but whatever works for you.

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@Serkan Yes, from your condition "for any finite group $H$, $|\text{Hom}(H,G_1)|=|\text{Hom}(G_2,H)|$ and $|\text{Hom}(H,G_2)|=|\text{Hom}(G_1,H)|$" it follows that $G_1$ and $G_2$ are groups of order $1$, and so they are isomorphic. – bof Dec 22 '13 at 9:18
up vote 10 down vote accepted

It seems to me that this question can be answered in the affirmative using the same idea (of Lovász) as that other question.

For finite groups $G$ and $H$, let $h(G,H)$ be the number of homomorphisms and $s(G,H)$ the number of surjective homomorphisms from $G$ to $H$.

Consider a finite group $H$. Let $H_1,\dots,H_n$ be all the proper subgroups of $H$ (or just the maximal ones). For $I\subseteq[n]=\{1,\dots,n\}$ let $H_I=\bigcap_{i\in I}H_i$ if $I\ne\emptyset$, and let $H_\emptyset=H$. By the in-and-out principle, for any finite group $G$ we have$$s(G,H)=\sum_{I\subseteq[n]}(-1)^{|I|}h(G,H_I).$$Therefore, if $G_1$ and $G_2$ are finite groups such that $h(G_1,H)=h(G_2,H)$ for every finite group $H$, it follows that $s(G_1,H)=s(G_2,H)$ for every finite group $H$, and in particular that $s(G_1,G_2)=s(G_2,G_2)\ge1$ and $s(G_2,G_1)=s(G_1,G_1)\ge1$, whence $G_1\cong G_2$.

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Very nice argument! – Igor Rivin Dec 22 '13 at 14:31

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