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Let $G$ and $H$ be finite groups. Consider the ratio

$$r_{G, H} \equiv {|Hom(G, H)| \over{|Hom(H,G)|}}$$

My question is

When is $r_{G, H} = 1$? Can we characterize the pairs of groups $(G, H)$ for which this occurs?

One can answer the corresponding problem in $Set$. Let $S$ and $T$ be finite sets. A straightforward counting argument shows that

$$r_{S, T} = 1 \iff |T|^{|S|} = |S|^{|T|}$$

In the case of groups one can use another counting argument to obtain the bounds

$$|H|^{-|G|} \leq r_{G, H} \leq |G|^{|H|}$$

It seems that these bounds are not very useful though; as $|G|$ and $|H|$ grow the lower bound goes to 0 and the upper bound goes to $\infty$.

So far we have:

  • $r_{G, H} = 1$ if $G$ and $H$ are abelian (Tom Goodwillie).
  • $r_{G, H} = 1$ if $G$ and $H$ are simple and not subgroups of one another (Alex Bartel and Richard Kent).

My guess is that this is a big problem in the general case, since counting homomorphisms seems to involve heavy machinery even in special cases. So I'd appreciate necessary and sufficient conditions for special classes of groups, or references to relevant literature.

Edit: The original question was "When is $r_{G, H} < 1$?". I've changed it in light of the comments, with the hope that the new version is more interesting and tractable.

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If they are both abelian then the numbers are equal. – Tom Goodwillie Nov 12 '10 at 1:46
I think the question was more in the spirit of, "If someone hands you groups $G$ and $H$, is there a nice way of determining which hom-set is larger?" – Kevin Ventullo Nov 12 '10 at 2:43
Another easy case is when $G$ is simple and not a subgroup of $H$. Then $|Hom(G,H)|=1$, while a quotient of $H$ might still inject into $G$. But I think this example is indicative of the fact that the result will depend so much on the exact structure of the groups that you are unlikely to get a nice result in full generality. – Alex B. Nov 12 '10 at 3:05
This problem looks extraordinarily difficult. Is there a chance of solving it in the case of solvable groups? – Todd Trimble Nov 12 '10 at 4:17
If this was a question in combinatorics then you'd have the option of a probabilistic answer. Does anyone know or have any ideas on how to apply this sort of technique to `random groups'? – James Griffin Nov 12 '10 at 13:27

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