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Let $G,H$ be two groups , let $h(G,H)$ be the number of all group homomorphisms from $G$ to $H$ and $i(G,H)$ be the no. of all injective group homomorphisms from $G$ to $H$ . I know the relation $h(G,H)=\sum_{N \lhd G } i (G/N , H)$ . Now let $H,K$ be finite groups , if for any finite group $G$ , $h(G,K)=h(G,H)$ holds , then is it true that $i(G,H)=i(G,K)$ for any finite group $G$ ?

  • Do you know any examples where $H$ and $K$ are different, but $h(G,K)=h(G,H)$ for all $G$? – Gerry Myerson Mar 12 '15 at 11:44
  • @GerryMyerson : No ... –  Mar 12 '15 at 11:47
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    http://math.stackexchange.com/questions/1161733 is similar, but is based on counting homomorphisms in the reversed direction i.e. from $H$ and $K$ to $G$. In that case, it turns out that there are no finite examples of nonisomorphic groups with the same homomorphism counts, and I would guess that the same is true in your case. – Derek Holt Mar 12 '15 at 11:59
  • I think here the words "for any finite group $G$" means a finite group $G$ which has the property $h(G,K)=h(G,H),$ not for all finite group $G.$ (I may be wrong also) – Krish Mar 12 '15 at 12:01
  • @I mean that for any finite group $G$ , $h(G,H)=h(G,K)$ holds ... –  Mar 12 '15 at 12:04
  • $i(G,H) = h(G,H)-\sum_{N \lhd G, N\ne 1 } i (G/N , H)$, so induct on the order of $G/N$. – j.p. Mar 12 '15 at 14:29
  • Rereading your question I wondered if you considered the implications of $i(G,H) = i(G,K)$ for $G$ being $H$ or $K$? – j.p. Mar 12 '15 at 15:09
  • @j.p. : yup , as $i(H,K)=i(H,H) \ne 0$ and $i(K,H)=i(K,K) \ne 0$ , so there exist a monomorphism from $H$ to $K$ and also a monomorphism from $K$ to $H$ –  Mar 12 '15 at 15:20
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    @j.p. : yes off-course ; $H=K$ :) –  Mar 13 '15 at 13:07
  • @SaunDev: So you can rephrase your question slightly simpler in the style of http://math.stackexchange.com/questions/1161733/can-i-recover-a-group-by-its-homomorphisms – j.p. Mar 13 '15 at 13:16

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