Timeline for The number of group elements whose squares lie in a given subgroup
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Jun 6, 2012 at 17:59 | comment | added | Anton Klyachko | Is this because the number of tuples $(g_1,\dots,g_n)$ such that $w(g_1,\dots,g_n)=x$ is a generalised character $\theta(x)$ for ANY word $w\in F$ or this is not so easy? | |
| Jun 5, 2012 at 22:12 | comment | added | Marty Isaacs | It seems to be easy to prove via character theory that if |W| = 1, then number of homomorphisms f such that f(W) <= H is a multiple of |H|. I don't see a proof along these lines if W has cardinality exceeding 1. | |
| Jun 4, 2012 at 23:41 | comment | added | Anton Klyachko | Thank you, Marty! Your proof is shorter than ours but less elementary. However, I would be happy if someone provides a reference proving that the fact is known. Actually, the paper cited in the question contains a more general fact (Corollary 5). <i>Suppose that $H$ is a subgroup of a group $G$ and $W$ is a subgroup (or a subset) of a free group $F$. Then the number of homomorphisms $f\colon F\to G$ such that $f(W)\subseteq H$ is divisible by $|H|$.</i> (Taking $F=\Bbb Z$, we obtain the statement from the question.) Does there exist a short character-theoretic proof for this too? | |
| Jun 4, 2012 at 23:04 | history | edited | Marty Isaacs | CC BY-SA 3.0 |
Fixed typo: replaced chi with theta_k
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| Jun 4, 2012 at 20:36 | history | answered | Marty Isaacs | CC BY-SA 3.0 |