The cohomology of the Mathieu group M23
The cohomology of the Mathieu group M23
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马蒂厄群 M23 的上同调
DOI：
10.1515/jgth.2000.008
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发表时间：
2000
影响因子：
0.5
通讯作者：
R. Milgram
中科院分区：
文献类型：
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作者：
R. Milgram
并且，作为我们计算的结果，对于i < 5，Hi（M23;Z）= 0。特别是M23是第一个已知的反例猜想，如果G是一个有限群，Hi（G;Z）= 0，i = 1，2，3，那么G = {1}。(M11，第一Mathieu群和J1，第一Janko群也满足Out（G）= Mult（G）= 1，但对于这两个群H3（G;Z）= 0。这将是诱人的修正猜想。它很可能只对简单群中的极少数零星个体失效。因此，人们很可能会怀疑存在一个（小）有限数n，使得H（G;Z）= 0，0 < i ≤ n意味着如果G是有限的，则G = {1}。但我不知道有没有合适的人选。
and, as a result of our calculation Hi(M23;Z) = 0 for i < 5. In particular M23 is the first known counterexample to the conjecture that if G is a finite group with Hi(G;Z) = 0, i = 1, 2, 3, then G = {1}‡. (M11, the first Mathieu group and J1, the first Janko group also satisfy Out(G) = Mult(G) = 1, but for both of these groups H3(G;Z) = 0.) It would be tempting to amend the conjecture. It is very likely that it only fails for a very small number of the sporadics among the simple groups. So one might well suspect that there is a (small) finite number n so that H(G;Z) = 0 for 0 < i ≤ n implies that G = {1} if G is finite. But I have no idea as to a suitable candidate for n.