Skip to main content
MathOverflow

Return to Answer

deleted 13 characters in body
Source Link
Yakov
Yakov
  • 103
  • 3

Let E be the elementary abelian group of order p^n. Then the order of its Schur multiplier has rank n(n-1}/2 (Issai Schur). Therefore, the representation group of E (that group is special) does not satisfies ($*$). It is possible to construct infinite set of such examples of arbitrary exponents. Therefore, assertion that (*) fulfilled in the most cases, is sinceless (in any case, I do not know what it means).

Let E be the elementary abelian group of order p^n. Then the order of its Schur multiplier has rank n(n-1}/2 (Issai Schur). Therefore, the representation group of E (that group is special) does not satisfies ($*$). It is possible to construct infinite set of such examples of arbitrary exponents. Therefore, assertion that (*) fulfilled in the most cases, is sinceless (in any case, I do not know what it means).

Let E be the elementary abelian group of order p^n. Then its Schur multiplier has rank n(n-1}/2 (Issai Schur). Therefore, the representation group of E (that group is special) does not satisfies ($*$). It is possible to construct infinite set of such examples of arbitrary exponents. Therefore, assertion that (*) fulfilled in the most cases, is sinceless (in any case, I do not know what it means).

Source Link
Yakov
Yakov
  • 103
  • 3

Let E be the elementary abelian group of order p^n. Then the order of its Schur multiplier has rank n(n-1}/2 (Issai Schur). Therefore, the representation group of E (that group is special) does not satisfies ($*$). It is possible to construct infinite set of such examples of arbitrary exponents. Therefore, assertion that (*) fulfilled in the most cases, is sinceless (in any case, I do not know what it means).