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Looking at this Example of group cohomology not annihilated by exponent of $G$? I stumbled upon one question I couldn't solve (probably because it's hard), so I post it here.

Using Lyndon resolvent, we can always find $G$-module $M$ of rank equal to number of relators in some presentation such that $H^2(G, M)$ has maximal exponent (and is equal to $\Bbb Z/|G|$). Because one-relator finite group is cyclic, I asked myself

  1. Is it true that if there's ideal $R$ with $H^2(G, \Bbb Z[G]/R)$ of exponent $|G|$, then $G$ cyclic?

    Is it true that if there's ideal $R$ with $H^2(G, \Bbb Z[G]/R)$ of exponent $|G|$, then $G$ cyclic?

  2. If there's $2$-generated $G$-module $M$ with $H^2$ of exponent $|G|$, what can we say about $G$?

  1. If there's $2$-generated $G$-module $M$ with $H^2$ of exponent $|G|$, what can we say about $G$?

Of course, we can look not only on second cohomology, and obtain series of filtrations on category of finite groups: $G \in LC_{n, k}$ (LC for "large cohomology") if exists a $k$-generated module s. t. $exp(H^n(G, M) = |G|$. (or maybe $H^{\leq n}$, or $H^{\geq n}$,..). Looks interesting.

Looking at this Example of group cohomology not annihilated by exponent of $G$? I stumbled upon one question I couldn't solve (probably because it's hard), so I post it here.

Using Lyndon resolvent, we can always find $G$-module $M$ of rank equal to number of relators in some presentation such that $H^2(G, M)$ has maximal exponent (and is equal to $\Bbb Z/|G|$). Because one-relator finite group is cyclic, I asked myself

  1. Is it true that if there's ideal $R$ with $H^2(G, \Bbb Z[G]/R)$ of exponent $|G|$, then $G$ cyclic?
  1. If there's $2$-generated $G$-module $M$ with $H^2$ of exponent $|G|$, what can we say about $G$?

Of course, we can look not only on second cohomology, and obtain series of filtrations on category of finite groups: $G \in LC_{n, k}$ (LC for "large cohomology") if exists a $k$-generated module s. t. $exp(H^n(G, M) = |G|$. (or maybe $H^{\leq n}$, or $H^{\geq n}$,..). Looks interesting.

Looking at this Example of group cohomology not annihilated by exponent of $G$? I stumbled upon one question I couldn't solve (probably because it's hard), so I post it here.

Using Lyndon resolvent, we can always find $G$-module $M$ of rank equal to number of relators in some presentation such that $H^2(G, M)$ has maximal exponent (and is equal to $\Bbb Z/|G|$). Because one-relator finite group is cyclic, I asked myself

  1. Is it true that if there's ideal $R$ with $H^2(G, \Bbb Z[G]/R)$ of exponent $|G|$, then $G$ cyclic?

  2. If there's $2$-generated $G$-module $M$ with $H^2$ of exponent $|G|$, what can we say about $G$?

Of course, we can look not only on second cohomology, and obtain series of filtrations on category of finite groups: $G \in LC_{n, k}$ (LC for "large cohomology") if exists a $k$-generated module s. t. $exp(H^n(G, M) = |G|$. (or maybe $H^{\leq n}$, or $H^{\geq n}$,..). Looks interesting.

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Denis T
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Looking at this Example of group cohomology not annihilated by exponent of $G$? I stumbled upon one question I couldn't solve (probably because it's hard), so I post it here.

Using Lyndon resolvent, we can always find $G$-module $M$ of rank equal to number of relators in some presentation such that $H^2(G, M)$ has maximal exponent (and is equal to $\Bbb Z/|G|$). Because one-relator finite group is cyclic, I asked myself

  1. Is it true that if there's ideal $R$ with $H^2(G, \Bbb Z[G]/R)$ of exponent $|G|$, then $G$ cyclic?
  1. If there's $2$-generated $G$-module $M$ with $H^2$ of exponent $|G|$, what can we say about $G$?

Of course, we can look not only on second cohomology, and obtain series of filtrations on category of finite groups: $G \in BC_{n, k}$$G \in LC_{n, k}$ (BCLC for "big"large cohomology") if exists a $k$-generated module s. t. $exp(H^n(G, M) = |G|$. (or maybe $H^{\leq n}$, or $H^{\geq n}$,..). Looks interesting.

Looking at this Example of group cohomology not annihilated by exponent of $G$? I stumbled upon one question I couldn't solve (probably because it's hard), so I post it here.

Using Lyndon resolvent, we can always find $G$-module $M$ of rank equal to number of relators in some presentation such that $H^2(G, M)$ has maximal exponent (and is equal to $\Bbb Z/|G|$). Because one-relator finite group is cyclic, I asked myself

  1. Is it true that if there's ideal $R$ with $H^2(G, \Bbb Z[G]/R)$ of exponent $|G|$, then $G$ cyclic?
  1. If there's $2$-generated $G$-module $M$ with $H^2$ of exponent $|G|$, what can we say about $G$?

Of course, we can look not only on second cohomology, and obtain series of filtrations on category of finite groups: $G \in BC_{n, k}$ (BC for "big cohomology") if exists a $k$-generated module s. t. $exp(H^n(G, M) = |G|$. (or maybe $H^{\leq n}$, or $H^{\geq n}$,..). Looks interesting.

Looking at this Example of group cohomology not annihilated by exponent of $G$? I stumbled upon one question I couldn't solve (probably because it's hard), so I post it here.

Using Lyndon resolvent, we can always find $G$-module $M$ of rank equal to number of relators in some presentation such that $H^2(G, M)$ has maximal exponent (and is equal to $\Bbb Z/|G|$). Because one-relator finite group is cyclic, I asked myself

  1. Is it true that if there's ideal $R$ with $H^2(G, \Bbb Z[G]/R)$ of exponent $|G|$, then $G$ cyclic?
  1. If there's $2$-generated $G$-module $M$ with $H^2$ of exponent $|G|$, what can we say about $G$?

Of course, we can look not only on second cohomology, and obtain series of filtrations on category of finite groups: $G \in LC_{n, k}$ (LC for "large cohomology") if exists a $k$-generated module s. t. $exp(H^n(G, M) = |G|$. (or maybe $H^{\leq n}$, or $H^{\geq n}$,..). Looks interesting.

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Denis T
  • 4.6k
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  • 32

Small modules over finite group with large cohomology

Looking at this Example of group cohomology not annihilated by exponent of $G$? I stumbled upon one question I couldn't solve (probably because it's hard), so I post it here.

Using Lyndon resolvent, we can always find $G$-module $M$ of rank equal to number of relators in some presentation such that $H^2(G, M)$ has maximal exponent (and is equal to $\Bbb Z/|G|$). Because one-relator finite group is cyclic, I asked myself

  1. Is it true that if there's ideal $R$ with $H^2(G, \Bbb Z[G]/R)$ of exponent $|G|$, then $G$ cyclic?
  1. If there's $2$-generated $G$-module $M$ with $H^2$ of exponent $|G|$, what can we say about $G$?

Of course, we can look not only on second cohomology, and obtain series of filtrations on category of finite groups: $G \in BC_{n, k}$ (BC for "big cohomology") if exists a $k$-generated module s. t. $exp(H^n(G, M) = |G|$. (or maybe $H^{\leq n}$, or $H^{\geq n}$,..). Looks interesting.