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I think I have a counterexample. Let $\operatorname{char}(k)=3$ and let $G$ be the symmetric group $S_{3}$. Then $kG$ has two simple modules: the trivial module $k$ and another one-dimensional module …

In Alperin, J. L., Large abelian subgroups of p-groups, Trans. Am. Math. Soc. 117, 10-20 (1965). ZBL0132.27204, the second part of Theorem 1 gives a group of order $2^{50}$ with no abelian subgroups o …

Assuming that by "sub-algebra" you mean "unital sub-algebra": Every group algebra has a one-dimensional module (the trivial module), so any subalgebra has a one-dimensional module. But many finite-dim …

Let $k$ be a field of characteristic $2$, and let $A$ be the path algebra over $k$ of the quiver with two vertices, $v_1$ and $v_2$, and arrows $a:v_1\to v_2$ and $b:v_2\to v_1$, modulo the relations …

SmallGroup(128,227) and SmallGroup(128,731)) are counterexamples. gap> S:=List([227,731],n->SmallGroup(128,n));; gap> for g in S do L:=List(ConjugacyClasses(g),c->Size(c));; Sort(L);; Print(L); od; [ …

No. Let $G=C_2\times C_2$, generated by elements $g$ and $h$, and let $K$ be any finite field of characteristic $2$. Let $V$ be a finite dimensional $K$-vector space of dimension greater than one wi …

No. $C_1$ is a retract of $C_2$, so $M(C_2,C_1)\simeq C_2$ would have to be a retract of $M(C_2,C_2)\simeq C_4$, which it isn't.

If a finite group has a presentation with $g$ generators and $r$ relations, then the Schur multiplier is generated by $r-g$ elements. There's been lots of work studying groups and presentations where …

For a finite group algebra, $T_H(g)=g^2$ and $S_H(g)=g^{-1}$ for any group element $g$. So for any two finite groups of exponent three and the same order, the quantities will be the same. For example …

I don't think it's true for $G=\mathbb{F}_2^3$ and $a=b=3$. If there were such sets $A$ and $B$, they must have exactly three elements each. By applying a translation and a group automorphism, we ma …

So long as $G$ is nontrivial, the augmentation ideal of $\mathbb{Z}[G]$ still works. If $I$ is any submodule of $\mathbb{Z}[G]$ then there is a short exact sequence of $\mathbb{Z}[G]$-modules $$0\to\ …

The condition on Brauer characters is not sufficient. Let $G$ be a $p$-group, $\pi$ any nontrivial representation over $\mathbb{F}_p$, and $\sigma$ the trivial representation over $\mathbb{Q}_p$ of t …

I think this example answers both questions. Let $k$ have characteristic $3$, and let $G=C_3\times S_3$. Then $kG$ has two simple modules, both one-dimensional, and for each simple module $S$, $\tex …

I think it's easiest to break this up into the composition of three simpler isomorphisms. First, for any ring $A$, if $M_A$ is a right $A$-module and $_AX$ is a left $A$-module, there is a natural ma …

This won't be true if $G=\mathbb{Z}/4\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$. If $G$ has a local ring structure with maximal ideal $\mathfrak{m}$, and quotient field $G/\mathfrak{m}$ isomorphic to $\ …