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A subgroup of $SL(2,\mathbf{Z}$) is free iff it's torsion-free; this is a useful trick (and it's not an immediately obvious fact: it's because $SL(2,\mathbf{Z})$ acts on a tree with finite stabilisers …

My standard rant about "what can we say about $G$": what we can say about $G$ is that the two partitions are the same. If the questioner doesn't find that a helpful answer then they might want to cons …

Yes, the dual of $SL_2$ is $PGL_2$. But you're not going down the right track with $PSL_2$. The problem with $PSL_2$ is that it's not a variety at all! You can quotient out the variety $SL_2$ by the …

I don't think so. Consider the $\mathbb{Z}$-module $M$ be the additive subgroup of the rationals consisting of rationals with square-free denominator.

Let $A$ denote the subgroup of $S_\infty$ consisting of permutations that only move finitely many elements, and have even signature. Then $A$ is a normal subgroup of $S_\infty$, and the quotient $S_\i …

Let $G$ be a finite group. If $L$ is a finite free $\mathbf{Z}$-module with an action of $G$, say $L$ is induced if it's isomorphic as a $G$-module to $Ind_H^G(\mathbf{Z})$ with $H$ a subgroup of $G$ …

$\newcommand{\Z}{\mathbf{Z}}$ Given a nice infinite collection of groups, for example the symmetric groups, one can ask whether any finite group is a subgroup of one of them. Of course any finite grou …

I think it's irreducible in $R$ as you suggest. Here's a sketch which I think works. If the polynomial factored in a non-trivial way, then because of the $t^n$ term the factors must have degree less t …