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The alternating group $A_4$ is a counterexample: It has order $2^2\cdot 3$, so $O^2(A_4)$ will contain an order $3$ element. But any order $3$ element of $A_4$ generates the whole group as a normal di …

EDIT: In an earlier version, I claimed erroneously that the following statement holds for general $I$, and can be formally reduced to the case of countable $I$. I currently do not see how to perform t …

If you continue to the end of Section 2.7.2, there are actually two different double covers of $S_n$ (abstractly this comes from the fact that $S_n$ is not perfect, both of these double covers restric …

The $\cup_1$-product is not always a coboundary. For example, for a cocycle $x\in Z^2$, $x\cup_1 x\in Z^3$ is a representative of the class $\operatorname{Sq}^1 x$. This can be non-trivial already for …

The answer to your question is actually "yes", but maybe not in the way you wanted. Indeed, the full (oriented) symmetry group of the icosahedron is isomorphic to $A_5$. The stabilizer of a vertex is …

I'm not aware of any widely used term for this in the context of groups, but in analogy with the usage in algebraic geometry and commutative algebra, you could say that a morphism of groups $G\to H$ i …

Okay, this happens precisely in the obvious case, namely if all primes dividing the order $|G|$ are invertible in $R$. To see this, note that $\operatorname{Ext}^*_{R[G]}(R,R)$ is group cohomology of …

YCor's example in the comments shows that you can have continuous characters with image an arbitrarily large finite group. In fact, there are even injective homomorphisms $\mathbb{Z}_p^\times \to \mat …

This is an answer to problem 1 and problem 3: As noted in the comments, $\operatorname{Poly}(G)$ with pointwise multiplication is a group for finite $G$. Consider the homomorphism $\operatorname{Poly} …

Here is a stupid approach: Fill in a $n\times n$ multiplication table randomly, then check whether if satisfies the properties of a group. (The hard part is associativity, apart from that we basical …