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 1d comment Squaring is linear in Galois Field $2$ It is not true that squaring is linear in all fields on characteristic two -- only that it is additive. 2d comment Making sense of the term $H^1(N,A)^{G/N}$ in the inflation-restriction exact sequence. See Benson - Representations and Cohomology II section 3.5 for a description of how this action is defined. Briefly, you take a $kG$-projective resolution of $k$ which is a $kN$-projective resolution by restriction, so can be used to calculate $H^1(N,A|_N)$. Then the space of $N$-homs from this resolution to $A$ is a $G$-module ($(g\cdot f)(q)=gf(g^{-1}q)$), but $N$ acts trivially, hence it becomes a $G/N$-module. This induces an action on the Ext-groups. Jul29 comment A problem about isomorphism in module theory Standard counterexample for questions like this: $R=\mathbb{Z}$, $B= \mathbb{Z} \oplus \mathbb{Z} \oplus \cdots$, $\ker g = \mathbb{Z} \oplus \mathbb{Z} \oplus 0 \oplus 0 \oplus \cdots$, $\operatorname{im} f = \mathbb{Z} \oplus 0 \oplus 0 \oplus \cdots$ Jul29 comment In general, how do you construct a nontrivial representation of a group? Every group has a regular representation, which is nontrivial so long as the group is nontrivial. Jul28 comment Hint to find the order of the group of $2\times 2$ matrices under multiplication Jul28 revised Is “Categories and Sheaves” a good followup to Aluffi's “Algebra: Chapter 0”? added 20 characters in body Jul23 comment MAGMA and groups of order 0 If your object is called g, call Type(g); It will say something like GrpPerm. Go to the handbook section for the function Order : GrpPerm -> RingIntElt (or index) and see if it specifies what 0 means. If not only people with inside Magma knowledge can help you (it's closed source). Jul22 comment MAGMA and groups of order 0 Certainly for some types of group 0 is to be interpreted as infinity: see magma.maths.usyd.edu.au/magma/handbook/text/779#8622 On the other hand, sometimes it returns 0 if it can't either find the order or prove it is infinite magma.maths.usyd.edu.au/magma/handbook/text/797#8917 You need to check the handbook for the particular type of group on which you are calling Order or # Jul21 awarded Nice Answer Jul14 answered Is it possible to put an equilateral triangle onto a square grid so that all the vertices are in corners? Jul11 comment What are “instantaneous” rates of change, really? @Henning NSA doesn't require you to shift back and forth between reals and hyperreals, for example there are no hyperreals in the IST approach. Jul10 reviewed Approve Evaluating an indefinite integral with a square root in the denominator Jul8 comment Recommendation for books on topology (light reads) A topological picturebook by George Francis? Jul8 comment arithmetic with quantum integers The relationship is that the first $[n]_q$ is equal to $q^{-n+2}\{ n\}_{q^2}$ where the curly brackets denote the second kind of $q$-integer. This means you can derive multiplication formulas for the first kind from ones for the second. Jul8 comment Prove or disprove $A_5$ has a subgroup that isomorphic to $\mathbb{Z}_6$ math.stackexchange.com/questions/1246662/… Jul8 comment Noncyclic Abelian Group of order 51 There is no need for the Sylow theorems here! The result you mention is Cauchy's theorem, much simpler than Sylow's theorems. Jul4 comment Acting algebraically Is G an algebraic group? Is V a variety? Jul1 comment Relations between $R^fG$ and either $\mathbb{C}^fG$ or $\mathbb{Z}^fG$. What do $\bar f$ and $\tilde f$ really mean? The obvious interpretation of $f$ is as (the class of) a map $G \times G \to R^\times$ -- how does this give a map $G \times G \to \{\pm 1\}$? Jun25 comment Is there a short symbol that denotes integration? People write $f^{(r)}$ for $f$ differentiated $r>0$ times (no indication of which variable you differentiate with respect to). You could just extend this to negative $r$, though it is definitely not standard notation. Jun24 comment How to determine non trivial homomorphisms There is no algorithm that will solve all of these problems, so the method you should use depends entirely on the groups or rings involved.