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 May 16 asked Baumslagâ€“Solitar $B(1,2)$ is not hyperbolic May 9 awarded Popular Question May 8 revised $\mathbb{1}\uparrow_H^{G}$ is the permutation representation on $G/H$ added 23 characters in body May 8 asked $\mathbb{1}\uparrow_H^{G}$ is the permutation representation on $G/H$ May 7 revised Showing that $g$ and $g^{-1}$ are conjugate iff $\chi(g)$ is real deleted 2 characters in body May 7 asked Showing that $g$ and $g^{-1}$ are conjugate iff $\chi(g)$ is real May 5 asked Question on Frobenius Reciprocity May 5 comment Proving that the tensor product is generated by $a\otimes b$ @Magdiragdag yes sorry that is what I meant, the stuff generated by the tensors $a\otimes b$. I think there is a problem with this approach though. Say we let $K$ be the module generated by stuff of the form $a\otimes b$. Then take $x,y\in A\otimes B-K$. Then I see no reason as to why we couldn't have $x+y\in K$ and $x+y\neq 0$ which would be a problem for the map. Apr 30 accepted Proving The Diamond Lemma Apr 30 comment $\mathbb{C}\otimes_\mathbb{C} \mathbb{C} \cong \mathbb{R}\otimes _\mathbb{R} \mathbb{C}$ Is there a slight subtly here in that we can write a map $f:\mathbb{C}\otimes \mathbb{C}\rightarrow \mathbb{C}$ such $f(a\otimes b)=ab$ and then show that this map is injectvive/surjective. However do we also need to check that this map is well defined, which is a problem as we don't know what $a\otimes b$ is and so we have to define $g:\mathbb{C}\times \mathbb{C}\rightarrow \mathbb{C}$ such that $g(a,b)=a\otimes b$ and then use the universal property to show that the map from before actually exists and is well defined? Apr 30 accepted Example of a functor on products Apr 30 accepted Proving that $V(R^*)=V(R)-1$ Apr 30 comment Showing that $\mathbb{Z}/2\mathbb{Z}[\mathbb{Z}/2\mathbb{Z}]$ is semi-simple @JackSchmidt Are you saying that $\{0,g\}$ is an ideal? I am now confused, don't we have $\{0,g\}g=\{0,e\}$ hence this is not an ideal? Apr 30 accepted Proving that the tensor product is generated by $a\otimes b$ Apr 30 asked Proving that the tensor product is generated by $a\otimes b$ Apr 29 comment Showing that $\mathbb{Z}/2\mathbb{Z}[\mathbb{Z}/2\mathbb{Z}]$ is semi-simple Thanks! Sorry for the pretty trivial question Apr 29 accepted Showing that $\mathbb{Z}/2\mathbb{Z}[\mathbb{Z}/2\mathbb{Z}]$ is semi-simple Apr 29 comment Showing that $\mathbb{Z}/2\mathbb{Z}[\mathbb{Z}/2\mathbb{Z}]$ is semi-simple So $\{0,e\}$ is not an ideal as $\{0,e\}g\not\subset \{0,e\}$ but $\{0,g\}$ is an ideal right as is $\{0,e+g\}$ so does this choice work? Apr 29 comment Showing that $\mathbb{Z}/2\mathbb{Z}[\mathbb{Z}/2\mathbb{Z}]$ is semi-simple oh yeah, oops. Yeah if I just leave the last summand off I think I am ok right? Apr 29 comment $\mathbb{C}\otimes_\mathbb{C} \mathbb{C} \cong \mathbb{R}\otimes _\mathbb{R} \mathbb{C}$ @FabioLucchini Yeah but I am just considering these as abelian groups so this is irrelevant right?