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 19h answered Is there an alternative proof of the abel-ruffini theorem? 19h comment Determine if a 4-tuple exists Suppose the 4-tuples we get are $a_0, a_1, a_2, a_3, ...$. At some point, two of them repeat, say $a_i = a_j$. We may assume this is the first repeat we encounter. If $i = 0$, then we have come back to the original. If $i > 0$, then we have $a_{i-1} = a_{j-1}$ (because the process is invertible), so we have an earlier repeat, contradiction. 19h comment Determine if a 4-tuple exists The process is invertible (given four numbers, we can determine the previous as well as the next); therefore, the first time it repeats, we must come back to the original. And it must repeat, since there are only finitely many possible 4-tuples. 1d comment Prove or disprove: For every integer $k\in \mathbb{Z}$, if $f(x)$ is additive, then $f(kz)=kf(z)$. In case 2, you want to split $kz = z+z+\ldots+z$ (with a total of $k$ $z$'s). Similarly in case 3, you want $-k$ $-z$'s. 1d answered Equating coefficients of binomial expansion modulo p Nov 25 comment What's wrong with my “proof” of the Cayley-Hamilton Theorem? Try $f(\lambda) = trace(A-\lambda I_n)$ instead. By the same logic, one would conclude $f(A) = 0$, but this is not true. Nov 24 accepted Minimal nontrivial coprime action - Kurzweil-Stellmacher Theorem 8.5.2 Nov 23 comment Show that multiplication $[(x, y)] * [(n,m)] = [(xn + ym, xm + yn)]$ is also well- defned. "which isn't coming out correctly" - can you explain what you are expecting to happen? Nov 23 comment Bayes Theorem: what is wrong in using counts instead, intuitively. It is mildly confusing to use "orange" as a fruit when the other dimension consists of colors. (I know, OP probably didn't make up the question.) Nov 23 asked Minimal nontrivial coprime action - Kurzweil-Stellmacher Theorem 8.5.2 Nov 23 comment Different series representation for the same function Also keep in mind that "the closest point at which the function is undefined" may be a complex number! If you expand $1/(1+x^2)$ around $x=0$, you will find the interval of convergence is $|x| < 1$ which might seem mysterious because the function is defined for all real numbers. But in the complex plane there are singularities at $\pm i$, which explains why there is no convergence around $x=0$ beyond radius 1. Nov 14 comment A group G with order $15$ is simple? @Mario How is this statement from Fraleigh related to your question or this answer? Nov 8 answered A definition in Character theory? Nov 8 comment Let $\gamma_1(G)=G$, $\gamma_i(G)=(\gamma_{i-1}(G),G)$. Show that $G$ is nilpotent iff $\gamma_m(G)=\langle e\rangle$ for some $m$. $C_{j+1}$ is defined as all elements that commute with all elements of $G$ up to an element of $C_j$. Another way of saying that: $C_{j+1}$ is the largest subgroup of $G$ satisfying $(C_{j+1}, G) \le C_j$. Does that help? Nov 8 revised Let $\gamma_1(G)=G$, $\gamma_i(G)=(\gamma_{i-1}(G),G)$. Show that $G$ is nilpotent iff $\gamma_m(G)=\langle e\rangle$ for some $m$. added 76 characters in body Nov 8 answered Let $\gamma_1(G)=G$, $\gamma_i(G)=(\gamma_{i-1}(G),G)$. Show that $G$ is nilpotent iff $\gamma_m(G)=\langle e\rangle$ for some $m$. Nov 8 comment Let $\gamma_1(G)=G$, $\gamma_i(G)=(\gamma_{i-1}(G),G)$. Show that $G$ is nilpotent iff $\gamma_m(G)=\langle e\rangle$ for some $m$. The two constructions are related by: $(C_i(G), G) \le C_{i-1}(G)$. This should help. Nov 3 answered What's the probability that hitting shuffle on an album of 5 tracks 10 times will allow you to see the original order at least once? Nov 2 comment Proving transitivity for a relation on Q You can't prove that $R$ is transitive, because it's not. Try to find a counterexample instead. Nov 2 answered Showing that the conjugates of a proper subgroup do not cover the group.