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 Mar 28 awarded Popular Question Dec 30 comment Homomorphism induced by a permutation group action must be injective By "the" action, I'm assuming you're referring to the one used in the solution on PCP. I think at one point I may have assumed that was the identity map -- in which case, yes, the action clearly has a trivial kernel. But why are we allowed to choose how $G$ acts on $A$? That is, if each action is in one-to-one correspondence with a homomorphism from $G$ to $S_A$, what's to stop me from choosing something like Dec 30 comment Homomorphism induced by a permutation group action must be injective @DerekHolt Just looked this up: D&F (p. 114) write: "If $G$ is a group, a permutation representation of $G$ is any homomorphism of $G$ into the symmetric group $S_A$ [...]. We shall say a given action of $G$ on $A$ affords or induces the associated permutation representation of $G$." Dec 30 comment Homomorphism induced by a permutation group action must be injective @DerekHolt So is my error in the terminology or in the computation? Dec 30 comment Homomorphism induced by a permutation group action must be injective @DerekHolt That I'm confusing myself seems totally plausible. However, I don't see how $\theta$ maps $\sigma$ to itself; I defined it as $\sigma \mapsto \sigma^2$. Unless I'm missing something -- like I said: totally plausible -- $\theta(\sigma^2)=\text{id}$. Dec 30 asked Homomorphism induced by a permutation group action must be injective Nov 29 asked On the distribution of the primes / the probability of a false positive from Miller-Rabin Nov 29 asked Conditions that suffice to show compositeness using Miller-Rabin test Nov 28 comment How do you prove null $A$={0} $\iff$ {$c_1,…,c_n$} is independent? @CoolKid, Apologies, I'm a bit rusty with linear algebra, so it was a bit more complicated than it needed to be. I've simplified the approach, so hopefully that will clarify things. (I also added a hidden solution.) Nov 28 revised How do you prove null $A$={0} $\iff$ {$c_1,…,c_n$} is independent? simplified path to a solution Nov 28 revised How do you prove null $A$={0} $\iff$ {$c_1,…,c_n$} is independent? simplified path to a solution Nov 28 revised How do you prove null $A$={0} $\iff$ {$c_1,…,c_n$} is independent? added 125 characters in body Nov 28 answered How do you prove null $A$={0} $\iff$ {$c_1,…,c_n$} is independent? Nov 22 comment Credit Analysis - Mean, Variance, Deviation Does your textbook define any of these terms? Nov 22 comment Credit Analysis - Mean, Variance, Deviation What characteristics must $P$ have in order to be a probability distribution? Of course, the entire question is pretty easy if the answer to (a) is no. So, just in case $P$ is a probability distribution: How do you define mean, variance and so forth? Nov 22 accepted Bounding the $p$th moment of a sum of random variables by the sum of the $p$th moments Nov 22 comment Bounding the $p$th moment of a sum of random variables by the sum of the $p$th moments Like I said (or almost did; I left out a word), it looked like Hoelder: I had seen his inequality written as a product of functions, not as a sum (or dot product). I'll look into Hoelder more, but this makes a lot more sense now. Thank you very much! Nov 22 comment Bounding the $p$th moment of a sum of random variables by the sum of the $p$th moments Thanks, I sorted out both parts, using your suggestion for (c) and proving (b) with what amounts to the same idea: If $f$ is convex, then $f[(x+y)/2]\le [f(x)+f(y)]/2$. The renditions of Minkowski that I've seen look like $||X+Y||_p \le ||X||_p +||Y||_p$; yours looks Hoelder's inequality, which I didn't expect to work. Could you expand on your ideas for (b)? Like I said, I got it one way, but I'm interested in learning how all this stuff works; I'm taking this class with a spotty background in analysis and non-existent background in measure theory, so I'm sure there's tons I've missed so far! Nov 22 comment Area between two curves. Split so that it is equal. @YunusSyed, I believe it's from the Briggs calculus text; I don't recall it being marked as a competition problem or anything, though... Nov 22 comment How many triangles are possible with no side greater than $4$ units? @Ben Longo, could you discuss how you got that?