# Search results

1. ### Challenge Math Challenge - May 2021

Proof: We have \begin{align*} \tau(h) \circ \text{Sym}(\varphi) &= \tau(h)\circ \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\ &= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(h) \circ \tau(g) \circ \varphi \circ \rho(g^{-1}) \\ &= \frac{1}{\vert G \vert}...
2. ### Challenge Math Challenge - May 2021

For ##\varphi \in \text{Hom}_{\mathbb{K}} ((\rho, V), (\tau, W))##, we have ##\tau(h) \text{Sym}(\varphi) = \text{Sym}(\varphi) \rho(h)## Proof: We have \begin{align*} \tau(h) \circ \text{Sym}(\varphi) &= \tau(h) \circ \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ...
3. ### Challenge Math Challenge - May 2021

I think the calculation is \begin{align*} \text{Sym}(\varphi) &= \frac{1}{\vert G \vert} \sum_{g \in G} \tau(g) \circ \varphi \circ \rho(g^{-1}) \\ &= \frac{1}{\vert G \vert} \sum_{g \in G} \varphi \\ &= \frac{1}{\vert G \vert} \cdot \vert G \vert \varphi \\ &= \varphi \\ \end{align*} Edit...

5. ### Challenge Math Challenge - May 2021

that clears things up, thank you!
6. ### Challenge Math Challenge - May 2021

One other question, what is ##\text{Hom}_{\mathbb{K}}((\rho, V), (\tau, W))##? I'm pretty sure ##\text{Hom}_{\mathbb{K}}(V, W)## is the set of all ##\mathbb{K}##-linear maps from ##V## to ##W##. And we can make ##V## into a ##\mathbb{K}G## module by defining ##g \cdot v = \rho(g)v## I think?? So...
7. ### Challenge Math Challenge - May 2021

I think I get it now, I had to look up the definition of GL(V). So, ##\rho(g^{-1})## is an automorphism of ##V## and ##\tau(g)## is an automorphism of ##W## and ##\varphi## is a k linear map from ##V## to ##W##.
8. ### Challenge Math Challenge - May 2021

Sorry for the dumb question but do the ##\circ##'s in 4) mean function composition or matrix multiplication? As I understand it, ##\tau(g)## and ##\rho(g^{-1})## are matrices in ##GL(W)## and ##GL(V)## resp. ?

10. ### Challenge Math Challenge - January 2021

Thank you for your time and feedback; and sorry for confusion! Here are two things I would add to make post #20 clearer: Lemma: Let ##G## be a group and ##H## a subgroup. If ##[G: H] = n##, then there exists a homomorphism ##f : G \rightarrow S_n## with ##\ker f \le H##. Proof: Let ##X## be...

Question 9
12. ### Challenge Math Challenge - December 2020

I think the only subgroup of index ##4## is the center ##\lbrace \pm1 \rbrace##. Any subgroup of order ##2## must be generated by an element of order ##2##. There is only one element of order ##2## in ##Q_8##, namely ##-1##, and so there can only be one subgroup of order ##2##. Equivalently...

15. ### Constructive Proofs (open) Little Fermat Proof

Proof: Let ##C## be the set of all ##p## colored chain with ##a## colors. Let ##c \in C##. We denote ##c = (c_1, c_2, \dots, c_p)## to be a ##p## colored chain where ##c_i## is any of the ##a## colors. Define ##\tau = (1 2 3 \dots p)##. Then ##G = \langle \tau \rangle## acts on ##C## by ##\tau^k...
16. ### Challenge Math Challenge - February 2020

For #5, Let ##R## be a commutative ring. To say a subset ##I \subseteq R## is an ideal of ##R## means that ##I## is nonempty, ##I## is closed under addition and that for any ##r \in R## and ##i \in I##, we have ##ri \in I##. And to say ##\operatorname{char} R \neq 2## means ##2## is not the...
17. ### Challenge Math Challenge - January 2020

Thank you for the correction and your insight!
18. ### Challenge Math Challenge - January 2020

I'll take a shot at #8...
19. ### Challenge Basic Math Challenge - June 2018

I sincerely mean this, I always appreciate your comments and had not thought of it from that perspective, that writing Euler's theorem is ambiguous and that writing FLT could save time for the reader. I definitely think the proof I wrote could be improved by being more specific.
20. ### Challenge Basic Math Challenge - June 2018

thanks! thanks for pointing that out, I guess since the moduli are all prime I did mean FLT to be more specific

For 6)