# 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. ### Finding the dimension of a subspace

We know ##y : V \rightarrow \mathbb{R}## is a linear transformation. By Rank-Nullity theorem we have ##\dim V = rank(y) + null(y)##. We note that ##null(y) = \dim W##. If ##rank(y) = 0##, then ##\dim V = \dim W## and so ##\dim W = n##. If ##rank(y) \neq 0##, then ##rank(y) = 1##. And so ##\dim...
13. ### Finding the dimension of a subspace

Thank you for your replies. Yes, I think we can use Rank-Nullity theorem, I will try it. And yeah, I'm happy to use a more standard notation if that's better; I was trying to mimic the textbook.
14. ### Finding the dimension of a subspace

I am stuck on finding the dimension of the subspace. Here's what I have so far. Proof: Let ##W = \lbrace x \in V : [x, y] = 0\rbrace##. We see ##[0, y] = 0##, so ##W## is non empty. Let ##u, v \in W## and ##\alpha, \beta## be scalars. Then ##[\alpha u + \beta v, y] = \alpha [u, y] + \beta [v...
15. ### Showing two groups are equal/Pontryagin duality

We want to show ##\vert (H^\perp)^\perp \vert = \vert H \vert##. We have ##\vert (H^\perp)^\perp \vert = [\widehat{\widehat{G}} : H^\perp] = \frac{\vert\widehat{\widehat{G}}\vert }{\vert H^\perp\vert} = \frac{\vert G \vert}{[G: H]} = \vert H \vert##
16. ### Showing two groups are equal/Pontryagin duality

I am confused because ##H## is a subgroup of ##G## and ##H^{\perp\perp}## is a set of homomorphisms. Are we trying to show ##f(H) = H^{\perp\perp}## where ##f## is the isomorphism defined in the definitions above? Proof: We want to show ##H = (H^\perp)^\perp##. ##(\subset):## Let ##h \in H##...
17. ### 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...

19. ### Lower central series

Thank you! I think I finally understand.
20. ### Lower central series

Thanks for the reply! So we're trying to prove ##(\gamma_{n} \lhd G) \Rightarrow (\gamma_{n+1} \le \gamma_n) \Rightarrow (\gamma_{n+1} \lhd G)##. For the second implication, we want to show ##\gamma_{n+1} \lhd G##. Let ##g \in G## and ##x \in \gamma_{n+1}##. Then ##gxg^{-1} \in \gamma_n##...

22. ### Lower central series

My attempt: If ##i = 1##, then ##\gamma_1 = G \rhd G' = \gamma_2##. We proceed by induction on ##i##. Consider an element ##xyx^{-1}y^{-1}## where ##x \in \gamma_i## and ##y \in G##. Since ##\gamma_i \rhd G##, we have ##yx^{-1}y^{-1} = x_0 \in \gamma_i##. So, ##xyx^{-1}y^{-1} = xx_0 \in \gamma_i...
23. ### Refining a normal series into a composition series

That makes sense; Maybe something like this would work? Let ##P(k)## say that if a group ##G## of order ##k## has a composition series, then it cannot have an infinite normal series. base case: If ##k = 1##, then ##G## has a composition series ##G \ge 1##. It's clear that any normal series of...
24. ### Refining a normal series into a composition series

I don't think I can find the ##H## refinement. Let ##G = G_0 \ge G_1 \ge \dots \ge G_n = 1## be a finite composition series and assume by contradiction there exists an infinite normal series ##G = H_0 \ge H_1 \ge \dots \ge 1##. I thought maybe to insert ##G_1, G_2, \dots , G_{n-1}## into the...
25. ### Refining a normal series into a composition series

Thanks for the reply! Let ##G = G_0 \ge G_1 \ge \dots \ge G_n = 1## be a normal series. 1) If ##G_0/G_1## is a non simple factor, then there exists a proper normal subgroup ##G_{0,0}/G_1 \triangleleft G_0/G_1##. By correspondence theorem, ##G_0 \triangleright G_{0,0} \triangleright G_1##...
26. ### Refining a normal series into a composition series

Attempt: Consider an arbitrary normal series ##G = G_0 \ge G_1 \ge G_2 \ge \dots \ge G_n = 1##. We will refine this series into a composition series. We start by adding maximal normal subgroups in between ##G_0## and ##G_1##. If ##G_0/G_1## is simple, then we don't have to do anything. Choose...
27. ### Normal series and composition series

That makes sense, thank you!
28. ### Normal series and composition series

I tried reading van der Waerden's proof and just want to make sure I understand... Suppose we have ##G_i/G_{i+1} = H_j/H_{j+1}##. There is a correspondence between the normal series for ##G_i/G_{i+1}## and the series' from ##G_{i}## to ##G_{i+1}##. Hence, for any refinement of a series from...
29. ### Normal series and composition series

Yes, I think you're right about the mapping taking place on the indices. So if ##G_i/G_{i+1} = H_j/H_{j+1}##, then for any normal series ##G_i \ge G_{i,1} \ge \dots \ge G_{i, k} \ge G_{i+1}## there exists equivalent normal series ##H_j \ge H_{j, 1} \dots \ge H_{j, k} \ge H_{j+1}##? I'm sorry if...
30. ### Normal series and composition series

Thanks for the reply! Proof: We are given that ##G## and ##H## are finite which implies they both have composition series. We are also given that there exists normal series $$G = G_0 \ge G_1 \ge \dots \ge G_n = 1$$ and $$H = H_0 \ge H_1 \ge \dots \ge H_n = 1$$ that are equivalent. Hence, for...