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...
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.
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...
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##
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##...
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##...
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...
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...
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...
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##...
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...
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...
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...
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...
Attempt so far: We're given that ##G## and ##H## have equivalent normal series
$$G = G_0 \ge G_1 \ge \dots \ge G_n = 1$$
and
$$H = H_0 \ge H_1 \ge \dots \ge H_n = 1$$
We can assume they have the same length because they are equivalent. I think from here I need to construct two composition...
Response to post #4
I think I can do the element of order ##2## part, but showing ##N_G(x)## has index ##5## has me stuck, unless we use post #2.
Proof: Continuing from the OP, we have ##n_2 = 5## or ##15##. Suppose ##n_2 = 5## and let ##P## be a Sylow 2 subgroup of ##G##. Then ##N_G(P)## has...
Thank you! My only question is, how did we rule out ##C_G(x) \neq 10##?
Proof: Let ##P, Q## be distinct Sylow ##2## subgroups of ##G##. Suppose ##P, Q## do not intersect trivially. Then there exists non identity element ##x \in P \cap Q##. Since ##P, Q## have order ##4##, they are abelian...
Proof: We note ##60 = 2^2\cdot3\cdot5##. By Sylow's theorem, ##n_5 = 1## or ##6##. Since ##G## is simple, we have ##n_5 = 6##. By Sylow's theorem, ##n_3 = 1, 4, ## or ##10##. Since ##G## is simple, ##n_3 \neq 1##. Let ##H## be a Sylow ##3## subgroup and suppose ##n_3 = 4##. Then ##[N_G(H) : G] =...
I think I got it?
Proof of ii): Let ##X = \lbrace gPg^{-1} : g \in G \rbrace##. Then ##G## acts on ##X## by conjugation. If ##g \in G##, then ##gPg^{-1} \le gKg^{-1} = K## since ##K## is a normal subgroup of ##G##. Hence, ##gPg^{-1}## is a Sylow p-subgroup of ##K##. So there is ##k \in K## such...
Attempt at solution:
Proof of i): Let ##x \in X##. Its clear ##G \supseteq HG_x##. Let ##g \in G##.Then there is ##y \in X## such that ##g \cdot x = y##. Since ##H## acts transitively on ##X##, there is ##h \in H## such that ##h \cdot x = y##. So, ##g \cdot x = h \cdot x##. This gives...
I see using post #3:
Proof: First we show ##\operatorname{null}(ST) = T^{-1}(\operatorname{null}(S))##.
##(\subseteq)## Let ##x \in \operatorname{null}(ST)##. Then ##S(Tx) = 0## which means ##Tx \in \operatorname{null}(S)## i.e. ##x \in T^{-1}(\operatorname{null}(S))##.
##(\supseteq)## Let...
Thank you, I think I got it using what you said:
Proof: Since ##U, V## are finite dimensional, we have ##\operatorname{null} S## and ##\operatorname{null}T## are finite dimensional. Let ##u_1, \dots, u_n## be a basis for ##\operatorname{null} T## and ##Tx_1, \dots, Tx_k## be a basis for...
Proof: Since ##U, V## are finite dimensional, we have that ##\operatorname{null} S, \operatorname{null} T## are finite dimensional. Let ##v_1, \dots v_m## be a basis of ##\operatorname{null}S## and ##u_1, \dots, u_n## be a basis of ##\operatorname{null} T##. It is enough to show there are ##m +...
Proof: Consider any finite ##\sigma##-algebra ##\mathcal{A}## such that for all ##A \in \mathcal{A}##, either ##A## or ##A^c## is finite, but not both. By definition of ##\nu##, ##\nu(\emptyset) = 0##. Suppose ##A, B \in \mathcal{A}## are infinite sets. Then ##\vert (A\cap B)^c \vert = \vert A^c...
Edit to OP: ##\nu## is a measure on ##\lbrace \emptyset, A , A^c, \mathbb{R} \rbrace## where ##A## or ##A^c## is finite but not both.
I'm sorry I have to come back in an hour or two.
Suppose ##\nu## is a measure on some ##\sigma##-algebra ##\mathcal{A}##. Then we must have for all ##A \in \mathcal{A}## either ##A## or ##A^c## is finite, but not both. Because otherwise ##\nu(A)## is undefined or not well defined.
I've verified that ##\lbrace \emptyset, X \rbrace## and...
Thank you, that makes sense.
I think things are cleared up. Carathéodory's theorem (the one in OP) gives us sufficient but not necessary conditions to extend a pre measure. So we don't necessarily have to meet the requirement of the theorem in order to extend a pre measure to a measure. And...