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5

So $\phi$ is defined by $\phi(g) = (1,\omega)$. Hence we must have $$\phi(1) = \phi(g^3) = (1, \omega)^3 = (1,1)$$ (Perhaps $(1,0)$ is a typo). For your second objection, note that $1+g$ mapsto $(2, 1+\omega)$, and $$\phi(1 + g)^2 = (2,1+\omega)^2 = (4, 1 + 2\omega + \bar\omega)$$ On the other hand \begin{align*} \phi(1 + 2g + g^2) &= (1, 1) + ...

3

You ought to say what $\phi:\mathbb{R}\to SO_2(\mathbb{R})$ is. Define $\phi(\theta)=\begin{pmatrix}\cos\theta&-\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix}$ and show that $\phi(\alpha+\beta)=\phi(\alpha)\phi(\beta)$ (so then it is a homomorphism). It is surjective by definition of $SO_2(\mathbb{R})$. Since $\phi(\theta)$ is the identity ...

2

The straight-forward approach is Gram-Schmidt. Start with the standard (canonical) basis $\{e_1,e_2\}$. Our goal is to get an orthonormal basis $\{v_1,v_2\}$. First, we take $$v_1 = \frac {1}{\sqrt{\langle e_1,e_1 \rangle}} e_1 = \frac 1 {\sqrt 2}e_1$$ Now, take $$w_2 = e_2 - \langle e_2,v_1 \rangle v_1 = e_2 - \frac 1 2 \langle e_2,e_1 \rangle e_1 = ... 2 Maschke's theorem holds if the characteristic of the field doesn't divide the order of the group (you missed a very important hypothesis there!). The exercise is asking you to show that this representation is decomposable but not reducible (meaning it has a subrepresentation but it can't be written as a direct sum of irreducible representations) Let's look ... 2 You're absolutely right that (ii) does not obviously follow from (i) as stated. What (i) should say is not just that I_i is ring-isomorphic to M(n_i,F) but that it is isomorphic to M(n_i,F) as an F-algebra. Concretely, this means that if you take an element a\in F, consider it as an element of FG, project it to I_i, and then map it to ... 2 W^{⊥⊥} contains W is trivial by definition. Equality follows from dimensionality since taking perp gives a new dimension of n minus the old one. This doesn't work in infinite dimensions but that's good because it's not true. 1 Generally, if W_1,\cdots,W_h are irreducible representations of G and h is the number of conjugacy classes of G (which equals the number of irreducible representations), letting n_i \overset{def}= \dim W_i, then one has$$ \sum_{i=1}^h n_i^2 = g. $$The proof is somewhat lengthy and involves proving many little intermediate results ; I suggest you ... 1 The first thing to note is that you have the wrong eigenvalues. The characteristic polynomial for g is \lambda^2-\lambda+1 and has eigenvalues$$\lambda=\frac{1+i\sqrt{3}}{2}=e^{2\pi i/6}$$and$$\overline{\lambda}=\frac{1-i\sqrt{3}}{2}=e^{-2\pi i/6}=\lambda^{-1}.$$Now, let \{x,y\} be a basis for \mathbb{C}^2 for which g acts by the matrix$$ ...

1

The first step is to adapt the proof that $$End_R(V_1\oplus V_2)\cong End_R(V_1)\oplus End_R(V_2)$$ to the case of arbitrary $V_1$ and $V_2$ satisfying $Hom_R(V_1,V_2)=0$. Then, prove the following: For $\phi\in\mathrm{End}_R(S^n)$, define $$\phi_{ij}=\pi_j\circ\phi\circ \iota_i$$ where $\pi_j:S^n\to S$ is the projection onto the $j$th factor and ...

1

We can write $$c_\lambda =\sum_{g\in P_\lambda ,g^\prime\in Q_\lambda}sgn(g^\prime )\,e_{gg^\prime}$$ It is enough to show that coefficient of $e_{id}$ (where $id$ is the identity) in the above sum is non zero; in fact, we will show that the coefficient is $1$. This will follow from the fact that $gg^\prime =id$, for $g\in P_\lambda , g^\prime\in ... 1 Note that one version of Maschke's theorem tells us even more, namely that the group algebra$\mathbb{C} G$is semisimple. This implies that every$\mathbb{C} G$-module (finite dimensional or not) is a direct sum of simple modules. If$V$is a$\mathbb{C} G$-module and$M_1,\dots, M_n$are representatives for the isomorphy classes of the simple$\mathbb{C} ...

1

The document only says that $$\phi(g)=(1, \omega)$$ As you observed you must have $$\phi(1)=(1,1)$$ Fixing this, the second relation becomes: $$((1,1) + (1,\omega))((1,1)+(1,\omega)) = (2,1+\omega)(2,1+\omega) = (4,1+2\omega+\bar{w}) \\ =(1,0) + 2(1,\omega) + (1,\overline \omega)$$

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A direct sum of $FH$-modules is also a direct sum of vector spaces, so it is enough to consider a single term $M \otimes t_i$ in the decomposition. Now $M \otimes t_i = \{ m \otimes t_i : m \in M \}$. So, if $a_1,\ldots,a_n$ is a basis of $M$ as a vector space over $F$, then every element of $M$ can be written uniquely as $\sum_{i=1}^n f_i a_i$ with $f_i ... 1 The isomorphism you have written down is just the Chinese remainder theorem. You have an isomorphism $$\phi:\mathbb{Q}[x]/(x^3-1)\to\mathbb{Q}[x]/(x-1)\oplus\mathbb{Q}[x]/(x^2+x+1),\;\;\; f+(x^3-1)\mapsto (f+(x-1),f+(x^2+x+1)).$$ I will omit the coset notation for convenience. Under this map, we have $$\phi(x^2+x+1)=(3,0)$$ since$x^2+x+1\equiv 1^2+1+1=3$... 1 Consider internal decomposition$\mathbb{Q}[G]=I\oplus J$. In$I$there is an element$e=\frac{1+g+g^2}{3}$(it is a Primitive central idempotent), and$I$is two-sided ideal generated by this element. We can write$I=Ie$. Then$Ie$becomes an algebra, in which additive identity is$0$but multiplicative identity element is$e$; we can show that$Ie$is ... 1 No,$\pi_0$is not linear (unless$G$is trivial so$U=0$and$\pi_0=0$). For instance, let$u\in U$and$v\in V\setminus U$be any two nonzero vectors. Then$u+v\not\in U$(otherwise$v=(u+v)-u$would be in$U$), so$\pi_0(u+v)=0\neq u=\pi_0(u)+\pi_0(v)\$.

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