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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: $$((1,1) + (1,\omega))((1,1)+(1,\omega)) = (2,1+\omega)(2,1+\omega) = (4,1+2\omega+\bar{w}) \\ =(1,0) + 2(1,\omega) + (1,\overline \omega)$$


1

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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