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6

"He wrote something about $\rho $ being an isomorphism but as far as I know, linear representations usually are only (Group-)Homomorphisms but not necessarily Isomorphisms." The problem is you're being a little sloppy about exactly what a representation is. If $\rho $ is a representation of $G$ on $V$ there's no such thing as $\rho W$. In fact $\rho $ is ...


3

Let $G$ be a Lie group and $H$ a closed subgroup of $G$. In general, the unitarity of a representation of $H$ is not preserved when inducing unless there is a choice of invariant positive density (positive Haar measure) on $G/H$. To avoid this problem, we tensor the original representation with the one-dimensional representation given by the square root of ...


3

$[G:G']$ is the number of degree $1$ irreducible representation. So your last case is impossible. There are only $5$ groups of order $12$, two of which is abelian (and hence $12 = 12\times 1$). The three non-ablian groups are $D_{12},A_4$ and $T$ (up to isomorphism). You can try to find the explicit representations by the given degrees. Don't forget $12 = 3\...


2

By definition the trace of $g$ acting on $U$ is $\psi(g)$. So the trace of $\frac1{|G|}\sum_{g\in G}\chi(g^{-1})g$ acting on $U$ is $$\frac1{|G|}\sum_{g\in G}\chi(g^{-1})\psi(g).$$ This is an integer, the number of copies of the irreducible module $U$ within the module $V$.


2

Firstly, in your solution it seems you're looking at $V\otimes V$ (where $V=\mathbb{C}^2$). If instead we consider $V\otimes V^*$, this is isomorphic to $M_2(\mathbb{C})$, via $$ v\otimes \langle w,\cdot\rangle \longleftrightarrow vw^\top, $$ where $\langle v,w\rangle = v^\top\overline{w}$ denotes the Hermitian inner product on $\mathbb{C}^2$. The ...


2

Your expectation that $v_{+}\otimes v_{+}$ ought to correspond to the identity is incorrect. The tensor product of two vectors $v$ and $w$ is (as a matrix) $$\begin{bmatrix} v_1w_1 & v_1w_2 \\ v_2w_1 & v_2w_2 \end{bmatrix}$$ Second, if you're talking about $M_2(\mathbb{R})$ there are the three listed irreps, but if you're talking about $M_2(\mathbb{...


2

Let $I\subset M$ be a submodule with $I/M^2$ a hyperplane of $M/M^2$. Then $I$ is in particular an ideal of $A$ and the annihilator of $A/I$ is precisely $I$. So if $A/I\simeq A/J$, then their annihilators are the same, so $I=J$: they are all pairwise nonisomorphic. Now the "geometry" part is that since $M/M^2$ is of dimension at least $2$, it has ...


2

Recall the irreducible characters forms a basis of the $\mathbb{C}$-vector space of class functions, so to prove you have all the irreducible characters, it suffices to show there are no class functions not represented by linear combinations of your set of irreducible characters. Since we have an inner product, we just need to show the orthogonal complement ...


2

This is an immediate consequence of Sylow's Theorem. $G$ has a $p$-Sylow subgroup $H$ of order $p$. Because $H$ is a Sylow subgroup, all other $p$-Sylow subgroups are necessarily conjugate to $H$, and the number $n$ of such conjugates satisfies $n \equiv 1 \pmod{p}$. Since $n \vert \vert G \vert$ and $p \geq 5$, this forces $n=1$ so the unique $p$-Sylow ...


1

Hint: if you have proved that $|G:G'|=4$, then $|G'|=p$. Now let $H \leq G$ with $|H|=p$. If $H \cap G'=1$ then $|HG'|=|H|\cdot|G'|=p^2 \leq |G|=4p$, whence $p \leq 4$, a contradiction. It follows that $H=G'$, and $H$ is normal.


1

Write $W=\bigoplus_{j=1}^m W_j$ with the $W_j$ irreducible. Then $$\text{Hom}(V,W)=\text{Hom}\left(V,\bigoplus_{j=1}^m W_j\right) \cong\bigoplus_{j=1}^m \text{Hom}(V,W_j)$$ and so $$\dim\text{Hom}(V,W)=\sum_{j=1}^m \dim\text{Hom}(V,W_j).$$ By Schur's lemma, $\dim\text{Hom}(V,W_j)=1$ or $0$ according to whether or not $W_j$ is isomorphic to $V$ or not. ...


1

Let $\pi_i:G\to\mathcal{H}_i$ be the two unitary representations and $A\colon \mathcal{H}_0 \to \mathcal{H}_1$ be the linear equivalence you have, i.e. $A\pi_0A^{-1} = \pi_1$, then $$ \pi_1^{-1} = \pi_1^{*} = (A\pi_0A^{-1})^* = A^{*-1}\pi_0^{*}A^* = A^{*-1}\pi_0^{-1}A^* \Rightarrow A^*\pi_0A^{*-1} = \pi_1 $$ Therefore $A^*$ is also a linear equivalence, and ...


1

$G=\{g_1,\cdots , g_k\}$ so you can apply the Grassman formula to have that $dim(V)\geq dim(F(g_1)+\cdots +F(g_k))=$ $\sum_{s=1}^k dim(F(g_s))-dim(\cap_{s=1}^kF(g_s))\geq$ $ \frac{k}{2}dim(V)-dim(\cap_{s=1}^kF(g_s) )$ So $dim(\cap_{s=1}^kF(g_s))\geq (\frac{k}{2}-1)dim(V)>1$ when the order of $G$ is at least $3$


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