# Tag Info

7

There are infinite groups which cannot be represented by finite-dimensional matrices over any commutative ring. If a group can be represented by matrices in such a way then it is called linear (perhaps this definition really requires field not commutative ring, but everything in this answer works for the more general commutative ring definition). Not all ...

4

The parity of an element $sgn(\sigma)$ gives you a linear (irreducible) character (one-dimensional representation). Hence, by Frobenius orthogonality relations, the inner product with the other characters is $0$, and this is exactly your sum! (Note that conjugate permutations have the same sign)

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I find this a natural question. Particularly because I recently heard that I would be teaching group theory for advanced undergrads next year (or the year after), so I want to test my motivational skills here :-) Indeed, one of the aspects of representation theory is to study the groups being represented. It is nice that we can take some abstract group, and ...

3

A representation of a group $G$ is a morphism $\rho:G\rightarrow\text{GL}(V)$ where $V$ is a vector space over a field $K$. There are always at least the following two representations: the trivial representation, where $\rho(g)=\text{id}_V$ for all $g\in G$. the regular representation, where $V$ is the space of $K$-valued functions on $G$ and ...

3

For 1 Yes, it's true. The trick is to remember that the simple modules of $A$ are the same as the simple modules of $A/J(A)$, where $J(A)$ is the Jacobson radical of $A$. Since $A$ is a finite dimensional algebra, it is a right and left Artinian and Noetherian ring. As such, it has a composition as a left module over itself (and as a right module over ...

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However, if $K = \mathbb{Q}$, the problem is different, I don't think I can construct $A$ for every $m$. This is correct. the following result is (I think) well known Let $m = p_1^{\alpha_1}\cdots p_k^{\alpha_k}$ be the prime factorization with $p_1<\cdots < p_k$. The group $\mathrm{GL}_n(\mathbb Q)$ has an element of order $m$ if and only if ...

1

You need to work over an algebraically closed field to state Schur's Lemma in the form given in Serre's book. Your generalization is basically correct (in that situation). It is probably easiest to break the proof into two cases : if $\rho^{1}$ and $\rho^{2}$ have no common irreducible constituent, then ${\rm Hom}(\rho^{1},\rho^{2}) = 0.$ And if $\rho^{1}$ ...

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A faithful representation $f\colon G \to GL(V)$ leads to infinitely many faithful representations, one each on $V, V\oplus V, V\oplus V\oplus V$ etc simply by replicating it in each factor. These extra things provide no new information. Irreducibilty would have eliminated this infinite unnecessary explosion.

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I won't calculate the exact result but let me give you some ingredients: You should know how to transform a matrix to another via a change of basis. This way we can work with a preferred one. Mine is the standard basis. Since you used $e_i$ already lets call them $v_1,v_2,v_3$. You hopefully know how the permutation matrices that you called ...

1

As you stated, every (finite) permutation group can be represented by matrices (take $G=S_n$ and represent the permutations by permutation matrices (the permutations act naturally on a vector space of dimension $n$ over some field $K$)). Subsequently, every finite group can be embedded in a permutation group (a Theorem of Cayley), since it acts on itself by ...

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The branching rule of $\mathrm O_n \subset \mathrm{GL}_n$ gives $$V_\lambda = \bigoplus_{\mu} \Big(\sum_{\delta} c_{2\delta,\mu}^\lambda \Big) W_\mu$$ where $W_\mu$ are the irreducible $\mathrm O_n$ representations and $c$ the Littlewood-Richardson coefficients. (Note that $2\delta$ are "even partitions", i.e. all summands are even.) Your question is now, ...

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They don't have to be... recall that you only get a matrix when you fix a basis... For example, let $p=a_3x^2+a_2x+a_1$ be in $V$ the vector space of quadratic polynomials and $\Phi$ the action of $S_3$ on $V$ given by $$\Phi(p,\pi)=\pi(a_3)x^2+\pi(a_2)x+\pi(a_1).$$ This thing is a representation (unless I am acting on the wrong side... not sure) Of ...

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