Abstract Nonsense

Crushing one theorem at a time

University of Maryland College Park Qualifying Exams (Group Theory and Representation Theory) (August-2004)


Point of Post: This is the August 2004 part of the post started here.

NOTE: Instead of doing this post by post I thought it would make more sense to have it all consolidated. See here for the up-to-date PDF of the parts of the exam I’ve finished.

Algebra Qual(Ph.D. Version)-August 2004

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Problem 1:

a) Suppose that A\leqslant B and (B:A)=2. Let C\leqslant B and |C| be odd. Show that C\subseteq A.

b) Let G be a finite group and suppose there exists subgroups

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G=G_0\supseteq\cdots\supseteq G_j=H

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with \left(G_i:G_{i+1}\right)=2 for i=0,\cdots,j. Suppose that H has odd order, show that H\unlhd G.

c) Let G be a group of order 2^j k where k is odd. Suppose that G contains a normal subgroup H of order k. Show that there exists subgroups

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G=G_0\supseteq\cdots\supseteq G_j=H

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with \left(G_i:G_{i+1}\right)=2 for i=0,\cdots,j.

Proof:  

a) Since \left(B:A\right)=2 we know that A\unlhd B and thus B/A\cong\mathbb{Z}_2. Let then c\in C, since |C| is odd and we know  |c|\mid |C| we know that |c| is odd and so |c|=2k+1 for some k\in\mathbb{Z}. Thus, c^{2(k+2)}=c that said if \pi is the canonical projection onto B/A then \pi(c)=\pi\left(c^{k+1}\right)^2=0 and thus by definition c\in \ker \pi=A. Since c\in C was arbitrary the conclusion follows.

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b) By definition we have that gHg^{-1}\subseteq G of course and since \left(G:G_1\right)=2 and \left|gHg^{-1}\right|=|H| is odd we have by a) that gHg^{-1}\subseteq G_1. Similarly, since \left(G_1:G_2\right)=2 and \left|gHg^{-1}\right| is odd we may conclude that gHg^{-1}\subseteq G_2. Continuing in this way we clearly get that gHg^{-1}\subseteq H. Since g\in G was arbitrary the conclusion follows.

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c) By Sylow’s Theorems we know that G has a subgroup Q_1 of order 2^{j-1}, and by Sylow’s theorems Q_1 has a subgroup Q_2 of order 2^{j-2}, and continuing we get a chain of subgroups

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Q_j\leqslant\cdots\leqslant Q_1\leqslant G

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such that \left|Q_r\right|=2^{j-r}. We recall then that since H\unlhd G we have that Q_1H\leqslant G and since \left(|Q_1|,|H|\right)=1 so that Q_1\cap H is trivial we may conclude from a common result that \left|Q_1 H\right|=2^jk. Similarly, we know that Q_rH\leqslant G and since \left(|Q_r|,|H|\right)=1 so that Q_r\cap H is trivial we may conclude that \left|Q_rH\right|=2^{j-r}k. Lastly, noting that since Q_i\geqslant Q_{i+1} so that Q_iH\geqslant Q_{i+1}H we note that

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G=Q_0\supseteq Q_1\supseteq\cdots\supseteq Q_j

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and

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\displaystyle \left(Q_i:Q_{i+1}\right)=\frac{2^{j-i}}{2^{j-(i+1)}}=2

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Remark: Of course this proves that every group of order 2^j k where k is odd which has a normal subgroup of order k is solvable.

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Problem 6: Let \rho:G\to\text{GL}_n\left(\mathbb{C}\right) be a representation of the finite group G. Define \widetilde{\rho}:G\times\mathbb{Z}_2\to\text{GL}_{2n}\left(\mathbb{C}\right) by the 2n\times 2n block matrix

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\widetilde{\rho}(g,0)=\left(\begin{array}{c|c}\rho_g & 0\\ \hline 0 & \rho_g\end{array}\right),\quad\quad\widetilde{\rho}(g,1)=\left(\begin{array}{c|c}0 & \rho_g\\ \hline \rho_g & 0\end{array}\right)

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a) Show that \widetilde{\rho} is a representation of G\times\mathbb{Z}_2.

b) Show that the number of times that the trivial representation of G occurs in \rho equals the number of times that the trivial representation occurs in the decomposition of \rho.

Proof:

a) Consider the representation \left(\rho\boxtimes \tau\right)\oplus\left(\rho\boxtimes\psi\right) where \tau is the trivial representation on \mathbb{Z}_2 and \psi is the other non-trivial representation on \mathbb{Z}_2. It’s evident that \widetilde{\rho}=\left(\rho\boxtimes\tau\right)\oplus\left(\rho\boxtimes\psi\right) and so trivially a homomorphism.

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b) We need only compute that \left\langle \chi_{\rho},\chi^{\text{triv}}\right\rangle=\left\langle\chi_{\widetilde{\rho}},\chi^{\text{triv}}\right\rangle where the inner product is taken on the group algebras \mathcal{A}\left(G\right) and \mathcal{A}\left(G\times\mathbb{Z}_2\right) respectively.  That said, from a) we see that \chi_{\widetilde{\rho}}=\chi_\rho\boxtimes\chi^{\text{triv}}+\chi_\rho\boxtimes\chi^- where \chi^-=\chi_\psi. It’s easy then to see that:

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\displaystyle \begin{aligned}\left\langle \chi_{\widetilde{\rho}},\chi^{\text{triv}}\right\rangle &=\frac{1}{|G\times\mathbb{Z}_2|}\sum_{x\in G\times\mathbb{Z}_2}\left(\chi_\rho\boxtimes\chi^{\text{triv}}+\chi_\rho\boxtimes\chi^{-}\right)(x)\\ &=\frac{1}{2|G|}\sum_{(g,0)\in G\times\mathbb{Z}_2}\left(\chi_\rho\boxtimes\chi^{\text{triv}}+\chi_\rho\boxtimes\chi^{-}\right)(g,0)\\ &=\frac{1}{|G|}\sum_{g\in G}\chi(g)\\ &=\left\langle\chi_\rho,\chi^{\text{triv}}\right\rangle\end{aligned}

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References:

1. Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

2. Simon, Barry. Representations of Finite and Compact Groups. Providence, RI: American Mathematical Society, 1996. Print.

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May 6, 2011 - Posted by | Algebra, Group Theory, Representation Theory, UMaryland Qualifying Exams | , , , , ,

3 Comments »

  1. (Q,+) has uncoutable sub groupes

    Comment by mohamad | May 13, 2011 | Reply

    • Indeed friend, it does. What does that have to do with this post though?

      Comment by Alex Youcis | May 14, 2011 | Reply

      • Lol, mohamad is confused.

        Nice post!

        Comment by Bruno Joyal | September 5, 2011


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