Abstract Nonsense

Crushing one theorem at a time

University of Maryland College Park Qualifying Exams (Group Theory and Representation Theory) (August-2003)

Point of Post: This is the August 2003 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 2003

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Problem 1:

a) Let G be a group and H\leqslant G be a cyclic normal subgroup. Show that if K\leqslant H then K\unlhd G.

b) Let G be the n^{\text{th}} dihedral group with presentation

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\left\langle a,b\mid a^n=b^2=e,ba=a^{-1}b\right\rangle

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Let p be an odd prime dividing n. Prove that G contains a unique Sylow p-subgroup.


a) Let H=\langle h\rangle then we know that K is cyclic and so K=\left\langle h^k\right\rangle for some k\in\mathbb{Z}. So let g\in G be arbitrary, to show that i_g\left(K\right)\subseteq K (where i_g is the inner automorphism associated to g) it obviously suffices to show that i_g\left(h^k\right)=gh^kg^{-1}\in K. To do this we first note that since H\unlhd G and h^k\in H we have that gh^kg^{-1}\in H. We then recall that order is invariant under conjugation and so in particular \left|\left\langle gh^kg^{-1}\right\rangle\right|=|K| but since a cyclic group has precisely one subgroup of every order dividing it we may conclude that \left\langle gh^kg^{-1}\right\rangle=K and so in particular gh^kg^{-1}\in K. The conclusion follows.

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b) Consider \left\langle a\right\rangle this is a cyclic subgroup of D_n of order n, moreover it’s normal since \left(D_n:\left\langle a\right\rangle\right)=2 to which we can appeal to the common fact that a subgroup of index two is always normal (this is a corollary of a more general fact). But, we note that if the maximum power of p that divides 2n is p^m that p^m\mid n (since p is an odd prime) and so by Sylow’s theorems that \left\langle a\right\rangle must contain a Sylow p-subgroup P of order p^m. But, by part a) we have that P is a normal Sylow p-subgroup of G and since for Sylow p-subgroup being normal and unique are equivalent the conclusion follows.

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Problem 6: The character table of a group G is given below. 

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\begin{array}{c|ccccc}G & \mathcal{C}_1 & \mathcal{C}_2 & \mathcal{C}_3 & \mathcal{C}_4 & \mathcal{C}_5\\ \hline \chi_1 & 1 & 1 & 1 & 1 & 1\\ \chi_2 & 3 & 0 & \varphi & 1-\varphi & -1\\ \chi_3 & 3 & 0 & 1-\varphi & \varphi & B\\ \chi_4 & 4 & 1 & -1 & -1 & 0\\ \chi_5 & 5 & C & 0 & 0 & 1\end{array}

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where \varphi is the golden ratio. Suppose then that \#(\mathcal{C}_1)=1, \#(\mathcal{C}_2)=20, \#(\mathcal{C}_3)=\#(\mathcal{C}_4)=12 and \#(\mathcal{C}_5)=A

a) Compute A,B,C.

b) Use the character table of G to show that G is not solvable.

c) Use the character table of G to show that there does not exist a homomorphism \rho:G\to\text{GL}_4\left(\mathbb{C}\right) such that 

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\displaystyle \rho(g)=\begin{pmatrix}1 & 26 & -3 & 56\\ 0 & -30 & 0 & -67\\ 1 & 0 & -2 & -2\\ 0 & 13 & 0 & 29\end{pmatrix}

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for some g\in\mathcal{C}_2.


a) We know that \displaystyle \sum_{j=1}^{5}\chi_j\left(\mathcal{C}_1\right)^2=|G| (we know that \mathcal{C}_1 is equal to the conjugacy class of e since it is the only trivial conjugacy class). Thus, |G|=60. But, by the orbit-decomposition theorem that \displaystyle \sum_{j=1}^{5}\#(\mathcal{C}_j)=|G|=60 and so doing the simple computation we find that A=15. To find C we recall (considering the inner product of \chi_5 and \chi_1) that

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\displaystyle 0=\frac{1}{60}\sum_{j=1}^{5}\#\left(\mathcal{C}_j\right)\chi_3\left(\mathcal{C}_j\right)=\frac{1}{60}\left(20C+5+15\right)

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and thus C=-1. Then, considering \left\langle \chi_3,\chi_5\right\rangle we see that

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\displaystyle 0=\frac{1}{60}\sum_{j=1}^{5}\#\left(\mathcal{C}_j\right)\chi_3\left(\mathcal{C}_j\right)\overline{\chi_5\left(\mathcal{C}_j\right)}=\frac{1}{60}\left(15+15 B\right)

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and so B=-1.

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b) We recall that the dual group \widehat{G_\mathfrak{L}} has order equal the the abelianization G^{\text{ab}}. But, by inspection we see that \left|\widehat{G_\mathfrak{L}}\right|=1 and so \left|G^{\text{ab}}\right|=1 and so G=\left[G,G\right] (the commutator subgroup). It evidently follows that G is not solvable.

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c) Suppose that there were such a \rho. We know then that there exists m_1\cdots,m_5\in\mathbb{N}\cup\{0\} such that \displaystyle \text{tr}(\rho(g))=\sum_{j=1}^{5}m_j \chi_j. But, this implies that \displaystyle 4=\text{tr}\left(\rho(e)\right)=m_1+3m_2+3m_3+4m_4+5m_5 and so in particular we know that m_5=0. But, by assumption we also have that if g is the element of \mathcal{C}_2 that is in the problem statement then -2=\text{tr}(\rho(g))=m_1+m_2-m_5=m_1+m_2 which is impossible.

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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.


May 1, 2011 - Posted by | Algebra, Group Theory, Representation Theory, UMaryland Qualifying Exams | , , , , ,

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