# Abstract Nonsense

## 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.

Proof:

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 $g\in\mathcal{C}_2$.

Proof:

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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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.