# Abstract Nonsense

## University of Maryland College Park Qualifying Exams (Group Theory and Representation Theory) ( January 2003))

Point of Post: A friend of mine who attended UMD CP for his Ph.D. has told me that they are famous for their brutally hard qualifying exams. Thus, to keep sharp and as a challenge I’d like to complete all of the qual exams for Algebra, Analysis, and Topology/Geometry eventually–although this is an ambitious venture. So, I plan to break them up into subsections and do all of the ____ subject (ring, field,…) in the Algebra quals and then maybe do all the ____  subject( measure theory, complex analysis, …) in the Analysis quals etc. Since what I’m currently doing on my blog has been representation theory and group theory mostly I thought I’d start with those two. Thus, this will be the first post in a series to solve the Group Theory/Representation Theory problems on the quals.

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NOTE: Just because I am posting references (which are really more like further reading) I will not use any book while taking these.

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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.

Remark: The set of all the algebra exams may be found here.

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Algebra Qual(Ph.D. Version)-January 2003

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Problem 1:

a) Let $H$ be a group of order $9$. Show that $\left|\text{Aut}(H)\right|\mid 48$ (Hint: you may assume that the group $\text{GL}_2\left(\mathbb{F}_3\right)$ has order $48$).

b) Let $G$ be a group of order $153=3^2\cdot17$. Show that $\mathcal{Z}(G)$ contains a subgroup of order $9$.

c) Find all groups of order $153$.

(Do not use theorems of the form “every group of order $p^2q$…”)

Proof:

a) Since $|H|=3^2$ we know that $H\cong\mathbb{Z}_9$ or $H\cong\mathbb{Z}_3^2$. Thus, it assumes to prove this is true for those two groups. So, first define $F:\text{Aut}\left(\mathbb{Z}_3^2\right)\to\text{Mat}_2\left(\mathbb{Z}_3\right)$ by

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$\displaystyle F\left(\varphi\right)=\begin{pmatrix}\pi_1(\varphi((1,0)) & \pi_1(\varphi(0,1))\\ \pi_2\left(\varphi(1,0)\right) & \pi_2\left(\varphi((0,1))\right)\end{pmatrix}$

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$\pi_i$ is the $i^{\text{th}}$ canonical projection. Evidently $\mathbb{Z}_3^2$ is a vector space over $\mathbb{Z}_3$ and evidently every group endomorphism will be a linear endomorphism. Thus, since by definition $F(\varphi)$ is by definition $\left[\varphi\right]_{\mathcal{B}}$ where $\mathcal{B}$ is the ordered basis $\left((1,0),(0,1)\right)$ we have from basic linear algebra that $\text{im}(F)\subseteq\text{GL}\left(\mathbb{F}_3,2\right)$ and $F$ is a monomorphism. Thus $F\cong\text{im}(F)\leqslant \text{GL}_2\left(\mathbb{F}_3\right)$ and so by Lagrange’s theorem $\left|\text{Aut}\left(\mathbb{Z}_3^2\right)\right|\mid \left|\text{GL}\left(\mathbb{F}_3\right)\right|=48$.

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For $\mathbb{Z}_9$ we appeal to the common fact that $\text{Aut}\left(\mathbb{Z}_n\right)\cong\mathbb{Z}_n^\times$ and so $\left|\text{Aut}\left(\mathbb{Z}_9\right)\right|=\varphi\left(9\right)=6\mid 48$ where here $\varphi$ is Euler’s totient function.

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b) Let $n_3=\#\left(\text{Syl}_3\left(G\right)\right)$ denote the number of Sylow $3$-subgroups of $G$. By Sylow’s theorems we must have that $n_3\mid 153$ and so $n_3=1,3,9,19,17\cdot3,153$ but since $n_3\equiv1\text{ mod }3$ we must conclude that $n_3=1$. Thus, by a common theorem we may conclude that if $H$ is the unique Sylow $3$-subgroup that $H\unlhd G$. Thus, consider the map $\Phi:G\to\text{Aut}(H)$ by $g\mapsto i_g$ where $i_g$ is the inner automorphism (this is well defined since $H$ is normal). This is evidently a homomorphism and so what we have by the first isomorphism theorem is that $G/\ker\Phi\cong\text{im}(\Phi)\leqslant \text{Aut}(H)$ and so in particular $\left|G/\ker\Phi\right|\mid \text{Aut}(H)$. But, by definition $\ker\Phi=\bold{C}_G(H)$ the centralizer of $H$ in $G$ and so by part a) we may conclude that $\left|G/\bold{C}_G(H)\right|\mid 48$. But, since $\left(48,153\right)=1$ we may conclude that $G/\bold{C}_G(H)$ is trivial and so $\bold{C}_G(H)=G$. Thus, $H\leqslant \mathcal{Z}(G)$.

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c) We use the fact that since $\left|\mathcal{Z}\left(G\right)\right|=9,153$ that $G/\mathcal{Z}(G)$ is inevitably cyclic and so by a common theorem we must have that $G$ is abelian. Then, by the structure theorem we may conclude that either $G\cong\mathbb{Z}_9\times\mathbb{Z}_{17}$ or $G\cong\mathbb{Z}_3\times\mathbb{Z}_3\times\mathbb{Z}_{17}$.

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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.

May 1, 2011 -