# Abstract Nonsense

## 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 -

## 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