# Abstract Nonsense

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

Point of Post: This is the January 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)-January 2004

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Problem 1:

a) Let $p$ be a prime and $H$ a non-trivial finite abelian $p$-group. Let $N$ be the number of elements of $H$ of order $p$. Show that $N\equiv -1\text{ mod }p$.

b) Let $p$ be a prime and let $G$ be a finite group. Let $S$ be a Sylow $p$-subgroup of $G$. Suppose that $x\in G$ has order $p$ and commutes with all elements of $S$. Show that $x\in S$.

c) Let the notation be as in b). Let $X$ be the set of elements of $G$ of order $p$. Then, $S$ acts on $X$ by conjugation. Let $x\in X$. Show that the orbit of $x$ has only one element if and only if $x\in\mathcal{Z}(s)$. Also, show that if the orbit of $x$ has more than one element, then the number of elements in the orbit is a multiple of $p$.

d) Let $G$ be a finite group and let $p$ be a prime dividing $|G|$. Let $M$ be the number of elements of $G$ of order $p$. Show that $M\equiv -1\text{ mod }p$.

Proof:

a) We know by the structure theorem that $G\cong\mathbb{Z}_{p^{a_1}}\times\cdots\times\mathbb{Z}_{p^{a_m}}$ for some $m\in\mathbb{N}$ and some $a_1,\cdots,a_m\in\mathbb{N}$. Thus, it suffices to prove this for groups of that form. We begin by recalling that the order of $(g_1,\cdots,g_m)\in\mathbb{Z}_{p^{a_1}}\times\cdots\times\mathbb{Z}_{p^{a_m}}$ has order $\text{l.c.m.}\left(|g_1|,\cdots,|g_m|\right)$. Thus, for the order of $(g_1,\cdots,g_m)$ to have order $p$ it is necessary and sufficient that for each $j\in[m]$ we have that $|g_j|=1,p$ and there exists at least one $j_0\in[m]$ such that $|g_{j_0}|=p$. Thus, of course we see that

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$\displaystyle \#\left\{g\in\mathbb{Z}_{p^{a_1}}\times\cdots\times\mathbb{Z}_{p^{a_m}}:|g|=p\right\}=\prod_{j=1}^{m}\#\left\{g\in\mathbb{Z}_{p^{a_j}}:|g|=1,p\right\}-1\quad\mathbf{(1)}$

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What we claim though is that $\#\left\{g\in\mathbb{Z}_{p^{a_j}}:|g|=1,p\right\}=p$ for each $j\in[m]$. Indeed, we note firstly that if $g\in\mathbb{Z}_{p^{a_j}}$ is of order $p$ then $p^{a_j}\mid pg$ and so $p^{a_j-1}\mid g$. Thus, the only such possible $g$ are $p^{a_j-1},\cdots,(p-1)p^{a_j-1}$ and evidently these are of order $p$. Thus, the number of elements of $\mathbb{Z}_{p^{a_j}}$ of order $p$ is $p-1$ and thus the number of elements of $\mathbb{Z}_{p^{a_j}}$ of order $1$ or $p$ is $p$ as claimed. Thus, by $\mathbf{(1)}$ we see that

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$\displaystyle \#\left\{g\in\mathbb{Z}_{p^{a_1}}\times\cdots\times\mathbb{Z}_{p^{a_m}}:|g|=p\right\}=p^m-1=\left(p-1\right)\sum_{j=0}^{m-1}p^j\equiv -1\text{ mod }p$

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We implicitly used that $m>0$ which we know is true since $H\ne \{e\}$.

b) We claim that if $x\notin S$ then $\left|\left\langle x,S\right\rangle\right|=p^{m+1}$ where $\left|S\right|=p^m$ which is by definition impossible. Indeed, by definition we have that

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$\displaystyle \left\langle x,S\right\rangle=\left\{\prod_{j}g_j:g_j=x\text{ or }g_j\in S\right\}$

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But, since $x$ commutes with everything in $S$ this may be rewritten as

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$\displaystyle \left\{x^ks:k\in\mathbb{Z}\text{ and }s\in S\right\}$

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What we claim though is that if $x^ks_1=x^\ell s_2$ where  $\ell\leqslant k$ then $x^{k-\ell}=s_2s_1^{-1}$ and so $x^{k-\ell}\in S$. But, since $|x|$ is prime we know that $\left\langle x^{k-\ell}\right\rangle=\left\langle x\right\rangle$ if $k\ne\ell$ and thus if $k\ne\ell$ then $x\in S$ contradictory to assumption. Thus, $k=\ell$ and consequently $x^k=x^\ell$ and $s_1=s_2$. It evidently follows that $\left|\left\langle x,S\right\rangle\right|=p^{m+1}$ which is a contradiction since $p^m$ is the maximal power of $p$ dividing $G$. Thus, $x\in S$.

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c)  It’s evident that $S$ acts on $X$ since order is preserved under conjugation. The rest is simple. Namely, let $x\in X$. If $\mathcal{O}_x$ (the orbit of $x$ under the $S$-action) has one element then evidently $x$ commutes with every element of $S$ and since $|x|=p$ we have by b) that $x\in S$. Recall though that by the orbit-stabilizer theorem that $\#\left(\mathcal{O}_x\right)=\left(S:\text{stab}(x)\right)$ and since $\left(S:\text{stab}(x)\right)\mid |S|=p^m$ we have that $\#\left(\mathcal{O}_x\right)=p^k$ for some $0\leqslant k\leqslant m$, and so if $\#\left(\mathcal{O}_x\right)>1$ we must have that $p\mid\#\left(\mathcal{O}_x\right)$.

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d) By the orbit decomposition theorem we may write $\displaystyle X=\mathcal{O}_1\sqcup\mathcal{O}_2$ where $\mathcal{O}_1$ is the union of the distinct one-element orbits of $X$ under the $S$-action described in 2) and $\mathcal{O}_2$ is the union of the distinct orbits which have more than one element and thus

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$\#\left(X\right)=\#\left(\mathcal{O}_1\right)+\#\left(\mathcal{O}_2\right)$

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What we claim though is that $\mathcal{O}_1=\left\{s\in\mathcal{Z}(S):|s|=p\right\}$. Indeed, if $s\in\mathcal{O}_1$ then by fact that it has a trivial orbit under the conjugation $S$-action tells us that it commutes with everything in $S$ and since as we proved $s\in S$ it must be that $s\in\mathcal{Z}(S)$, and since $s\in X$ so that $|s|=p$ we have that $s\in\left\{s\in\mathcal{Z}(S):|s|=p\right\}$. Conversely, if $s\in\left\{s\in\mathcal{Z}(S):|s|=p\right\}$ then evidently $|s|=p$ and so $s\in X$ and by definition its orbit under the $S$-action is trivial and thus $s\in\mathcal{O}_1$. Note though that $\mathcal{Z}(S)$ is a non-trivial (since we know the center of $p$-groups is non-trivial) abelian group and so by a) we know that $\#(\mathcal{O}_1)\equiv -1\text{ mod }p$. And, since $p\mid\#\left(\mathcal{O}_2\right)$ by b) we may conclude that

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$\displaystyle \#\left(X\right)=\#\left(\mathcal{O}_1\right)+\#\left(\mathcal{O}_2\right)\equiv -1+0=-1\text{ mod }p$

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Problem 6: Let $G$ be a finite group and $H\unlhd G$. Let $a_1H,\cdots,a_nH$ be the distinct cosets of $H$. Let $\mathscr{V}$ be thefree complex vector space over the symbols $\{e_1,\cdots,e_n\}$. Then, for $g\in G$ define $\rho_g$ on $\mathscr{V}$ by $\rho_g(e_i)=e_j$ if $ga_iH=a_j H$. The map $\rho:G\to\text{GL}\left(\mathscr{V}\right)$ is a representation of $G$ (you do not need to prove this). Show that in the decomposition of $\rho$ into irreps the trivial representation occurs only once.

Proof: We first note that if $g\in G$ and $ga_iH=a_iH$ then by definition $ga_i\in a_iH$ and so $ga_i=a_ih$ and so $g=a_iha_i^{-1}$ and since $H$ is normal this implies that $g\in H$. Conversely, if $g\in H$ then and $ga_iH=gHa_i=Ha_i=a_iH$. We thus note that if $g\notin H$ then $\rho_(e_i)=e_k$ for $k\ne i$ for all $i\in[n]$. Thus, if $\left[\rho_g\right]_{\mathcal{B}}$ is the matrix of $\rho_g$ with respect to the ordered basis $(e_1,\cdots,e_n)$ then the diagonal entries of $[\rho_g]_{\mathcal{B}}$ are zero. Moreover, if $g\in H$ then $\rho_g(e_i)=e_i$ for every $e_i$ and thus $\left[\rho_g\right]_{\mathcal{B}}=I_{(G:H)}$ where $I_{(G:H)}$ is the $(G:H)\times (G:H)$ identity matrix. Thus, using the obvious fact that the number of times that the trivial representation shows up in the decomposition of $\rho$ is equal to $\left\langle \chi_\rho,\chi^{\text{triv}}\right\rangle$ where the inner product is taken in the group algebra. But, the above shows that

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$\displaystyle \left\langle \chi_\rho,\chi^{\text{triv}}\right\rangle=\frac{1}{|G|}\sum_{g\in G}\chi_\rho(g)\overline{\chi^{\text{triv}}(g)}=\frac{1}{|G|}\sum_{h\in H}\chi_\rho(h)=\frac{|H|(G:H)}{|G|}=1$

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where we have evidently used Lagrange’s theorem.

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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.