Abstract Nonsense

Crushing one theorem at a time

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.

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May 6, 2011 - Posted by | Algebra, Group Theory, Representation Theory, UMaryland Qualifying Exams | , , , , ,

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