8.3: Two-Hour Examination

Two-Hour Examination

1. Which of the following statements are true (give brief justifications for each of (a), (b), (c), (d); give a correct set of implications for (e)).

1. If \(a\) and \(b\) are elements of a group, then \(a^{2}=1,\quad b^{3}=1\implies(ab)^{6}=1\).

2. The following two elements are conjugate in \(S_{7}\):

   \[% \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7\\ 3 & 4 & 5 & 6 & 7 & 2 & 1 \end{pmatrix} ,\quad% \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7\\ 2 & 3 & 1 & 5 & 6 & 7 & 4 \end{pmatrix} . \nonumber \]

3. If \(G\) and \(H\) are finite groups and \(G\times A_{594}\approx H\times A_{594},\) then \(G\approx H\).

4. The only subgroup of \(A_{5}\) containing \((123)\) is \(A_{5}\) itself.

5. \(\text{Nilpotent}\implies\text{cyclic}\implies\text{commutative}% \implies\text{solvable}\) (for a finite group).

2. How many Sylow \(11\)-subgroups can a group of order \(110=2\cdot5\cdot11\) have? Classify the groups of order \(110\) containing a subgroup of order 10. Must every group of order 110 contain a subgroup of order 10?

3. Let \(G\) be a finite nilpotent group. Show that if every commutative quotient of \(G\) is cyclic, then \(G\) itself is cyclic. Is the statement true for nonnilpotent groups?

4. (a) Let \(G\) be a subgroup of \(\Sym(X)\), where \(X\) is a set with \(n\) elements. If \(G\) is commutative and acts transitively on \(X\), show that each element \(g\neq1\) of \(G\) moves every element of \(X\). Deduce that \((G:1)\leq n\).

(b) For each \(m\geq1\), find a commutative subgroup of \(S_{3m}\) of order \(3^{m}\).

(c) Show that a commutative subgroup of \(S_{n}\) has order \(\leq3^{\frac n3}\).

5. Let \(H\) be a normal subgroup of a group \(G\), and let \(P\) be a subgroup of \(H\). Assume that every automorphism of \(H\) is inner. Prove that \(G=H\cdot N_{G}(P)\).

6. (a) Describe the group with generators \(x\) and \(y\) and defining relation \(yxy^{-1}=x^{-1}\).

(b) Describe the group with generators \(x\) and \(y\) and defining relations \(yxy^{-1}=x^{-1}\), \(xyx^{-1}=y^{-1}\).

You may use results proved in class or in the notes, but you should indicate clearly what you are using.

Solutions

1. (a) False: in \(\langle a,b|a^{2},b^{3}\rangle\), \(ab\) has infinite order.

(b) True, the cycle decompositions are (1357)(246), (123)(4567).

(c) True, use the Krull-Schmidt theorem.

(d) False, the group it generates is proper.

(e) Cyclic \(\implies\) commutative \(\implies\) nilpotent \(\implies\) solvable.

2. The number of Sylow \(11\)-subgroups \(s_{11}% =1,12,\ldots\) and divides \(10\). Hence there is only one Sylow \(11\)-subgroup \(P\). Have

\[G=P\rtimes_{\theta}H,\quad P=C_{11},\quad H=C_{10}\text{ or }D_{5}. \nonumber \]

Now have to look at the maps \(\theta:H\rightarrow\Aut(C_{11})=C_{10}\). Yes, by the Schur-Zassenhaus lemma.

3. Suppose \(G\) has class \(>1\). Then \(G\) has quotient \(H\) of class \(2\). Consider

\[1\rightarrow Z(H)\rightarrow H\rightarrow H/Z(H)\rightarrow1. \nonumber \]

Then \(H\) is commutative by (4.17), which is a contradiction. Therefore \(G\) is commutative, and hence cyclic.

Alternatively, by induction, which shows that \(G/Z(G)\) is cyclic.

No! In fact, it’s not even true for solvable groups (e.g., \(S_{3}\)).

4. (a) If \(gx=x\), then \(ghx=hgx=hx\). Hence \(g\) fixes every element of \(X\), and so \(g=1\). Fix an \(x\in X\); then \(g\mapsto gx:G\rightarrow X\) is injective. [Note that Cayley’s theorem gives an embedding \(G\hookrightarrow S_{n}\), \(n=(G:1)\).]

(b) Partition the set into subsets of order \(3\), and let \(G=G_{1}\times \cdots\times G_{m}\).

(c) Let \(O_{1},\ldots,O_{r}\) be the orbits of \(G\), and let \(G_{i}\) be the image of \(G\) in \(\Sym(O_{i})\). Then \(G\hookrightarrow G_{1}\times\cdots\times G_{r}\), and so (by induction),

\[(G:1)\leq(G_{1}:1)\cdots(G_{r}:1)\leq3^{\frac{n_{1}}3}\cdots3^{\frac{ n_{r}}% 3}=3^{\frac n3}. \nonumber \]

5. Let \(g\in G\), and let \(h\in H\) be such that conjugation by \(h\) on \(H\) agrees with conjugation by \(g\). Then \(gPg^{-1}% =hPh^{-1}\), and so \(h^{-1} g\in N_{G}(P)\).

6. (a) It’s the group .

\[G=\langle x\rangle\rtimes\langle y\rangle=C_{\infty}\rtimes_{\theta}C_{\infty}% \nonumber \]

with \(\theta\colon C_{\infty}\rightarrow\Aut(C_{\infty})=\pm1\). Alternatively, the elements can be written uniquely in the form \(x^{i}y^{j}\), \(i,j\in \mathbb{Z}\), and \(yx=x^{-1}y\).

(b) It’s the quaternion group. From the two relations get

\[yx=x^{-1}y,\quad yx=xy^{-1}% \nonumber \]

and so \(x^{2}=y^{2}\). The second relation implies

\[xy^{2}x^{-1}=y^{-2},=y^{2}, \nonumber \]

and so \(y^{4}=1\).

Alternatively, the Todd-Coxeter algorithm shows that it is the subgroup of \(S_{8}\) generated by \((1287)(3465)\) and \((1584)(2673)\).

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1. A family should be distinguished from a set. For example, if \(f\) is the function \(\mathbb{Z}% {}\rightarrow\mathbb{Z}{}/3\mathbb{Z}{}\) sending an integer to its equivalence class, then \(\{f(i)\mid i\in\mathbb{Z\}}\) is a set with three elements whereas \((f(i))_{i\in\mathbb{Z}{}}\) is family with an infinite index set.↩

2. “Abelian group” is more common than “commutative group”, but I prefer to use descriptive names.↩

3. When we use the basis to identify \(V\) with \(F^{n}\), the pairing \(\phi\) becomes

   \[\left( \begin{smallmatrix} \vphantom{b}a_{1}\\ \vdots\\ \vphantom{b}a_{n}% \end{smallmatrix} \right) ,\left( \begin{smallmatrix} b_{1}\\ \vdots\\ b_{n}% \end{smallmatrix} \right) \mapsto(a_{1},\ldots,a_{n})\cdot P\cdot\left( \begin{smallmatrix} b_{1}\\ \vdots\\ b_{n}% \end{smallmatrix} \right) . \nonumber \]

   If \(A\) is the matrix of \(\alpha\) with respect to the basis, then \(\alpha\) corresponds to the map \(\left( \begin{smallmatrix} a_{1}\\ \vdots\\ a_{n}% \end{smallmatrix} \right) \mapsto A\left( \begin{smallmatrix} a_{1}\\ \vdots\\ a_{n}% \end{smallmatrix} \right) .\)Therefore, ([e8]) becomes the statement that

   \[(a_{1},\ldots,a_{n})\cdot A^{\mathrm{T}}\cdot P\cdot A\cdot\left( \begin{smallmatrix} b_{1}\\ \vdots\\ b_{n}% \end{smallmatrix} \right) =(a_{1},\ldots,a_{n})\cdot P\cdot\left( \begin{smallmatrix} b_{1}\\ \vdots\\ b_{n}% \end{smallmatrix} \right) \text{ for all }\left( \begin{smallmatrix} \vphantom{b}a_{1}\\ \vdots\\ \vphantom{b}a_{n}% \end{smallmatrix} \right) ,\left( \begin{smallmatrix} b_{1}\\ \vdots\\ b_{n}% \end{smallmatrix} \right) \in F^{n}. \nonumber \]

   On examining this statement on the standard basis vectors for \(F^{n}\), we see that it is equivalent to ([e9]).↩

4. This group is denoted \(D_{2n}\) or \(D_{n}\) depending on whether the author is viewing it abstractly or concretely as the symmetries of an \(n\)-polygon (or perhaps on whether the author is a group theorist or not; see mo48434).↩

5. More formally, \(D_{n}\) can be defined to be the subgroup of \(S_{n}\) generated by \(r\colon i\mapsto i+1\) (mod \(n)\) and \(s\colon i\mapsto n+2-i\) (mod \(n\)). Then all the statements concerning \(D_{n}\) can proved without appealing to geometry.↩

6. In fact Besche et al. did not construct the groups of order 1024 individually, but it is known that there are 49487365422 groups of that order. The remaining 423164062 groups of order up to 2000 (of which 408641062 have order 1536) are available as libraries in GAP and Magma. I would guess that 2048 is the smallest number such that the exact number of groups of that order is unknown (Derek Holt, mo46855; Nov 21, 2010).↩

7. More formally, \((G:H)\) is the cardinality of the set \(\{aH\mid a\in G\}\).↩

8. Some authors say “ factor”  instead of “quotient”, but this can be confused with “direct factor”.↩

9. John Stillwell tells me that, for finite commutative groups, this is similar to the first proof of the theorem, given by Kronecker in 1870.↩

10. Following Bourbaki, I require locally compact spaces to be Hausdorff.↩

11. I don’t know who found this beautiful proof. Apparently the original proof of G.A. Miller is very complicated; see mo24913.↩

12. This property of \(Q\) is unusual. In fact, the only noncommutative groups in which every subgroup is normal are the groups of the form \(Q\times A\times B\) with \(Q\) the quaternion group, \(A\) a commutative group whose elements have finite odd order, and \(B\) a commutative group whose elements have order \(2\) (or \(1\)). See Hall 1959, 12.5.4.↩

13. Nielsen (1921) proved this for finitely generated subgroups, and in fact gave an algorithm for deciding whether a word lies in the subgroup; Schreier (1927) proved the general case.↩

14. For example, the commutator subgroup of the free group on two generators has infinite rank.↩

15. Strictly speaking, I should say the relations \(a^{2^{n-1}}\), \(a^{2^{n-2}}b^{-2}\), \(bab^{-1}a\).↩

16. Each element of \(R\) represents an element of \(FX\), and the condition requires that the unique extension of \(\alpha\) to \(FX\) sends each of these elements to \(1\).↩

17. This is Bourbaki’s terminology (Algèbre, I §6); some authors call ([e27]) an extension of \(N\) by \(Q\).↩

18. It has been shown that every group on the list can be generated by two elements, and so this is true for all finite simple groups. If a proof of this could be found that doesn’t use the classification, then the proof of the classification would be greatly simplified (mo59213).↩

19. Here is a direct proof that the theorem holds for an abelian group \(Z\). We use induction on the order of \(Z\). It suffices to show that \(Z\) contains an element whose order is divisible by \(p,\) because then some power of the element will have order exactly \(p\). Let \(g\neq1\) be an element of \(Z\). If \(p\) doesn’t divide the order of \(g\), then it divides the order of \(Z/\langle g\rangle\), in which case there exists (by induction) an element of \(G\) whose order in \(Z/\langle g\rangle\) is divisible by \(p\). But the order of such an element must itself be divisible by \(p\).↩

20. To solve a problem, an algorithm must always terminate in a finite time with the correct answer to the problem. The Todd-Coxeter algorithm does not solve the problem of determining whether a finite presentation defines a finite group (in fact, there is no such algorithm). It does, however, solve the problem of determining the order of a finite group from a finite presentation of the group (use the algorithm with \(H\) the trivial subgroup 1.)↩

21. Equivalently, the usual map \(G\rightarrow\Sym(G/N)\).↩

22. Some authors write “normal series” where we write “subnormal series” and “ invariant series” where we write “normal series”.↩

23. Jordan showed that corresponding quotients had the same order, and Hölder that they were isomorphic.↩

24. Burnside (1897, p. 379) wrote:No simple group of odd order is at present known to exist. An investigation as to the existence or non-existence of such groups would undoubtedly lead, whatever the conclusion might be, to results of importance; it may be recommended to the reader as well worth his attention. Also, there is no known simple group whose order contains fewer than three different primes….Significant progress in the first problem was not made until Suzuki, M., The nonexistence of a certain type of simple group of finite order, 1957. However, the second problem was solved by Burnside himself, who proved using characters that any group whose order contains fewer than three different primes is solvable (see , p. 182).↩

25. Strictly, this should be called the Wedderburn-Remak-Schmidt-Krull-Ore theorem — see the Wikipedia: Krull-Schmidt theorem.↩

26. Some authors call it a generalized character, but this is to be avoided: there is more than one way to generalize the notion of a character.↩