Groups: Sylow Theory

Fall 2019 #1 #algebra/qual/completed

Let \(G\) be a finite group with \(n\) distinct conjugacy classes. Let \(g_1 \cdots g_n\) be representatives of the conjugacy classes of \(G\). Prove that if \(g_i g_j = g_j g_i\) for all \(i, j\) then \(G\) is abelian.

  • \(Z(g) = G \iff g\in Z(G)\), i.e. if the centralizer of \(g\) is the whole group, \(g\) is central.

  • If \(H\leq G\) is a proper subgroup, then \(\bigcup_{g\in G} hGh^{-1}\) is again a proper subgroup (subset?) I.e. \(G\) is not a union of conjugates of any proper subgroup.

  • So if \(G\) is a union of conjugates of \(H\), then \(H\) must not be proper, i.e. \(H= G\).

  • We have \(g_j \subseteq Z(g_k)\) for all \(k\) by assumption.
  • If we can show \(Z(g_k) = G\) for all \(k\), then \(g_k \in Z(G)\) for all \(k\).
    • Then each conjugacy class is size 1, and since \(G = {\textstyle\coprod}_{i=1}^n [g_i] = {\textstyle\coprod}_{i=1}^n \left\{{g_i}\right\}\), every \(g\in G\) is some \(g_i\). So \(G \subseteq Z(G)\), forcing \(G\) to be abelian.
  • If we can show \(G \subseteq \bigcup_{h\in H} h Z(g_k) h^{-1}\) for some \(k\), this forces \(Z(g_k) = G\) and \(g_k \in Z(G)\).
    • If we can do this for all \(k\), we’re done!
  • Since \(g\in G\) is in some conjugacy class, write \(g=hg_j h^{-1}\) for some \(h\in G\) and some \(1\leq j\leq n\).
  • Now use \(g_j \in Z(g_k)\) for all \(k\): \begin{align*} g\in G &\implies g = hg_j h^{-1}&& \text{for some } h\in H \\ g_j \in Z(g_k) \forall k &\implies g\in hZ(g_k)h^{-1}&&\text{for some }h, \, \forall 1\leq k \leq n \\ &\implies g\in \bigcup_{h\in G} h Z(g_k) h^{-1} &&\forall 1\leq k \leq n \\ .\end{align*}
    • Note that it’s necessary to get rid of the \(h\) dependence, since now now every \(g\in G\) is in \(\bigcup_{h\in G} hZ(g_k)h^{-1}\).
  • Now \begin{align*} G \subseteq \bigcup_{h\in G} hZ(g_k) \subseteq G \,\,\forall k \implies Z(g_k) = G\,\, \forall k ,\end{align*} and we’re done.

Fall 2019 Midterm #2 #algebra/qual/work

Let \(G\) be a finite group and let \(P\) be a sylow \(p{\hbox{-}}\)subgroup for \(p\) prime. Show that \(N(N(P)) = N(P)\) where \(N\) is the normalizer in \(G\).

Fall 2013 #2 #algebra/qual/work

Let \(G\) be a group of order 30.

  • Show that \(G\) has a subgroup of order 15.

  • Show that every group of order 15 is cyclic.

  • Show that \(G\) is isomorphic to some semidirect product \({\mathbf{Z}}_{15} \rtimes{\mathbf{Z}}_2\).

  • Exhibit three nonisomorphic groups of order 30 and prove that they are not isomorphic. You are not required to use your answer to (c).

Spring 2014 #2 #algebra/qual/work

Let \(G\subset S_9\) be a Sylow-3 subgroup of the symmetric group on 9 letters.

  • Show that \(G\) contains a subgroup \(H\) isomorphic to \({\mathbf{Z}}_3 \times{\mathbf{Z}}_3 \times{\mathbf{Z}}_3\) by exhibiting an appropriate set of cycles.

  • Show that \(H\) is normal in \(G\).

  • Give generators and relations for \(G\) as an abstract group, such that all generators have order 3. Also exhibit elements of \(S_9\) in cycle notation corresponding to these generators.

  • Without appealing to the previous parts of the problem, show that \(G\) contains an element of order 9.

Fall 2014 #2 #algebra/qual/work

Let \(G\) be a group of order 96.

  • Show that \(G\) has either one or three 2-Sylow subgroups.

  • Show that either \(G\) has a normal subgroup of order 32, or a normal subgroup of order 16.

Spring 2016 #3 #algebra/qual/work

  • State the three Sylow theorems.

  • Prove that any group of order 1225 is abelian.

  • Write down exactly one representative in each isomorphism class of abelian groups of order 1225.

Spring 2017 #2 #algebra/qual/work

  • How many isomorphism classes of abelian groups of order 56 are there? Give a representative for one of each class.

  • Prove that if \(G\) is a group of order 56, then either the Sylow-2 subgroup or the Sylow-7 subgroup is normal.

  • Give two non-isomorphic groups of order 56 where the Sylow-7 subgroup is normal and the Sylow-2 subgroup is not normal. Justify that these two groups are not isomorphic.

Fall 2017 #2 #algebra/qual/completed

  • Classify the abelian groups of order 36.

    For the rest of the problem, assume that \(G\) is a non-abelian group of order 36. You may assume that the only subgroup of order 12 in \(S_4\) is \(A_4\) and that \(A_4\) has no subgroup of order 6.

  • Prove that if the 2-Sylow subgroup of \(G\) is normal, \(G\) has a normal subgroup \(N\) such that \(G/N\) is isomorphic to \(A_4\).

  • Show that if \(G\) has a normal subgroup \(N\) such that \(G/N\) is isomorphic to \(A_4\) and a subgroup \(H\) isomorphic to \(A_4\) it must be the direct product of \(N\) and \(H\).

  • Show that the dihedral group of order 36 is a non-abelian group of order 36 whose Sylow-2 subgroup is not normal.

  • Classifying abelian groups of order \(n\): factor \(n = \prod_{i=1}^k p_i^{n_i}\), then there are \(p(n_1)p(n_2)\cdots p(n_k)\) abelian groups of that order, where \(p(\ell)\) is the integer partition function.
  • Transitive subgroups of \(S_n\):
    • \(n=3 \leadsto S_3, A_3\)
    • \(n=4 \leadsto S_4, A_4, D_4, C_4, C_2^2\) where \(C_n\) is a cyclic group.
    • \(n=5\leadsto S_5, A_5, F_{20}, D_{10}, C_5\).
  • Background for this question: there is a theorem that if \({\sharp}G = p^2 q^2\) with \(p < q\) then \(G\) must have a normal \(q{\hbox{-}}\)Sylow subgroup unless \({\sharp}G = 36\), the only counterexample, in which case \(G\) has either a normal Sylow \(2{\hbox{-}}\)subgroup or a normal Sylow \(3{\hbox{-}}\)subgroup. The counterexample is evidenced by \(C_3 \times A_4\), which has \(n_3 = 4\).
    • The reason this happens: \(p = 2, q =3\) are consecutive primes!

This is slightly more difficult/lengthy than the average group theory problem in recent years.

Part a: write \(36 = 2^2\cdot 3^2\), then the distinct groups correspond to all combinations of integer partitions of the exponents \(2\) and \(2\):

  • \((2, 2) \leadsto ({\mathbf{Z}}/4{\mathbf{Z}}) \times ({\mathbf{Z}}/9{\mathbf{Z}})\)
  • \((1+1, 2) \leadsto ({\mathbf{Z}}/2{\mathbf{Z}})^2 \times ({\mathbf{Z}}/9{\mathbf{Z}})\)
  • \((2, 1+1) \leadsto {\mathbf{Z}}/4{\mathbf{Z}}\times ({\mathbf{Z}}/3{\mathbf{Z}})^2\)
  • \((1+1, 1+1) \leadsto ({\mathbf{Z}}/2{\mathbf{Z}})^2 \times ({\mathbf{Z}}/3{\mathbf{Z}})^2\)

Part b: let \(H_2 \leq G\) be the unique Sylow \(2{\hbox{-}}\)subgroup, so \({\sharp}H_2 = 2^2\). Then \(H_2{~\trianglelefteq~}G\), so \(G/H\) is a group of size \([G:H] = 3^2\). Note that by pure numeric observation (and the hint), it is necessary to find a normal subgroup \(N{~\trianglelefteq~}G\) of size 3 and thus index 12, since \({\sharp}A_4 = 4!/2 = 12 = 36/3\). Since normal subgroups are kernels of homomorphisms, we can look for a morphism \(\phi: G\to S_4\) where \(\ker(\phi)\) is order 3, anticipating applying the first isomorphism theorem to get \(G/\ker(\phi) \cong \operatorname{im}(\phi) \leq S_4\), where we hope that \(\operatorname{im}(\phi) \cong A_4\). So we look for an action of \(G\) on a set with 4 elements.

Applying the Sylow theorems, we have \(n_3 \divides 2^2 \implies n_3\in\left\{{1,2,4}\right\}\) and \(n_3\equiv 1\operatorname{mod}4\implies n_3\in \left\{{1, 4}\right\}\). Note that \(n_3=1\) implies the Sylow \(3{\hbox{-}}\)subgroup is normal, in which case \(G \cong H_2 \times H_3\) splits into a product of groups of order \(2^2, 3^2\), and thus \(G\) is abelian since any group of order \(p^2\) is abelian for any prime \(p\) and products of abelian groups are abelian. So we can conclude that \(n_3 = 4\) and consider the action of \(G\) on the set of four Sylow \(3{\hbox{-}}\)subgroups of \(G\) to get a map \(\phi: G\to S_4\).

It is a theorem that this action is transitive, and so \(\operatorname{im}(\phi)\) is a transitive subgroup of \(S_4\). By standard facts in Galois theory, the only such subgroups are \(S_4, A_4, D_4, C_2^2, C_4\), and so \({\sharp}\operatorname{im}(\phi) \in \left\{{24 ,12, 8, 4}\right\}\). By the hint it suffices to show that \({\sharp}\operatorname{im}(\phi) = 12\).

  • \({\sharp}\operatorname{im}(\phi)\neq 24\), since then \(\phi\) surjects and \(G/\ker(\phi) \cong S_4\), but the RHS has order 24 and the LHS order \(36/k\) for some \(k\in {\mathbf{Z}}\), which is impossible.
  • \({\sharp}\operatorname{im}(\phi) \neq 8\) since then \(36/k = 8\) for some \(k\), again impossible.
  • \({\sharp}\operatorname{im}(\phi) \neq 4\): if so, \({\sharp}\ker(\phi) = 9\) so \(\ker(\phi)\) is a Sylow \(3{\hbox{-}}\)subgroup. Since kernels are normal and all Sylows are conjugate, this would force \(n_3 = 1\), a contradiction.

Summarizing, to produce \(N\), consider \(G\curvearrowright{\operatorname{Syl}}_3(G)\) to get a map \(\phi: G\to S_4\). Then \(\operatorname{im}(\phi)\leq S_4\) must have order 12 with kernel \(N\coloneqq\ker(\phi)\) of order 3, and since the only subgroup of order 12 in \(S_4\) is \(A_4\), we have \(G/N\cong A_4\).

Part 3: Since \(N{~\trianglelefteq~}G\), we need to show:

  • \(H \cap N = \left\{{e}\right\}\),
  • \(NH = G\),
  • \(H{~\trianglelefteq~}G\)

The first two conditions will give a semidirect product, and the third will show that the semidirect product is actually direct.

Note that \(L \coloneqq H \cap N\) is a subgroup of both \(H\) and \(N\), so if \(L\neq \left\{{e}\right\}\) then \({\sharp}L = 3\) and \(L\) is a subgroup of both \(N\) and \(A_4\). So \(L = N\) and thus \(N \subseteq A_4\) is a cyclic subgroup of order 3 generated by (say) \(n\); however since \(2\divides {\sharp}A_4 = 12\), Cauchy’s theorem produces an element \(m\) of order two. But then the subgroup \(\left\langle{n, m}\right\rangle\) is an order 6 subgroup of \(A_4\), which by the hint does not exist, so by contradiction we must have \(L \coloneqq N \cap H = \left\{{e}\right\}\).

That \(NH = G\) now follows by counting elements: \(NH \leq G\) is a subgroup, and \({\sharp}NH = ({\sharp}N)\cdot({\sharp}H) = 3\cdot 12 = {\sharp}G\).

Finally, to see that \(H{~\trianglelefteq~}G\), we’ll instead show that the semidirect product is trivial directly. By the first two conditions above, we already have \(G\cong N\rtimes_\phi H\) for some \(\phi: H\to \mathop{\mathrm{Aut}}(N)\). Since \(N\cong C_3\) is cyclic, \(\mathop{\mathrm{Aut}}(N)\cong N^{\times}\) which is of order \(\phi(3) = 2\), so \(\phi\) is a group morphism from a group of order 12 to one of order 2. If \(\phi\) is nontrivial, \({\sharp}\ker \phi = 12/2 = 6\) is a subgroup of order 6 in \(H\cong A_4\), contradicting the hint and forcing \(\phi\) to be trivial and \(G\cong H\times N\).

Part d If \(G\coloneqq D_{18}\) has a normal Sylow \(2{\hbox{-}}\)subgroup, by part 2 \(\exists N{~\trianglelefteq~}G\) with \(G/N\cong A_4\), and (claim) since \(G\) has a subgroup \(H\cong A_4\) we must have \(G = N\times H\), a product of groups of orders 3 and 12 respectively. Somehow this is a contradiction??

That \(H\cong A_4 \leq G\) exists: unclear, maybe even not true. Not sure what the intended approach is, so here’s an alternative.

A general fact: for \(D_{2n}\), for any odd prime \(p\divides 2n\), the Sylow \(p{\hbox{-}}\)subgroup \(H_p\) is cyclic and normal since \(D_{2n}\cong C_n \rtimes C_2\) and \(p > 2\) implies \(H_p\) descends to a Sylow \(p{\hbox{-}}\)subgroup of \(C_n\) and all subgroups of cyclic groups are abelian and cyclic. Here \(n=18\), so take \(p=3\) to get \(H_{3} {~\trianglelefteq~}D_{18}\) a normal subgroup of order 9. If \(n_2=1\), so there is one single normal Sylow \(2{\hbox{-}}\)subgroup, then \(H_2, H_3\) are both normal and we get \(D_{18} \cong H_2 \times H_3\) as a direct product. But \(H_2, H_3\) are abelian and \(D_{18}\) is nonabelian, so this is a contradiction.

Fall 2012 #2 #algebra/qual/work

Let \(G\) be a group of order 30.

  • Show that \(G\) contains normal subgroups of orders 3, 5, and 15.

  • Give all possible presentations and relations for \(G\).

  • Determine how many groups of order 30 there are up to isomorphism.

Fall 2018 #1 #algebra/qual/completed

Let \(G\) be a finite group whose order is divisible by a prime number \(p\). Let \(P\) be a normal \(p{\hbox{-}}\)subgroup of \(G\) (so \({\left\lvert {P} \right\rvert} = p^c\) for some \(c\)).

  • Show that \(P\) is contained in every Sylow \(p{\hbox{-}}\)subgroup of \(G\).

  • Let \(M\) be a maximal proper subgroup of \(G\). Show that either \(P \subseteq M\) or \(|G/M | = p^b\) for some \(b \leq c\).

  • Sylow 2: All Sylow \(p{\hbox{-}}\)subgroups are conjugate.
  • \({\left\lvert {HK} \right\rvert} = {\left\lvert {H} \right\rvert} {\left\lvert {K} \right\rvert} / {\left\lvert {H\cap K} \right\rvert}\).
  • Lagrange’s Theorem: \(H\leq G \implies {\left\lvert {H} \right\rvert} \divides {\left\lvert {G} \right\rvert}\)

  • Every \(p{\hbox{-}}\)subgroup is contained in some Sylow \(p{\hbox{-}}\)subgroup, so \(P \subseteq S_p^i\) for some \(S_p^i \in \mathrm{Syl}_p(G)\).

  • \(P {~\trianglelefteq~}G \iff gPg^{-1}= P\) for all \(g\in G\).

  • Let \(S_p^j\) be any other Sylow \(p{\hbox{-}}\)subgroup,

  • Since Sylow \(p{\hbox{-}}\)subgroups are all conjugate \(gS_p^i g^{-1}= S_p^j\) for some \(g\in G\).

  • Then \begin{align*} P = gPg^{-1}\subseteq gS_p^i g^{-1}= S_p^j .\end{align*}

  • If \(P\) is not contained in \(M\), then \(M < MP\) is a proper subgroup

  • By maximality of \(M\), \(MP = G\)

  • Note that \(M\cap P \leq P\) and \({\left\lvert {P} \right\rvert} = p^c\) implies \({\left\lvert {M\cap P} \right\rvert} = p^a\) for some \(a\leq c\) by Lagrange

  • Then write \begin{align*} G = MP &\iff {\left\lvert {G} \right\rvert} = \frac{{\left\lvert {M} \right\rvert} {\left\lvert {P} \right\rvert}}{{\left\lvert {M\cap P} \right\rvert}} \\ \\ &\iff { {\left\lvert {G} \right\rvert} \over {\left\lvert {M} \right\rvert}} = {{\left\lvert {P} \right\rvert} \over {\left\lvert {M\cap P} \right\rvert}} = {p^c \over p^a} = p^{c-a} \coloneqq p^b \end{align*}

    where \(a\leq c \implies 0 \leq c-b \leq c\) so \(0\leq b \leq c\).

Fall 2019 #2 #algebra/qual/completed

Let \(G\) be a group of order 105 and let \(P, Q, R\) be Sylow 3, 5, 7 subgroups respectively.

  • Prove that at least one of \(Q\) and \(R\) is normal in \(G\).

  • Prove that \(G\) has a cyclic subgroup of order 35.

  • Prove that both \(Q\) and \(R\) are normal in \(G\).

  • Prove that if \(P\) is normal in \(G\) then \(G\) is cyclic.

  • The \(pqr\) theorem.

  • Sylow 3: \({\left\lvert {G} \right\rvert} = p^n m\) implies \(n_p \divides m\) and \(n_p \cong 1 \operatorname{mod}p\).

  • Theorem: If \(H, K \leq G\) and any of the following conditions hold, \(HK\) is a subgroup:

    • \(H{~\trianglelefteq~}G\) (wlog)
    • \([H, K] = 1\)
    • \(H \leq N_G(K)\)
  • Theorem: For a positive integer \(n\), all groups of order \(n\) are cyclic \(\iff n\) is squarefree and, for each pair of distinct primes \(p\) and \(q\) dividing \(n\), \(q - 1 \neq 0 \operatorname{mod}p\).

  • Theorem: \begin{align*} A_i{~\trianglelefteq~}G, \quad G = A_1 \cdots A_k,\quad A_k \cap\prod_{i\neq k} A_i = \emptyset \implies G = \prod A_i .\end{align*}

  • The intersection of subgroups is a again a subgroup.

  • Any subgroups of coprime order intersect trivially?

  • We have

  • \(n_3 \divides 5\cdot 7, \quad n_3 \cong 1 \operatorname{mod}3 \implies n_3 \in \left\{{1, 5, 7, 35}\right\} \setminus \left\{{5, 35}\right\}\)

  • \(n_5 \divides 3\cdot 7, \quad n_5 \cong 1 \operatorname{mod}5 \implies n_5 \in \left\{{1, 3, 7, 21}\right\}\setminus \left\{{3, 7}\right\}\)

  • \(n_7 \divides 3\cdot 5, \quad n_7 \cong 1 \operatorname{mod}7 \implies n_7 \in \left\{{1, 3, 5, 15}\right\}\setminus\left\{{3, 5}\right\}\)

  • Thus \begin{align*} n_3 \in \left\{{1, 7}\right\} \quad n_5 \in \left\{{1, 21}\right\} \quad n_7 \in \left\{{1, 15}\right\} .\end{align*}

  • Toward a contradiction, if \(n_5\neq 1\) and \(n_7 \neq 1\), then \begin{align*} {\left\lvert {{\operatorname{Syl}}(5) \cup{\operatorname{Syl}}(7)} \right\rvert} = (5-1)n_5 + (7-1)n_7 + 1 &= 4(21) + 6(15) = 174 > 105 \text{ elements} \end{align*} using the fact that Sylow \(p{\hbox{-}}\)subgroups for distinct primes \(p\) intersect trivially (?).

  • By (a), either \(Q\) or \(R\) is normal.
  • Thus \(QR \leq G\) is a subgroup, and it has order \({\left\lvert {Q} \right\rvert} \cdot {\left\lvert {R} \right\rvert} = 5\cdot 7 = 35\).
  • By the \(pqr\) theorem, since \(5\) does not divide \(7-1=6\), \(QR\) is cyclic.
\todo[inline]{Part (b) not finished!}

  • We want to show \(Q, R{~\trianglelefteq~}G\), so we proceed by showing \(\textbf{not }\qty{n_5 = 21 \text{ or } n_7 = 15}\), which is equivalent to \(\qty{n_5 = 1 \text{ and } n_7 = 1}\) by the previous restrictions.

  • Note that we can write \begin{align*} G = \left\{{\text{elements of order } n}\right\} {\textstyle\coprod}\left\{{\text{elements of order not } n}\right\} .\end{align*} for any \(n\), so we count for \(n=5, 7\):

    • Elements in \(QR\) of order not equal to 5: \({\left\lvert {QR - Q\left\{{\operatorname{id}}\right\} + \left\{{\operatorname{id}}\right\}} \right\rvert} = 35 - 5 + 1 = 31\)
    • Elements in \(QR\) of order not equal to 7: \({\left\lvert {QR - \left\{{\operatorname{id}}\right\}R + \left\{{\operatorname{id}}\right\}} \right\rvert} = 35 - 7 + 1 = 29\)
  • Since \(QR \leq G\), we have

    • Elements in \(G\) of order not equal to 5 \(\geq 31\).
    • Elements in \(G\) of order not equal to 7 \(\geq 29\).
  • Now both cases lead to contradictions:

    • \(n_5 = 21\): \begin{align*} {\left\lvert {G} \right\rvert} &= {\left\lvert {\left\{{\text{elements of order } 5}\right\} {\textstyle\coprod}\left\{{\text{elements of order not } 5}\right\}} \right\rvert} \\ &\geq n_5(5-1) + 31 = 21(4) + 31 = 115 > 105 = {\left\lvert {G} \right\rvert} .\end{align*}

    • \(n_7 = 15\): \begin{align*} {\left\lvert {G} \right\rvert} &= {\left\lvert {\left\{{\text{elements of order } 7}\right\} {\textstyle\coprod}\left\{{\text{elements of order not } 7}\right\}} \right\rvert} \\ &\geq n_7(7-1) + 29 = 15(6) + 29 = 119 > 105 = {\left\lvert {G} \right\rvert} .\end{align*}

Suppose \(P\) is normal and recall \({\left\lvert {P} \right\rvert} = 3, {\left\lvert {Q} \right\rvert} = 5, {\left\lvert {R} \right\rvert} = 7\).

  • \(P\cap QR = \left\{{e}\right\}\) since \((3, 35) = 1\)
  • \(R\cap PQ = \left\{{e}\right\}\) since \((5, 21) = 1\)
  • \(Q\cap RP = \left\{{e}\right\}\) since \((7, 15) = 1\)

We also have \(PQR = G\) since \({\left\lvert {PQR} \right\rvert} = {\left\lvert {G} \right\rvert}\) (???).

We thus have an internal direct product \begin{align*} G \cong P\times Q \times R \cong {\mathbf{Z}}_3 \times{\mathbf{Z}}_5 \times{\mathbf{Z}}_7 \cong {\mathbf{Z}}_{105} .\end{align*} by the Chinese Remainder Theorem, which is cyclic.

Spring 2021 #3 #algebra/qual/work

  • Show that every group of order \(p^2\) with \(p\) prime is abelian.

  • State the 3 Sylow theorems.

  • Show that any group of order \(4225 = 5^2 13^2\) is abelian.

  • Write down one representative from each isomorphism class of abelian groups of order 4225.

Fall 2020 #1 #algebra/qual/work

  • Using Sylow theory, show that every group of order \(2p\) where \(p\) is prime is not simple.

  • Classify all groups of order \(2p\) and justify your answer. For the nonabelian group(s), give a presentation by generators and relations.

Fall 2020 #2 #algebra/qual/work

Let \(G\) be a group of order 60 whose Sylow 3-subgroup is normal.

  • Prove that \(G\) is solvable.

  • Prove that the Sylow 5-subgroup is also normal.

#1 #algebra/qual/completed #2 #algebra/qual/work #3