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 Sylow3 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 2Sylow 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 Sylow2 subgroup or the Sylow7 subgroup is normal.
 Give two nonisomorphic groups of order 56 where the Sylow7 subgroup is normal and the Sylow2 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 nonabelian 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 2Sylow 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 nonabelian group of order 36 whose Sylow2 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^{ca} \coloneqq p^b \end{align*}
where \(a\leq c \implies 0 \leq cb \leq c\) 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} = (51)n_5 + (71)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 \(71=6\), \(QR\) 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(51) + 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(71) + 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 3subgroup is normal.

Prove that \(G\) is solvable.

Prove that the Sylow 5subgroup is also normal.