# 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 $$\geq 29$$.

• $$n_5 = 21$$: \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