MTH3003 Weekly Problems 10

Original Documents: mth3003 weekly problem sheet 10.pdf / mth3003 weekly problem sheet 10 handwritten solutions.pdf

Vibes: Final problem sheet, all on Sylow’s theorems. Mix of “find a subgroup of order $X$” (apply Sylow’s First), counting Sylow subgroups (apply Sylow’s Third), and structural deductions (uniqueness $\Rightarrow$ normality $\Rightarrow$ direct product). The last problem nudges into the classification of small-order groups.

Used Techniques:

• Sylow’s First Theorem: prime $p\mid |G|\Rightarrow$ Sylow $p$-subgroup of order $p^r$ exists.
• Sylow’s Third Theorem: $a(p)\mid t$ AND $a(p)\equiv 1(\mathrm{mod}p)$.
• Uniqueness $\Rightarrow$ normality: $a(p)=1\Rightarrow$ unique Sylow $p$-subgroup is normal in $G$.
• Internal direct product: two normal subgroups with trivial intersection and full product give $G\cong H\times K$.
• Counting $r$-cycles in $S_n$: $\frac{n!}{r\cdot (n-r)!}$ for the number of $r$-cycles.

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10.1. Existence of Subgroups via Sylow’s First Theorem

Question

Prove the following: (a) Every group of order $48,600$ contains a subgroup of order $25$. (b) Prove that $S_{12}$ has a subgroup of order $243$. (c) Prove that a group of order $726$ has a subgroup of order $121$.

(a) Factorise: $48600=2^3\cdot 3^5\cdot 5^2$. Take $p=5$, so $|G|=5^2\cdot t$ with $t=2^3\cdot 3^5=1944$ coprime to $5$. By Sylow’s First Theorem, $G$ has a Sylow $5$-subgroup of order $5^2=\boxed{25}$. $■$

(b) $|S_{12}|=12!$. The exponent of $3$ in $12!$ is

$$\lfloor \frac{12}{3}\rfloor +\lfloor \frac{12}{9}\rfloor =4+1=5.$$

So $3^5=243$ divides $|S_{12}|$, and the Sylow $3$-subgroup has order $\boxed{243}$. By Sylow’s First Theorem, $S_{12}$ has such a subgroup. $■$

(c) $726=2\cdot 3\cdot 11^2$. Take $p=11$: a Sylow $11$-subgroup has order $11^2=\boxed{121}$. By Sylow’s First Theorem, $G$ has such a subgroup. $■$

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10.2. Sylow Subgroups of $S_4$

Question

Find a $2$-Sylow subgroup and a $3$-Sylow subgroup of $S_4$.

$|S_4|=24=2^3\cdot 3$. Sylow $2$-subgroup has order $8$, Sylow $3$-subgroup has order $3$.

Sylow $3$-subgroup. $⟨(123)⟩=\{e,(123),(132)\}\cong C_3$.

Sylow $2$-subgroup. The dihedral group of the square (with vertices labelled $1,2,3,4$ in cyclic order):

$$D_8=⟨(1234),(13)⟩=\{e,(1234),(13)(24),(1432),(13),(24),(12)(34),(14)(23)\}$$

• eight elements, isomorphic to $D_8$. ✓

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10.3. Group of Order 24 with Exactly Four Sylow 2-Subgroups?

Question

Does there exist a finite group of order $24$ with precisely four distinct Sylow 2-subgroups? Give an example or a proof of nonexistence.

Solution. No such group exists.

$24=2^3\cdot 3$, so by Sylow’s Third Theorem $a(2)\mid 3$ and $a(2)\equiv 1(\mathrm{mod}2)$. Divisibility forces $a(2)\in \{1,3\}$; mod-2 forces $a(2)$ odd. Both conditions agree: $a(2)\in \{1,3\}$.

$a(2)=4$ is impossible, since $4\nmid 3$ and $4\not\equiv 1(\mathrm{mod}2)$. $■$

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10.4. $S_8$ Has at Least Eight Sylow 7-Subgroups

Question

Prove that $S_8$ has at least eight Sylow $7$-subgroups.

Solution. $|S_8|=8!=40320=2^7\cdot 3^2\cdot 5\cdot 7$. Sylow $7$-subgroups have order $7$ (so are cyclic, generated by $7$-cycles).

By Sylow’s Third Theorem, $a(7)\mid 5760$ and $a(7)\equiv 1(\mathrm{mod}7)$, so $a(7)\in \{1,8,15,22,29,36,\dots \}$.

Lower-bound by exhibiting two distinct Sylow 7-subgroups. $S_8$ contains many $7$-cycles, e.g. $(1234567)$ and $(1234568)$. The first generates $⟨(1234567)⟩$, which doesn’t contain the second (the first cycle moves $7$ but fixes $8$, while the second moves $8$). So these are two distinct Sylow $7$-subgroups. Hence $a(7)>1$.

Combined with $a(7)\equiv 1(\mathrm{mod}7)$ and $a(7)\ge 2$, the smallest possibility is $a(7)=8$. So $\boxed{a(7)\ge 8}$. $■$

Note

In fact $a(7)$ is much larger. The number of $7$-cycles in $S_8$ is $8\cdot 7!/7=5760$, and each Sylow $7$-subgroup contains $6$ generators (the $7$-cycle and its powers $1,2,\dots ,6$), so $a(7)=5760/6=960$.

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10.5. Unique Sylow $\Rightarrow$ Normal

Question

Prove that if $a(p)=1$, then the Sylow $p$-subgroup of $G$ is normal.

Solution. Let $H$ be the unique Sylow $p$-subgroup of $G$. For any $g\in G$, the conjugate $g^{-1}Hg$ is a subgroup of $G$ (Quick Subgroup Test) of order $|g^{-1}Hg|=|H|=p^r$ (the conjugation map $h\mapsto g^{-1}hg$ is a bijection). So $g^{-1}Hg$ is a Sylow $p$-subgroup. Since there’s only one of those (by hypothesis), $g^{-1}Hg=H$ for every $g\in G$. Hence $H⊴G$. $■$

Note

This is the workhorse Sylow trick. A key reason Sylow’s Third Theorem is so useful: if the divisibility constraints force $a(p)=1$, you get a normal subgroup for free.

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10.6. Subgroups of a Group of Order 35

Question

Let $|G|=35$. Find all subgroups of $G$.

Solution. $35=5\cdot 7$.

Possible orders. By Lagrange, subgroups have orders dividing $35$: $1,5,7,35$.

Order $1$. Just the trivial subgroup $\{e\}$.

Order $35$. Just $G$ itself.

Order $5$. Sylow $5$-subgroups. By Sylow’s Third Theorem, $a(5)\mid 7$ and $a(5)\equiv 1(\mathrm{mod}5)$, so $a(5)\in \{1,7\}\cap \{1,6,11,\dots \}=\{1\}$. So exactly one Sylow $5$-subgroup.

Order $7$. Sylow $7$-subgroups. $a(7)\mid 5$ and $a(7)\equiv 1(\mathrm{mod}7)$, so $a(7)\in \{1,5\}\cap \{1,8,15,\dots \}=\{1\}$. So exactly one Sylow $7$-subgroup.

Conclusion. $G$ has exactly four subgroups: $\{e\}$, the unique $C_5$, the unique $C_7$, and $G$. $■$

Note

Combined with the 10.7. Every Group of Order 221 is Cyclic’s argument: any such $G$ has $G\cong C_5\times C_7\cong C_{35}$.

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10.7. Every Group of Order 221 is Cyclic

Question

Let $|G|=221$. Prove $G$ is cyclic.

Solution. $221=13\cdot 17$ - a product of two distinct primes.

Sylow $13$-subgroup count. $a(13)\mid 17$ and $a(13)\equiv 1(\mathrm{mod}13)$. $a(13)\in \{1,17\}$. Now $17\equiv 4(\mathrm{mod}13)$, so $a(13)=1$.

Sylow $17$-subgroup count. $a(17)\mid 13$ and $a(17)\equiv 1(\mathrm{mod}17)$. $a(17)\in \{1,13\}$. Now $13\not\equiv 1(\mathrm{mod}17)$, so $a(17)=1$.

Let $H$ be the unique Sylow $13$-subgroup and $K$ the unique Sylow $17$-subgroup. Both are normal in $G$ (by problem 10.5).

Trivial intersection. $|H\cap K|$ divides both $|H|=13$ and $|K|=17$, hence divides $\gcd (13,17)=1$. So $H\cap K=\{e\}$.

Product is $G$. $|HK|=|H||K|/|H\cap K|=13\cdot 17/1=221=|G|$, so $HK=G$.

Internal direct product. By the internal direct product theorem (lecture 18),

$$G\cong H\times K\cong C_{13}\times C_{17}\cong C_{221},$$

where the last isomorphism uses $\gcd (13,17)=1$ (the direct product of two coprime cyclic groups is cyclic). $■$

Generalisation

The same argument applies to any group whose order is a product of two distinct primes $pq$ with $p<q$ and $p\nmid q-1$: the group must be cyclic. In particular $|G|=15,33,35,51,65,77,85,91,\dots$ all force $G$ cyclic. (When $p\mid q-1$ a non-cyclic example exists: $|G|=6$ allows $S_3$.)