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What are the orders of all Sylow $p$-subgroups where $G$ has order 18, 24, 54, 72, and 80?
If $|G| = 18 = 2 \cdot 3^2\text{,}$ then the order of a Sylow 2-subgroup is 2, and the order of a Sylow 3-subgroup is 9.
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What are the orders of all Sylow $p$-subgroups where $G$ has order 18, 24, 54, 72, and 80?
If $|G| = 18 = 2 \cdot 3^2\text{,}$ then the order of a Sylow 2-subgroup is 2, and the order of a Sylow 3-subgroup is 9.
Find all the Sylow 3-subgroups of $S_4$ and show that they are all conjugate.
The four Sylow 3-subgroups of $S_4$ are $P_1 = \{ (1), (123), (132) \}\text{,}$ $P_2 = \{ (1), (124), (142) \}\text{,}$ $P_3 = \{ (1), (134), (143) \}\text{,}$ $P_4 = \{ (1), (234), (243) \}\text{.}$
Show that every group of order 45 has a normal subgroup of order 9.
Let $H$ be a Sylow $p$-subgroup of $G\text{.}$ Prove that $H$ is the only Sylow $p$-subgroup of $G$ contained in $N(H)\text{.}$
Prove that no group of order 96 is simple.
Since $|G| = 96 = 2^5 \cdot 3\text{,}$ $G$ has either one or three Sylow 2-subgroups by the Third Sylow Theorem. If there is only one subgroup, we are done. If there are three Sylow 2-subgroups, let $H$ and $K$ be two of them. Therefore, $|H \cap K| \geq 16\text{;}$ otherwise, $HK$ would have $(32 \cdot 32)/8 = 128$ elements, which is impossible. Thus, $H \cap K$ is normal in both $H$ and $K$ since it has index 2 in both groups.
Prove that no group of order 160 is simple.
If $H$ is a normal subgroup of a finite group $G$ and $|H| = p^k$ for some prime $p\text{,}$ show that $H$ is contained in every Sylow $p$-subgroup of $G\text{.}$
Let $G$ be a group of order $p^2 q^2\text{,}$ where $p$ and $q$ are distinct primes such that $q \nmid p^2 - 1$ and $p \nmid q^2 - 1\text{.}$ Prove that $G$ must be abelian. Find a pair of primes for which this is true.
Show that $G$ has a normal Sylow $p$-subgroup of order $p^2$ and a normal Sylow $q$-subgroup of order $q^2\text{.}$
Show that a group of order 33 has only one Sylow 3-subgroup.
Let $H$ be a subgroup of a group $G\text{.}$ Prove or disprove that the normalizer of $H$ is normal in $G\text{.}$
False.
Let $G$ be a finite group divisible by a prime $p\text{.}$ Prove that if there is only one Sylow $p$-subgroup in $G\text{,}$ it must be a normal subgroup of $G\text{.}$
Let $G$ be a group of order $p^r\text{,}$ $p$ prime. Prove that $G$ contains a normal subgroup of order $p^{r-1}\text{.}$
Suppose that $G$ is a finite group of order $p^n k\text{,}$ where $k \lt p\text{.}$ Show that $G$ must contain a normal subgroup.
Let $H$ be a subgroup of a finite group $G\text{.}$ Prove that $g N(H) g^{-1} = N(gHg^{-1})$ for any $g \in G\text{.}$
Prove that a group of order 108 must have a normal subgroup.
Classify all the groups of order 175 up to isomorphism.
Let $G$ have order $p_1^{e_1} \cdots p_n^{e_n}$ and suppose that $G$ has $n$ Sylow $p$-subgroups $P_1, \ldots, P_n$ where $|P_i| = p_i^{e_i}\text{.}$ Prove that $G$ is isomorphic to $P_1 \times \cdots \times P_n\text{.}$
Let $P$ be a normal Sylow $p$-subgroup of $G\text{.}$ Prove that every inner automorphism of $G$ fixes $P\text{.}$
What is the smallest possible order of a group $G$ such that $G$ is nonabelian and $|G|$ is odd? Can you find such a group?
If $H$ is a normal subgroup of a finite group $G$ and $P$ is a Sylow $p$-subgroup of $H\text{,}$ for each $g \in G$ show that there is an $h$ in $H$ such that $gPg^{-1} = hPh^{-1}\text{.}$ Also, show that if $N$ is the normalizer of $P\text{,}$ then $G= HN\text{.}$
Show that if the order of $G$ is $p^nq\text{,}$ where $p$ and $q$ are primes and $p>q\text{,}$ then $G$ contains a normal subgroup.
Prove that the number of distinct conjugates of a subgroup $H$ of a finite group $G$ is $[G : N(H) ]\text{.}$
Define a mapping between the right cosets of $N(H)$ in $G$ and the conjugates of $H$ in $G$ by $N(H) g \mapsto g^{-1} H g\text{.}$ Prove that this map is a bijection.
Prove that a Sylow 2-subgroup of $S_5$ is isomorphic to $D_4\text{.}$
Suppose $p$ is prime and $p$ does not divide $m\text{.}$ Show that
Let ${\mathcal S}$ denote the set of all $p^k$ element subsets of $G\text{.}$ Show that $p$ does not divide $|{\mathcal S}|\text{.}$
Define an action of $G$ on ${\mathcal S}$ by left multiplication, $aT = \{ at : t \in T \}$ for $a \in G$ and $T \in {\mathcal S}\text{.}$ Prove that this is a group action.
Prove $p \nmid | {\mathcal O}_T|$ for some $T \in {\mathcal S}\text{.}$
Let $\{ T_1, \ldots, T_u \}$ be an orbit such that $p \nmid u$ and $H = \{ g \in G : gT_1 = T_1 \}\text{.}$ Prove that $H$ is a subgroup of $G$ and show that $|G| = u |H|\text{.}$
Show that $p^k$ divides $|H|$ and $p^k \leq |H|\text{.}$
Show that $|H| = |{\mathcal O}_T| \leq p^k\text{;}$ conclude that therefore $p^k = |H|\text{.}$
Let $G$ be a group. Prove that $G' = \langle a b a^{-1} b^{-1} : a, b \in G \rangle$ is a normal subgroup of $G$ and $G/G'$ is abelian. Find an example to show that $\{ a b a^{-1} b^{-1} : a, b \in G \}$ is not necessarily a group.
Let $a G', b G' \in G/G'\text{.}$ Then $(a G')( b G') = ab G' = ab(b^{-1}a^{-1}ba) G' = (abb^{-1}a^{-1})ba G' = ba G'\text{.}$