Use Sylow’s theorem and determine the number of $5$-Sylow subgroup of the group $G$.
Check out the post Sylow’s Theorem (summary) for a review of Sylow’s theorem.
Proof.
(a) When $|G|=100$.
The prime factorization of $100$ is $2^2\cdot 5^2$. Let us determine the number $n_5$ of $5$-Sylow subgroup of $G$.
By Sylow’s theorem, we know that $n_5 \equiv 1 \pmod{5}$ and $n_5$ divides $2^2$.
The only number satisfies both constraints is $n_5=1$. Thus there is only one $5$-Sylow subgroup of $G$. This implies that the $5$-Sylow subgroup is a normal subgroup of $G$.
Since the order of the $5$-Sylow subgroup is $25$, it is a proper normal subgroup. Thus, the group $G$ is not simple.
(b) When $|G|=200$
The prime factorization is $200=2^3\cdot 5^2$.
We again consider the number $n_5$ of $5$-Sylow subgroups of $G$.
Sylow’s theorem implies that $n_5 \equiv 1 \pmod{5}$ and $n_5$ divides $2^3$.
These two constraints has only one solution $n_5=1$.
Thus the group $G$ has a unique proper normal $5$-Sylow subgroup of order $25$. Hence $G$ is a simple group.
This is the 100th problems in this blog.
To post 100 problems were not so simple.
The next goal is to archive the 200th problem.
This problem suggests that this goal is again not simple.
(Update: On 11/25/2016 I achieved the 200th problem: Maximize the dimension of the null space of $A-aI$.)
Group of Order $pq$ Has a Normal Sylow Subgroup and Solvable
Let $p, q$ be prime numbers such that $p>q$.
If a group $G$ has order $pq$, then show the followings.
(a) The group $G$ has a normal Sylow $p$-subgroup.
(b) The group $G$ is solvable.
Definition/Hint
For (a), apply Sylow's theorem. To review Sylow's theorem, […]
Sylow Subgroups of a Group of Order 33 is Normal Subgroups
Prove that any $p$-Sylow subgroup of a group $G$ of order $33$ is a normal subgroup of $G$.
Hint.
We use Sylow's theorem. Review the basic terminologies and Sylow's theorem.
Recall that if there is only one $p$-Sylow subgroup $P$ of $G$ for a fixed prime $p$, then $P$ […]
Group of Order 18 is Solvable
Let $G$ be a finite group of order $18$.
Show that the group $G$ is solvable.
Definition
Recall that a group $G$ is said to be solvable if $G$ has a subnormal series
\[\{e\}=G_0 \triangleleft G_1 \triangleleft G_2 \triangleleft \cdots \triangleleft G_n=G\]
such […]
A Group of Order $pqr$ Contains a Normal Subgroup of Order Either $p, q$, or $r$
Let $G$ be a group of order $|G|=pqr$, where $p,q,r$ are prime numbers such that $p<q<r$.
Show that $G$ has a normal subgroup of order either $p,q$ or $r$.
Hint.
Show that using Sylow's theorem that $G$ has a normal Sylow subgroup of order either $p,q$, or $r$.
Review […]
Non-Abelian Group of Order $pq$ and its Sylow Subgroups
Let $G$ be a non-abelian group of order $pq$, where $p, q$ are prime numbers satisfying $q \equiv 1 \pmod p$.
Prove that a $q$-Sylow subgroup of $G$ is normal and the number of $p$-Sylow subgroups are $q$.
Hint.
Use Sylow's theorem. To review Sylow's theorem, check […]
Every Group of Order 12 Has a Normal Subgroup of Order 3 or 4
Let $G$ be a group of order $12$. Prove that $G$ has a normal subgroup of order $3$ or $4$.
Hint.
Use Sylow's theorem.
(See Sylow’s Theorem (Summary) for a review of Sylow's theorem.)
Recall that if there is a unique Sylow $p$-subgroup in a group $GH$, then it is […]
A Group of Order $20$ is Solvable
Prove that a group of order $20$ is solvable.
Hint.
Show that a group of order $20$ has a unique normal $5$-Sylow subgroup by Sylow's theorem.
See the post summary of Sylow’s Theorem to review Sylow's theorem.
Proof.
Let $G$ be a group of order $20$. The […]
If a Sylow Subgroup is Normal in a Normal Subgroup, it is a Normal Subgroup
Let $G$ be a finite group. Suppose that $p$ is a prime number that divides the order of $G$.
Let $N$ be a normal subgroup of $G$ and let $P$ be a $p$-Sylow subgroup of $G$.
Show that if $P$ is normal in $N$, then $P$ is a normal subgroup of $G$.
Hint.
It follows from […]
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