The Index of the Center of a Non-Abelian $p$-Group is Divisible by $p^2$
Problem 124
Let $p$ be a prime number.
Let $G$ be a non-abelian $p$-group.
Show that the index of the center of $G$ is divisible by $p^2$. Add to solve later
Sponsored Links
Proof.
Suppose the order of the group $G$ is $p^a$, for some $a \in \Z$.
Let $Z(G)$ be the center of $G$. Since $Z(G)$ is a subgroup of $G$, the order of the center is also a power of $p$, that is, $|Z(G)|=p^b$, for some $b \in \Z$.
Then we have the index $[G: Z(G)]=p^{a-b}$.
If $a-b=0$, then we have $G=Z(G)$ and $G$ is an abelian group. This contradicts with the assumption that $G$ is non-abelian. So $a-b \neq 0$.
Subgroup Containing All $p$-Sylow Subgroups of a Group
Suppose that $G$ is a finite group of order $p^an$, where $p$ is a prime number and $p$ does not divide $n$.
Let $N$ be a normal subgroup of $G$ such that the index $|G: N|$ is relatively prime to $p$.
Then show that $N$ contains all $p$-Sylow subgroups of […]
Equivalent Definitions of Characteristic Subgroups. Center is Characteristic.
Let $H$ be a subgroup of a group $G$. We call $H$ characteristic in $G$ if for any automorphism $\sigma\in \Aut(G)$ of $G$, we have $\sigma(H)=H$.
(a) Prove that if $\sigma(H) \subset H$ for all $\sigma \in \Aut(G)$, then $H$ is characteristic in $G$.
(b) Prove that the center […]
The Center of a p-Group is Not Trivial
Let $G$ be a group of order $|G|=p^n$ for some $n \in \N$.
(Such a group is called a $p$-group.)
Show that the center $Z(G)$ of the group $G$ is not trivial.
Hint.
Use the class equation.
Proof.
If $G=Z(G)$, then the statement is true. So suppose that $G\neq […]
Normal Subgroup Whose Order is Relatively Prime to Its Index
Let $G$ be a finite group and let $N$ be a normal subgroup of $G$.
Suppose that the order $n$ of $N$ is relatively prime to the index $|G:N|=m$.
(a) Prove that $N=\{a\in G \mid a^n=e\}$.
(b) Prove that $N=\{b^m \mid b\in G\}$.
Proof.
Note that as $n$ and […]
Group of Order $pq$ is Either Abelian or the Center is Trivial
Let $G$ be a group of order $|G|=pq$, where $p$ and $q$ are (not necessarily distinct) prime numbers.
Then show that $G$ is either abelian group or the center $Z(G)=1$.
Hint.
Use the result of the problem "If the Quotient by the Center is Cyclic, then the Group is […]
The Order of $ab$ and $ba$ in a Group are the Same
Let $G$ be a finite group. Let $a, b$ be elements of $G$.
Prove that the order of $ab$ is equal to the order of $ba$.
(Of course do not assume that $G$ is an abelian group.)
Proof.
Let $n$ and $m$ be the order of $ab$ and $ba$, respectively. That is,
\[(ab)^n=e, […]
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 […]
Non-Abelian Simple Group is Equal to its Commutator Subgroup
Let $G$ be a non-abelian simple group. Let $D(G)=[G,G]$ be the commutator subgroup of $G$. Show that $G=D(G)$.
Definitions/Hint.
We first recall relevant definitions.
A group is called simple if its normal subgroups are either the trivial subgroup or the group […]