A Group with a Prime Power Order Elements Has Order a Power of the Prime.

Group Theory Problems and Solutions in Mathematics

Problem 17

Let $p$ be a prime number. Suppose that the order of each element of a finite group $G$ is a power of $p$. Then prove that $G$ is a $p$-group. Namely, the order of $G$ is a power of $p$.

LoadingAdd to solve later

Sponsored Links

Hint.

You may use Sylow’s theorem.
For a review of Sylow’s theorem, please check out the post Sylow’s Theorem (summary).

Proof.

If $G$ is a trivial group, then the claim is trivial. So assume that $|G|>1$.

Seeking a contradiction, suppose that $|G|=p^nm$ for some $n,m \in \Z$ and $p$ and $m>1$ are relatively prime.
Let $l$ be a prime factor of $m$. Then by Sylow’s theorem, there exists a Sylow $l$-subgroup of $G$.
The order of a nontrivial element of this subgroup is divisible by the prime $l$ and this contradicts that each element has order power of $p$ since $l$ and $p$ are relatively prime.

Comment.

If we assume Sylow’s theorem, then the proof of this problem is straightforward.
How about proving it more directly (without using Sylow’s theorem)?


LoadingAdd to solve later

Sponsored Links

More from my site

  • Are Groups of Order 100, 200 Simple?Are Groups of Order 100, 200 Simple? Determine whether a group $G$ of the following order is simple or not. (a) $|G|=100$. (b) $|G|=200$.   Hint. 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 […]
  • If a Sylow Subgroup is Normal in a Normal Subgroup, it is a Normal SubgroupIf 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 […]
  • The Index of the Center of a Non-Abelian $p$-Group is Divisible by $p^2$The Index of the Center of a Non-Abelian $p$-Group is Divisible by $p^2$ 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 $p^2$. 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 […]
  • Subgroup Containing All $p$-Sylow Subgroups of a GroupSubgroup 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 […]
  • Finite Group and a Unique Solution of an EquationFinite Group and a Unique Solution of an Equation Let $G$ be a finite group of order $n$ and let $m$ be an integer that is relatively prime to $n=|G|$. Show that for any $a\in G$, there exists a unique element $b\in G$ such that \[b^m=a.\]   We give two proofs. Proof 1. Since $m$ and $n$ are relatively prime […]
  • If a Group is of Odd Order, then Any Nonidentity Element is Not Conjugate to its InverseIf a Group is of Odd Order, then Any Nonidentity Element is Not Conjugate to its Inverse Let $G$ be a finite group of odd order. Assume that $x \in G$ is not the identity element. Show that $x$ is not conjugate to $x^{-1}$.   Proof. Assume the contrary, that is, assume that there exists $g \in G$ such that $gx^{-1}g^{-1}=x$. Then we have \[xg=gx^{-1}. […]
  • Sylow’s Theorem (Summary)Sylow’s Theorem (Summary) In this post we review Sylow's theorem and as an example we solve the following problem. Show that a group of order $200$ has a normal Sylow $5$-subgroup. Review of Sylow's Theorem One of the important theorems in group theory is Sylow's theorem. Sylow's theorem is a […]
  • Every Group of Order 12 Has a Normal Subgroup of Order 3 or 4Every 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 […]

You may also like...

Please Login to Comment.

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Group Theory
Normal Subgroups Problems and Solutions in Group Theory
Any Subgroup of Index 2 in a Finite Group is Normal

Show that any subgroup of index $2$ in a group is a normal subgroup.

Close