Finite Group and Subgroup Criteria

Group Theory Problems and Solutions in Mathematics

Problem 160

Let $G$ be a finite group and let $H$ be a subset of $G$ such that for any $a,b \in H$, $ab\in H$.

Then show that $H$ is a subgroup of $G$.
 
LoadingAdd to solve later

Proof.

Let $a \in H$. To show that $H$ is a subgroup of $G$, it suffices to show that the inverse $a^{-1}$ is in $H$.
If $a=e$ is the identity element, this is trivial. So we assume that $a \neq e$.

Note that $a^2=a\cdot a\in H$, $a^3=a^2\cdot a\in H$, and repeating this we see that $a^n\in H$ for any positive integer $n$.
Since $G$ is finite, not all of $a^n$ can be different.
Thus there exists positive integers $m, n$ such that $a^m=a^n$ and $m>n$.

Note that we actually have $m>n+1$.
For if $m=n+1$, then we have $a^{n+1}=a^n$ and this implies that $a=e$.
This contradicts out choice of $a$. Thus we have $m>n+1$, or equivalently we have
\[m-n-1>0.\]

Since we have
\[a^{m-n}=e,\] multiplying by $a^{-1}$ we obtain
\[a^{-1}=a^{m-n-1}.\]

Since $m-n-1>0$, the element $a^{m-n-1}\in H$, hence the inverse $a^{-1}\in H$.
Therefore, $H$ is closed under the group operation and inverse, thus $H$ is a subgroup of $G$.

Remark.

In fact, the group $G$ itself can be an infinite group.
We just need that $H$ is a finite subset satisfying the closure property:for any $a,b \in H$, $ab\in H$.

The proof of this generalization is identical to the proof given above.


LoadingAdd to solve later

More from my site

  • Elements of Finite Order of an Abelian Group form a SubgroupElements of Finite Order of an Abelian Group form a Subgroup Let $G$ be an abelian group and let $H$ be the subset of $G$ consisting of all elements of $G$ of finite order. That is, \[H=\{ a\in G \mid \text{the order of $a$ is finite}\}.\] Prove that $H$ is a subgroup of $G$.   Proof. Note that the identity element $e$ of […]
  • Torsion Subgroup of an Abelian Group, Quotient is a Torsion-Free Abelian GroupTorsion Subgroup of an Abelian Group, Quotient is a Torsion-Free Abelian Group Let $A$ be an abelian group and let $T(A)$ denote the set of elements of $A$ that have finite order. (a) Prove that $T(A)$ is a subgroup of $A$. (The subgroup $T(A)$ is called the torsion subgroup of the abelian group $A$ and elements of $T(A)$ are called torsion […]
  • Pullback Group of Two Group Homomorphisms into a GroupPullback Group of Two Group Homomorphisms into a Group Let $G_1, G_1$, and $H$ be groups. Let $f_1: G_1 \to H$ and $f_2: G_2 \to H$ be group homomorphisms. Define the subset $M$ of $G_1 \times G_2$ to be \[M=\{(a_1, a_2) \in G_1\times G_2 \mid f_1(a_1)=f_2(a_2)\}.\] Prove that $M$ is a subgroup of $G_1 \times G_2$.   […]
  • Quotient Group of Abelian Group is AbelianQuotient Group of Abelian Group is Abelian Let $G$ be an abelian group and let $N$ be a normal subgroup of $G$. Then prove that the quotient group $G/N$ is also an abelian group.   Proof. Each element of $G/N$ is a coset $aN$ for some $a\in G$. Let $aN, bN$ be arbitrary elements of $G/N$, where $a, b\in […]
  • Abelian Group and Direct Product of Its SubgroupsAbelian Group and Direct Product of Its Subgroups Let $G$ be a finite abelian group of order $mn$, where $m$ and $n$ are relatively prime positive integers. Then show that there exists unique subgroups $G_1$ of order $m$ and $G_2$ of order $n$ such that $G\cong G_1 \times G_2$.   Hint. Consider […]
  • 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 […]
  • Sylow Subgroups of a Group of Order 33 is Normal SubgroupsSylow 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 $pq$ Has a Normal Sylow Subgroup and SolvableGroup 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, […]

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

More in Group Theory
Group Theory Problems and Solutions
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 $G=D(G)$.  

Close