Two Normal Subgroups Intersecting Trivially Commute Each Other

Group Theory Problems and Solutions in Mathematics

Problem 196

Let $G$ be a group. Assume that $H$ and $K$ are both normal subgroups of $G$ and $H \cap K=1$. Then for any elements $h \in H$ and $k\in K$, show that $hk=kh$.
 
LoadingAdd to solve later

Sponsored Links

Proof.

It suffices to show that $h^{-1}k^{-1}hk \in H \cap K$.
In fact, if this it true then we have $h^{-1}k^{-1}hk=1$, and thus $hk=kh$.

Since $h\in H$ and $H$ is a normal subgroup of $G$, we see that the conjugate $k^{-1}hk\in H$.
Thus we have
\[h^{-1}k^{-1}hk =h^{-1}(k^{-1}hk)\in H. \tag{*}\]

Also, since $k^{-1}\in K$ and $K$ is a normal subgroup of $G$, we have the conjugate $h^{-1}k^{-1}h\in K$.
Hence, we see that
\[h^{-1}k^{-1}hk =(h^{-1}k^{-1}h)k\in K. \tag{**}\]

From (*) and (**), we see that the element $h^{-1}k^{-1}hk$ is in both $H$ and $K$, hence in $H\cap K$ as claimed.


LoadingAdd to solve later

Sponsored Links

More from my site

  • Abelian Normal Subgroup, Intersection, and Product of GroupsAbelian Normal Subgroup, Intersection, and Product of Groups Let $G$ be a group and let $A$ be an abelian subgroup of $G$ with $A \triangleleft G$. (That is, $A$ is a normal subgroup of $G$.) If $B$ is any subgroup of $G$, then show that \[A \cap B \triangleleft AB.\]   Proof. First of all, since $A \triangleleft G$, the […]
  • 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 […]
  • Conjugate of the Centralizer of a Set is the Centralizer of the Conjugate of the SetConjugate of the Centralizer of a Set is the Centralizer of the Conjugate of the Set Let $X$ be a subset of a group $G$. Let $C_G(X)$ be the centralizer subgroup of $X$ in $G$. For any $g \in G$, show that $gC_G(X)g^{-1}=C_G(gXg^{-1})$.   Proof. $(\subset)$ We first show that $gC_G(X)g^{-1} \subset C_G(gXg^{-1})$. Take any $h\in C_G(X)$. Then for […]
  • Two Quotients Groups are Abelian then Intersection Quotient is AbelianTwo Quotients Groups are Abelian then Intersection Quotient is Abelian Let $K, N$ be normal subgroups of a group $G$. Suppose that the quotient groups $G/K$ and $G/N$ are both abelian groups. Then show that the group \[G/(K \cap N)\] is also an abelian group.   Hint. We use the following fact to prove the problem. Lemma: For a […]
  • Group Generated by Commutators of Two Normal Subgroups is a Normal SubgroupGroup Generated by Commutators of Two Normal Subgroups is a Normal Subgroup Let $G$ be a group and $H$ and $K$ be subgroups of $G$. For $h \in H$, and $k \in K$, we define the commutator $[h, k]:=hkh^{-1}k^{-1}$. Let $[H,K]$ be a subgroup of $G$ generated by all such commutators. Show that if $H$ and $K$ are normal subgroups of $G$, then the subgroup […]
  • Commutator Subgroup and Abelian Quotient GroupCommutator Subgroup and Abelian Quotient Group Let $G$ be a group and let $D(G)=[G,G]$ be the commutator subgroup of $G$. Let $N$ be a subgroup of $G$. Prove that the subgroup $N$ is normal in $G$ and $G/N$ is an abelian group if and only if $N \supset D(G)$.   Definitions. Recall that for any $a, b \in G$, the […]
  • The Preimage of a Normal Subgroup Under a Group Homomorphism is NormalThe Preimage of a Normal Subgroup Under a Group Homomorphism is Normal Let $G$ and $G'$ be groups and let $f:G \to G'$ be a group homomorphism. If $H'$ is a normal subgroup of the group $G'$, then show that $H=f^{-1}(H')$ is a normal subgroup of the group $G$.   Proof. We prove that $H$ is normal in $G$. (The fact that $H$ is a subgroup […]
  • Any Subgroup of Index 2 in a Finite Group is NormalAny Subgroup of Index 2 in a Finite Group is Normal Show that any subgroup of index $2$ in a group is a normal subgroup. Hint. Left (right) cosets partition the group into disjoint sets. Consider both left and right cosets. Proof. Let $H$ be a subgroup of index $2$ in a group $G$. Let $e \in G$ be the identity […]

You may also like...

Leave a Reply

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

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

More in Group Theory
Group Theory Problems and Solutions in Mathematics
Abelian Normal Subgroup, Intersection, and Product of Groups

Let $G$ be a group and let $A$ be an abelian subgroup of $G$ with $A \triangleleft G$. (That is,...

Close