Quotient Group of Abelian Group is Abelian

Abelian Group problems and solutions

Problem 340

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.

 
LoadingAdd to solve later

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 $a, b\in G$.

Then we have
\begin{align*}
(aN)(bN)&=(ab)N \\
&=(ba)N && \text{since $G$ is abelian}\\
&=(bN)(aN).
\end{align*}
Here the first and the third equality is the definition of the group operation of $G/N$.

Remark

Since $N$ is a normal subgroup of $G$, the set of left cosets $G/H$ becomes a group with group operation
\[(aN)(bN)=(ab)N\] for any $a, b\in G$.

Related Question.

As an application, try the following problem.

Problem.
Let $H$ and $K$ be normal subgroups of a group $G$. Suppose that $H < K$ and the quotient group $G/H$ is abelian. Then prove that $G/K$ is also an abelian group.

The proof of this problem is given in the post ↴
If quotient $G/H$ is abelian group and $H < K \triangleleft G$, then $G/K$ is abelian.


LoadingAdd to solve later

More from my site

  • 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 […]
  • 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 […]
  • 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 […]
  • Normal Subgroups, Isomorphic Quotients, But Not IsomorphicNormal Subgroups, Isomorphic Quotients, But Not Isomorphic Let $G$ be a group. Suppose that $H_1, H_2, N_1, N_2$ are all normal subgroup of $G$, $H_1 \lhd N_2$, and $H_2 \lhd N_2$. Suppose also that $N_1/H_1$ is isomorphic to $N_2/H_2$. Then prove or disprove that $N_1$ is isomorphic to $N_2$.   Proof. We give a […]
  • 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 […]
  • If Quotient $G/H$ is Abelian Group and $H < K \triangleleft G$, then $G/K$ is AbelianIf Quotient $G/H$ is Abelian Group and $H < K \triangleleft G$, then $G/K$ is Abelian Let $H$ and $K$ be normal subgroups of a group $G$. Suppose that $H < K$ and the quotient group $G/H$ is abelian. Then prove that $G/K$ is also an abelian group.   Solution. We will give two proofs. Hint (The third isomorphism theorem) Recall the third […]
  • If a Subgroup $H$ is in the Center of a Group $G$ and $G/H$ is Nilpotent, then $G$ is NilpotentIf a Subgroup $H$ is in the Center of a Group $G$ and $G/H$ is Nilpotent, then $G$ is Nilpotent Let $G$ be a nilpotent group and let $H$ be a subgroup such that $H$ is a subgroup in the center $Z(G)$ of $G$. Suppose that the quotient $G/H$ is nilpotent. Then show that $G$ is also nilpotent.   Definition (Nilpotent Group) We recall here the definition of a […]
  • Group of Order 18 is SolvableGroup 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 […]

You may also like...

1 Response

  1. 03/17/2017

    […] that [G/K cong (G/H)/(G/K).] Since the group $G/H$ is abelian by assumption, and in general a quotient group of an abelian group is abelian, it follows $(G/H)/(G/K)$ is an abelian […]

Leave a Reply

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

More in Group Theory
Group Theory Problems and Solutions in Mathematics
Special Linear Group is a Normal Subgroup of General Linear Group

Let $G=\GL(n, \R)$ be the general linear group of degree $n$, that is, the group of all $n\times n$ invertible...

Close