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Problem 88 (The Quotient Group $G/N$ Is A Group)

Let $G$ be a group and let $N$ be a normal subgroup of $G$. Prove that the set $G/N$ is actually a group under the operation of coset products.

Solution

We will show that $G/N$ is a group under the operation $(Na)(Nb)=N(ab)$ by using the definition of a group.

We will first show $G/N$ is closed under the operation. Pick $Na,Nb\in G/N$ such that $a,b\in G$. Thus we have $(Na)(Nb)=N(ab)$. Recall $G$ is a group, therefore we have $ab\in G$. Thus we know $N(ab)\in G/N$. Therefore we know $G/N$ is closed under the operation.

Next, we show $G/N$ is associative. Pick $Na,Nb,Nc\in G/N$ such that we have $a,b,c\in G$. Thus we have $$\begin{align} (NaNb)(Nc)&=(Nab)(Nc) \\ &= N(abc) \\ &= (Na)(Nbc) \\ &= (Na)(NbNc). \end{align}$$ Thus, we have $G/N$ is associative.

Next, we show there is an identity in $G/N$. Pick $Ne\in G/N$ such that $e\in G$ is the identity in $G$. Pick $Na\in G/N$. We compute $$\begin{align} (Ne)(Na)&=N(ea) \\ &= Na, \end{align}$$ and we also compute $$\begin{align} (Na)(Ne)&=N(ae) \\ &= Na. \end{align}$$ Thus we have $(Na)(Ne)=Na=(Ne)(Na)$. Thus $Ne$ is an identity in $G/N$.

Lastly, we show there exists inverses in $G/N$. Pick $Na\in G/N$ such that $a\in G$. Pick $Na^{-1}\in G/N$ such that $a^{-1}$ is the inverse of $a$ in $G$. We compute $$\begin{align} (Na^{-1})(Na)&=N(a^{-1}a) \\ &= Ne, \end{align}$$ and we also compute $$\begin{align} (Na)(Na^{-1})&=N(aa^{-1}) \\ &= Ne. \end{align}$$ Thus we have $(Na^{-1})(Na)=Ne=(Na)(Na^{-1})$. Thus $Na^{-1}$ is the inverse of $Na$ in $G/N$. Thus there exists inverses in $G/N$.

Thus, by definition of a group, we know $G/N$ is a group under the operation $(Na)(Nb)=N(ab)$.

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