Quotient Groups

# Quotient Groups

Recall from the Normal Subgroups page that if $(G, *)$ is a group and $(H, *)$ is a subgroup then $(H, *)$ is said to be a normal subgroup of $(G, *)$ if $gH = Hg$ for all $g \in G$.

We will now define a new type of group called a quotient group which requires that we have a group $(G, *)$ and a normal subgroup $(H, *)$.

Definition: Let $(G, *)$ be a group and let $(H, *)$ be a normal subgroup of $(G, *)$. Let $G / H = \{ gH : g \in G \}$ be the set of left cosets of $H$ in $G$. Then the Quotient of $G$ by $H$ is the group $(G/H, \cdot)$ where $\cdot$ is the binary operation on $G/H$ defined for all $g_1H, g_2H \in G/H$ by $(g_1H) \cdot (g_2H) = (g_1 * g_2)H$. |

At the moment is extremely unclear as to whether the definition above is well-defined. Fortunately it is as we prove below.

Theorem 1: Let $(G, *)$ be a group and let $(H, *)$ be a normal subgroup of $(G, *)$. Then the quotient of $G$ by $H$, $(G/H, \cdot)$ is a group. |

**Proof:**We must first verify that the binary operation $\cdot$ is well defined. Let $aH, bH \in G/H$. We want to show that if $a'H, b'H \in G/H$ and $aH = a'H$, $bH = b'H$ that then $aH \cdot bH = a'H \cdot b'H$. Since $aH = a'H$ and $a \in H$ we have that there exists an $n_1 \in H$ such that $a = a'n_1$. Similarly, since $bH = b'H$ and $b \in H$ we have that there exists an $n_2 \in H$ such that $b = b'n_2$. So by using normality and noting that $H n_2 = H$ and $n_1H = H$ we see that:

\begin{align} \quad (aH) \cdot (bH) &= (ab)H \\ &= (a'n_1b'n_2) H \\ &= (a'n_1b') H n_2 \\ &= (a'n_1b') H \\ &= (a'n_1) H b' \\ &= (a')(n_1H) b' \\ &= (a') H b' \\ &= (a'b')H \\ &= (a'H) \cdot (b'H) \end{align}

- So indeed the operation $\cdot$ is well-defined regardless of the representative used to indicate the left cosets of $H$ in $G$.

- Now for all $aH, bH, cH \in G/H$ we have that:

\begin{align} \quad [(aH) \cdot (bH)] \cdot (cH) = [(a * b)H] \cdot (cH)= [([a * b] * c)H]= [(a * [b * c])H] = (aH) \cdot [(b * c)]H = (aH) \cdot [(bH) \cdot c(H)] \end{align}

- So $\cdot$ is associative on $G/H$.

- The identity element in $G / H$ is $eH$ where $e$ is the identity in $(G, *)$ since for all $aH \in G/H$ we have that:

\begin{align} \quad (eH) \cdot (aH) = (e * a)H = (aH) \end{align}

(4)
\begin{align} \quad (aH) \cdot (eH) = (a * e)H = (aH) \end{align}

- For each element $aH \in G / H$ we have that $(aH)^{-1} = a^{-1}H$ since:

\begin{align} \quad (aH) \cdot (aH)^{-1} = (aH) \cdot (a^{-1}H) = (a * a^{-1})H = eH \end{align}

(6)
\begin{align} \quad (aH)^{-1} \cdot (aH) = (a^{-1}H) \cdot (aH) =(a^{-1} * a)H = eH \end{align}

- So $(G/H, \cdot)$ is a group. $\blacksquare$

What's next to investigate is the size of a quotient group. Fortunately, the answer is what we'd expect and follows directly from Lagrange's Theorem.

Theorem 2: Let $(G, *)$ be a finite group and let $(H, *)$ be a normal subgroup of $(G, *)$. Then $\displaystyle{\mid G / H \mid = \frac{\mid G \mid}{\mid H \mid}}$. |

**Proof:**By Lagrange's Theorem we have that $\mid G \mid = [G : H] \mid H \mid$. But $\mid G / H \mid = [G : H]$ and so by rearranging this quality we get $\displaystyle{\mid G / H \mid = \frac{\mid G \mid}{\mid H \mid}}$. $\blacksquare$