The Index of a Subgroup

# The Index of a Subgroup

If $(G, \cdot)$ is a group and $(H, \cdot)$ is a subgroup then we might want to know the number of left cosets and the number of right cosets of $H$. As the following theorem will show - the number of left cosets of $H$ will always equal the number of right cosets of $H$.

 Theorem 1: Let $(G, \cdot)$ be a group and let $(H, \cdot)$ be a subgroup. Then the number of left cosets of $H$ equals the number of right cosets of $H$.
• Proof: Let $L_H$ denote the set of all left cosets of $H$ and let $R_H$ denote the set of all right cosets of $H$. Define a function $f : L_H \to R_H$ for all $gH \in L_H$ by
(1)
\begin{align} \quad f(gH) = Hg^{-1} \end{align}
• If we can show that $f$ is bijective then $\mid L_H \mid = \mid R_H$. We first show that $f$ is injective. Let $g_1H, g_2H \in L_H$ and suppose that $f(g_1H) = f(g_2H)$. Then:
(2)
\begin{align} \quad Hg_1^{-1} = Hg_2^{-1} \end{align}
(3)
\begin{align} \quad g_1H = g_2H \end{align}
• Hence $f$ is injective. We now show that $f$ is surjective. Let $Hg \in R_H$. Then we have that for $g^{-1}H \in L_H$ that:
(4)
\begin{align} \quad f(g^{-1}H) = H(g^{-1})^{-1} = Hg \end{align}
• So $f$ is indeed surjective. Since $f$ is a function from a finite set to a finite set that is bijective we must have that $\mid L_H \mid = \mid R_H \mid$, i.e., the number of left cosets of $H$ equals the number of right cosets of $H$. $\blacksquare$

With the result above, we can unambiguously define the index of subgroup $(H, \cdot)$ in a group $(G, \cdot)$.

 Definition: Let $(G, \cdot)$ be a group and let $(H, \cdot)$ be a subgroup. The Index of $H$ in $G$ denoted $[G : H]$ is defined as the number of left (or right) cosets of $H$.

For example, consider the symmetric group $(S_3, \circ)$ and let $H = \{ \epsilon, (12) \}$. Let's find the index $S_3 : H ]$. Note that $S_3 = \{ \epsilon, (12), (13), (23), (123), (132) \}$. So the left cosets of $H$ in $S_3$ are:

(5)
\begin{align} \quad \epsilon H = \{ \epsilon \circ \epsilon, \epsilon \circ (12) \} = \{ \epsilon, (12) \} \\ \end{align}
(6)
\begin{align} \quad (12)H= \{ (12) \circ \epsilon, (12) \circ (12) \} = \{ (12), \epsilon \} \end{align}
(7)
\begin{align} \quad (13)H = \{ (13) \circ \epsilon, (13) \circ (12) \} = \{ (13), (123) \} \end{align}
(8)
\begin{align} \quad (23)H = \{ (23) \circ \epsilon, (23) \circ (12) \} = \{ (23), (132) \} \end{align}
(9)
\begin{align} \quad (123)H = \{ (123) \circ \epsilon, (123) \circ (12) \} = \{ (123), (13) \} \end{align}
(10)
\begin{align} \quad (132)H = \{ (132) \circ \epsilon, (132) \circ (12) \} = \{ (132), (23) \} \end{align}

So the set of left cosets of $H$ is:

(11)
\begin{align} \quad \{\{ \epsilon, (12) \}, \{ (13), (123) \}, \{ (23), (132) \} \} \end{align}

There are three such cosets so $[S_3 : H] = 3$.

For another example, consider the group $(\mathbb{Z}, +)$ and the subgroup $(3\mathbb{Z}, +)$. Then the left cosets of $(3\mathbb{Z}, +)$ are:

(12)
\begin{align} \quad 0 + 3\mathbb{Z} = \{ ..., -3, 0, 3, ... \} \end{align}
(13)
\begin{align} \quad 1 + 3\mathbb{Z} = \{ ..., -2, 1, 4, ... \} \end{align}
(14)
\begin{align} \quad 2 + 3\mathbb{Z} = \{ ..., -1, 2, 5, ... \} \end{align}

Note that if $m \equiv n \pmod 3$ then $m + 3\mathbb{Z} = n + 3\mathbb{Z}$. So in this example we have that $[\mathbb{Z} : 3\mathbb{Z}] = 3$.