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
- 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:
- But then by the theorem of equivalent statements presented on the Left and Right Cosets of Subgroups page we must have that:
- 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:
- 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)So the set of left cosets of $H$ is:
(11)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)Note that if $m \equiv n \pmod 3$ then $m + 3\mathbb{Z} = n + 3\mathbb{Z}$. So in this example we have that $[\mathbb{Z} : 3\mathbb{Z}] = 3$.