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**Formal Definition**

Let $H$ be a subgroup of a group $G$. The number of left cosets of $H$ in $G$ is the **index** $(G:H)$ **of** $H$ **in** $G$.

**Informal Definition**

The index is the number of cosets which can be enumerated by dividing the number of elements in the group by the number of elements in the subgroup.

**Example(s)**

Find the index of $\langle 3 \rangle$ in the group $\mathbb{Z_{24}}$.

$\langle 3 \rangle = \{1,3,6,9,12,15,18,21\}$ has 8 elements, so its index is $24/8 = 3$

**Non-example(s)**

**Claim:**

$A_4$ has no subgroup of order 6.

$|A_4| = 12$

**Proof:**

Suppose $A_4$ does have a subgroup of order 6, call it $H$. Then the cosets of $H$ partition $A_4$ into 2 cells. $A_4$ has 8 elements of order 3. $(a_1 a_2 a_3)$. How many distinct elements of this form are there? ($4 \times 3 \times 2 = 24$) but $(a_1 a_2 a_3) = (a_2 a_3 a_1) = (a_3 a_1 a_2)$ where $24/3 = 8$.

So we let $a$ be an element of order 3. Then either $a\in H$ or $a\notin H$. If $a\notin H$, then $A_4 = H \cup aH$. So $a^2 \in H$ or $a^2 \in aH$. If $a \in H$ then $(a^2)^{-1} \in H$ but $(a^2)^{-1} = a$ so $a \in H$.

If $a^2 \in aH$ then $a^2 =ah$ for some $h \in H$ \implies a^3=a^2h$ (by multiplication on the left)

$\implies 1 = a^2h \in H$

$\implies a^2 = h^{-1}$

$\implies a^2 \in H$

Therefore there is no subgroup of order 6.

**Additional Comments**

The index $(G:H)$ just defined may be finite or infinite. If $G$ is finite, then obviously $(G:H)$ is finite and $(G:H)=|G|/|H|$,since every coset of $H$ contains $|H|$ elements.