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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.


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 $24/8 = 3$


$A_4$ has no subgroup of order 6.
$|A_4| = 12$

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.

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