Let $H$ be a subgroup of a group $G$ and let $a \in G$. The left coset of $H$ containing $a$ is the set

The element $a$ is called a representative of the coset $aH$.

Similarly, the right coset of $H$ containing $a$ is the set

In general, left cosets and right cosets are different. However, when the group operation is written additively (as in abelian groups), we write $a + H = \{a + h : h \in H\}$, and left and right cosets coincide.

Consider the subgroup

of $(\mathbb{Z}, +)$. The cosets of $3\mathbb{Z}$ are:

These three cosets partition $\mathbb{Z}$, and each coset corresponds to a residue class modulo $3$.

Let $G = S_3$ and let $H = \{e, (12)\}$, a subgroup of order $2$. The left cosets of $H$ are:

Thus there are exactly three distinct left cosets: $H$, $(13)H$, and $(23)H$.

Let $H$ be a subgroup of a group $G$ and let $a, b \in G$. Then:

1. $a \in aH$
2. $aH = H$ if and only if $a \in H$
3. $aH = bH$ if and only if $a^{-1}b \in H$
4. $aH = bH$ or $aH \cap bH = \emptyset$
5. $|aH| = |H|$
6. The distinct left cosets of $H$ partition $G$

Two left cosets $aH$ and $bH$ are equal if and only if $b^{-1}a \in H$.

Let $H$ be a subgroup of a group $G$. The index of $H$ in $G$, denoted $[G:H]$, is the number of distinct left cosets of $H$ in $G$.

The index $[G:H]$ may be finite or infinite. When $G$ is finite, the index is always finite. When $G$ is infinite, the index may be finite (as with $[\mathbb{Z} : n\mathbb{Z}] = n$) or infinite.

Let $G$ be a finite group and let $H$ be a subgroup of $G$. Then $|H|$ divides $|G|$. More precisely:

The theorem is named after Joseph-Louis Lagrange (1736–1813), who proved a version of this result for permutation groups in 1770–1771, predating the abstract definition of a group by several decades.

In $S_3$, we have $|S_3| = 6$. The possible orders of subgroups are the divisors of $6$: namely $1, 2, 3, 6$. Indeed:

• Order $1$: the trivial subgroup $\{e\}$
• Order $2$: $\{e, (12)\}$, $\{e, (13)\}$, $\{e, (23)\}$
• Order $3$: $\{e, (123), (132)\} = A_3$
• Order $6$: $S_3$ itself

The converse of Lagrange's theorem is false in general: if $d$ divides $|G|$, there need not exist a subgroup of order $d$. The smallest counterexample is the alternating group $A_4$ of order $12$, which has no subgroup of order $6$.

Let $G$ be a finite group and let $a \in G$. Then the order of $a$ divides $|G|$.

Let $G$ be a finite group and let $a \in G$. Then $a^{|G|} = e$.

Let $G$ be a group of prime order $p$. Then $G$ is cyclic, and every non-identity element is a generator.

Let $p$ be a prime and let $a$ be an integer not divisible by $p$. Then:

Let $n$ be a positive integer and let $a$ be an integer with $\gcd(a, n) = 1$. Then:

where $\phi(n)$ is Euler's totient function.

Let $K \leq H \leq G$ be groups. Then:

In particular, if any two of the three indices are finite, so is the third.

If $H$ and $K$ are subgroups of a finite group $G$ with $K \leq H$, then $[G:K]$ is divisible by both $[G:H]$ and $[H:K]$.

A permutation $\sigma \in S_n$ is called even if it can be written as a product of an even number of transpositions, and odd if it can be written as a product of an odd number of transpositions. The alternating group $A_n$ is the set of all even permutations in $S_n$.

The alternating group $A_n$ is a subgroup of $S_n$ of index $2$. Consequently:

In $S_3$:

• Even permutations: $e$, $(123)$, $(132)$
• Odd permutations: $(12)$, $(13)$, $(23)$

Thus $A_3 = \{e, (123), (132)\}$ has order $3$.

The group $A_4$ has order $12$. Its elements are:

• The identity $e$
• The eight $3$-cycles: $(123), (132), (124), (142), (134), (143), (234), (243)$
• The three products of disjoint transpositions: $(12)(34), (13)(24), (14)(23)$