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# Section7.3The index of a subgroup and Lagrange's Theorem

##### Definition7.3.1

We define the index of $H$ in $G$, denoted $(G:H)\text{,}$ to be the cardinality of $G/H\text{.}$

##### Remark7.3.2

If $H\leq G\text{,}$ there are always have the same number of left cosets and right cosets of $H$ in $G\text{.}$ The proof of this is left as an exercise for the reader.

(Keep in mind your proof must apply even when there are infinitely many left and/or right cosets of $H$ in $G\text{.}$ Hint: Show there is a bijection between $G/H$ and the set of all right cosets of $H$ in $G\text{.}$)

In the cases in which $G/H$ contains infinitely many elements, we won't worry about specific cardinality, and may simply say that the index of $H$ in $G$ is infinite, and write $(G:H)=\infty\text{.}$ However, we can, of course, distinguish between countably infinite and uncountably infinite indices.

Note that if $G$ is finite then $(G:H)$ must be finite; however, we see below that if $G$ is infinite then $(G:H)$ could be finite or infinite.

##### Example7.3.3

If $(\R:\Z)$ were finite, then we'd be able to write $\R$ as a finite union of countable sets, implying that $\R$ is countable—which it isn't. Thus, $(\R:\Z)=\infty\text{.}$

##### Example7.3.4

Since $\langle i\rangle =\{i,-1,-i,1\}$ is a finite subgroup of $C^*$ and $C^*$ is infinite, we must have that $(C^*:\langle i\rangle )$ is infinite. However, $(C^*:C^*)=1\text{.}$

##### Example7.3.5

Referring to our previous examples, we have:

\begin{equation*} (S_3:\langle (12)\rangle )=3, \end{equation*} \begin{equation*} (D_4:\langle f\rangle )=4, \end{equation*} \begin{equation*} (\Z:5\Z)=5, \end{equation*} \begin{equation*} (4\Z:12\Z)=3, \end{equation*} \begin{equation*} \text{and }(\Z_{12}:\langle 4\rangle )=4. \end{equation*}

Notice that in the cases in which $G$ is finite, $(G:H)=|G|/|H|\text{.}$ This makes sense, since the left cosets of $H$ in $G$ partition $G\text{,}$ and each left coset has cardinality $|H|\text{.}$

Since the left cosets of a subgroup $H$ of a group $G$ partition $G$ and all have the same cardinality, we have the following two theorems. The first is known as Lagrange's Theorem (named for the French mathematician Joseph-Louis Lagrange).

##### Remark7.3.8

The converse of Lagrange's Theorem does not hold. Indeed $A_4$ is a group of order 12 which contains no subgroup of order $6$ even though $6$ divides $|A_4|=12\text{.}$

We end this chapter with two corollaries to Lagrange's Theorem.

Finally, we have the following corollary, whose proof is left as an exercise for the reader.