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


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


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.


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, \((\R:\Z)=\infty\text{.}\)


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, \((C^*:C^*)=1\text{.}\)


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).


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.