7.3: The Index of a Subgroup and Lagrange's Theorem
Definition: Index
We define the index of \(H\) in \(G\), denoted \((G:H)\text{,}\) to be the cardinality of \(G/H\text{.}\)
Remark
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.
Example \(\PageIndex{1}\)
If \((\mathbb{R}:\mathbb{Z})\) were finite, then we'd be able to write \(\mathbb{R}\) as a finite union of countable sets, implying that \(\mathbb{R}\) is countable—which it isn't. Thus, \((\mathbb{R}:\mathbb{Z})=\infty\text{.}\)
Example \(\PageIndex{2}\)
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, \((C^*:C^*)=1\text{.}\)
Example \(\PageIndex{3}\)
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*} (\mathbb{Z}:5\mathbb{Z})=5, \end{equation*} \begin{equation*} (4\mathbb{Z}:12\mathbb{Z})=3, \end{equation*} \begin{equation*} \text{and }(\mathbb{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).
Theorem \(\PageIndex{1}\): Lagrange's Theorem
If \(G\) is a finite group and \(H\leq G\text{,}\) then \(|H|\) divides \(|G|\text{.}\)
Theorem \(\PageIndex{2}\)
If \(G\) is a finite group and \(H\leq G\text{,}\) then \((G:H)=|G|/|H|\text{.}\)
Remark
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.
Corollary \(\PageIndex{1}\)
Let \(G\) be a finite group and let \(a\in G\text{.}\) Then \(a^{|G|}=e_G\text{.}\)
- Proof
-
Let \(d=o(a)\). By Lagrange's Theorem, \(d\) divides \(|G|\), so there exists \(k∈ \mathbb{Z}\) with \(|G|=dk\). Then \(a^{|G|}=a^{dk}=(a^d)^k=(e_G)^k=e_G\).
Finally, we have the following corollary, whose proof is left as an exercise for the reader.
Corollary \(\PageIndex{2}\)
Let \(G\) be a group of prime order. Then \(G\) is cyclic. It follows that for every prime \(p\text{,}\) there exists a unique group of order \(p\text{,}\) up to isomorphism.