2.11 Cosets and Lagrange’s Theorem
In this section, \(G\) is a group with identity element \(e\) and \(H \leq G\) is a subgroup.
A left coset of \(H\) is a subset of \(G\) of the from \(gH\). A right coset of \(H\) is a subset of the form \(Hg\). We write \(G/H\) for the set of all left cosets of \(G\), and we write \(H\backslash G\) for the set of all right cosets of \(G\). An element of a left (or right) coset is called a coset representative.
Note that \(H = eH = He\) is always both a left and right coset of \(H\). We call it the trivial coset.
For \(g_1\), \(g_2\in G\), write \(g_1L g_2\) if \(g_1\in g_2H\). Then \(g_1Lg_2 \Leftrightarrow g_1^{-1}g_2\in H\). Moreover, \(L\) is an equivalence relation on \(G\) and, for each \(g\in G\), \([g]_L = gH\).
Pick \(g_1\) and \(g_2\in G\). Then \(g_1Lg_2 \Leftrightarrow g_1 \in g_2 H \Leftrightarrow g_1 = g_2 h\) for some \(h\in H \Leftrightarrow g_1^{-1}g_2 = h\) for some \(h\in H \Leftrightarrow g_1^{-1}g_2\in H\).
To see that \(L\) is an equivalence relation, first note that, as \(H\leq G\), for all \(g\in G\), \(e = g^{-1}g \in H\). This shows that \(L\) is reflexive.
To see that \(L\) is symmetric, suppose \(g_1Lg_2\). Then \(g_1^{-1}g_2\in H\). But this implies that \(g_2^{-1}g_1 = (g_1^{-1}g_2)^{-1}\in H\) as well. So \(g_1Lg_1\).
To see that \(L\) is transitive, suppose that \(g_1Lg_2\) and \(g_2Lg_3\). Then \(g_1^{-1}g_2\) and \(g_2^{-1}g_3\) are both in \(H\). But this implies that \(g_1^{-1}g_3 = (g_1^{-1}g_2)(g_2^{-1}g_3) \in H\).
Given this, by definition \([g]_L = \{ x\in G: xLg\} = \{ x\in G: x\in gH\} = gH\).
The set \(G/H\) of left cosets of \(H\) form a partition of \(G\).
Obvious from Lemma 2.85 and the fact that equivalence classes of an equivalence relation form a partition.
The index \([G:H]\) of \(H\) in \(G\) is the number of left cosets of \(H\) in \(G\). We write \([G:H] = \infty \) if that number is infinite.
Suppose \(g\in G\). Then the map \(f:H\to gH\) given by \(f(h) = gh\) sets up a one-one correspondence between \(H\) and the left-coset \(gH\).
Trivial.
Every left coset \(gH\) of \(H\) has the same cardinality. More precisely, for any \(g\in G\), we have \(|gH| = |H|\).
Sets in one-one correspendence have the same cardinality by definition.
We have \(|G| = [G:H] |H|\).
Let’s prove this when \(G\) is finite. (The general proof requires explaining what we mean by the product of infinite sets, but is essentially the same.)
The set \(G/H\) is a partition of \(G\) into \([G:H]\) subsets each having cardinality \(|H|\).
Suppose \(|G| {\lt}\infty \). Then
The order \(|H|\) of any subgroup \(H\leq G\) divides \(|G|\).
The order \(|g|\) of any element \(g\in G\) divides \(|G|\).
2 Suppose \(g\in G\) and set \(H = \langle g\rangle \). If \(|g| = \infty \), then all the elements \(g^i\) for \(i\in \mathbb {Z}\) are distinct by Theorem 2.551. So \(H\) is infinite, which contradicts the assumption that \(G\) is finite.
So we can assume that \(|g| = d {\lt}\infty \), and then \(|H| = d\) by Theorem 2.552. So \(d\) divides \(|H|\) by 1.
Any group of prime order is cyclic. In fact, if \(G\) is a group of prime order than \(G\) is generated by any nonidentity element.
Suppose \(|G| = p\) is a prime number. Then \(|G| {\gt} 1\). So there exists a nonidentity element \(g\in G\). So \(|g| {\gt} 1\), but \(|g| | p\). It follows that \(|g| = p\). So \(|\langle g\rangle | = p\). So \(G\) is cyclic.