1.6: Cosets

For a subset \(S\) of a group \(G\) and an element \(a\) of \(G\), we let

\[\begin{aligned} aS & =\{as\mid s\in S\}\\ Sa & =\{sa\mid s\in S\}.\end{aligned} \nonumber \]

Because of the associativity law, \(a(bS)=(ab)S\,\), and so we can denote this set unambiguously by \(abS.\)

When \(H\) is a subgroup of \(G\), the sets of the form \(aH\) are called the left cosets of \(H\) in \(G\), and the sets of the form \(Ha\) are called the right cosets of \(H\) in \(G\). Because \(e\in H\), \(aH=H\) if and only if \(a\in H\).

[bd13]Let \(G=(\mathbb{R}^{2},+)\), and let \(H\) be a subspace of dimension \(1\) (line through the origin). Then the cosets (left or right) of \(H\) are the lines \(a+H\) parallel to \(H\).

[bd14] Let \(H\) be a subgroup of a group \(G\).

(a) An element \(a\) of \(G\) lies in a left coset \(C\) of \(H\) if and only if \(C=aH.\)

(b) Two left cosets are either disjoint or equal.

(c) \(aH=bH\) if and only if \(a^{-1}b\in H.\)

(d) Any two left cosets have the same number of elements (possibly infinite).

(a) Certainly \(a\in aH\). Conversely, if \(a\) lies in the left coset \(bH\), then \(a=bh\) for some \(h\), and so

\[aH=bhH=bH. \nonumber \]

(b) If \(C\) and \(C^{\prime}\) are not disjoint, then they have a common element \(a\), and \(C=aH\) and \(C^{\prime}=aH\) by (a).

(c) If \(a^{-1}b\in H\), then \(H=a^{-1}bH\), and so \(aH=aa^{-1}bH=bH\). Conversely, if \(aH=bH\), then \(H=a^{-1}bH\), and so \(a^{-1}b\in H\).

(d) The map \((ba^{-1})_{L}\colon ah\mapsto bh\) is a bijection \(aH\rightarrow bH.\)

The index \((G:H)\) of \(H\) in \(G\) is defined to be the number of left cosets of \(H\) in \(G\).⁷ For example, \((G:1)\) is the order of \(G\).

As the left cosets of \(H\) in \(G\) cover \(G\), ([bd14]b) shows that they form a partition \(G\). In other words, the condition “\(a\) and \(b\) lie in the same left coset” is an equivalence relation on \(G\).

[Lagrange][bd15] If \(G\) is finite, then

\[(G:1)=(G:H)(H:1). \nonumber \]

In particular, the order of every subgroup of a finite group divides the order of the group.

The left cosets of \(H\) in \(G\) form a partition of \(G\), there are \((G:H)\) of them, and each left coset has \((H:1)\) elements.

[bd16] The order of each element of a finite group divides the order of the group.

Apply Lagrange’s theorem to \(H=\langle g\rangle\), recalling that \((H:1)=\text{order}(g)\).

[bd17] If \(G\) has order \(p\), a prime, then every element of \(G\) has order \(1\) or \(p\). But only \(e\) has order \(1\), and so \(G\) is generated by any element \(a\neq e\). In particular, \(G\) is cyclic and so \(G\approx C_{p}\). This shows, for example, that, up to isomorphism, there is only one group of order \(1,000,000,007\) (because this number is prime). In fact there are only two groups of order \(1,000,000,014,000,000,049\) (see [ga16]).

[bd18]For a subset \(S\) of \(G\), let \(S^{-1}=\{g^{-1}\mid g\in S\}\). Then \((aH)^{-1}\) is the right coset \(Ha^{-1}\), and \((Ha)^{-1}=a^{-1}H\). Therefore \(S\mapsto S^{-1}\) defines a one-to-one correspondence between the set of left cosets and the set of right cosets under which \(aH\leftrightarrow Ha^{-1}\). Hence \((G:H)\) is also the number of right cosets of \(H\) in \(G.\) But, in general, a left coset will not be a right coset (see [bd22] below).

[bd18p] Lagrange’s theorem has a partial converse: if a prime \(p\) divides \(m=(G:1)\), then \(G\) has an element of order \(p\) (Cauchy’s theorem [ga13]); if a prime power \(p^{n}\) divides \(m\), then \(G\) has a subgroup of order \(p^{n}\) (Sylow’s theorem [st2]). However, note that the \(4\)-group \(C_{2}\times C_{2}\) has order \(4\), but has no element of order \(4\), and \(A_{4}\) has order \(12\), but has no subgroup of order \(6\) (see Exercise [x31]).

More generally, we have the following result.

[bd19] For any subgroups \(H\supset K\) of \(G\),

\[(G:K)=(G:H)(H:K) \nonumber \]

(meaning either both are infinite or both are finite and equal).

Write \(G=\bigsqcup\nolimits_{i\in I}g_{i}H\) (disjoint union), and \(H=\bigsqcup_{j\in J}h_{j}K\) (disjoint union). On multiplying the second equality by \(g_{i}\), we find that \(g_{i}H=\bigsqcup_{j\in J}g_{i}h_{j}K\) (disjoint union), and so \(G=\bigsqcup_{i,j\in I\times J}g_{i}h_{j}K\) (disjoint union). This shows that

\[(G:K)=|I||J|=(G:H)(H:K). \nonumber \]