Given a group \(G\) with subgroup \(H\text{,}\) we define \(\siml\) on \(G\) by
\begin{equation*} a\siml b \text{ if and only if } a^{-1}b\in H \end{equation*}and \(\simr\) on \(G\) by
\begin{equation*} a\simr b \text{ if and only if } ab^{-1}\in H. \end{equation*}\(\siml\) and \(\simr\) are equivalence relations on \(G\text{.}\)
Of course, different subgroups \(H\) and \(K\) in a group \(G\) will give rise to different relations \(\siml\) and \(\simr\) on \(G\text{;}\) that is, these relations are really defined with respect to a particular subgroup of \(G\text{.}\)
From now on, whenever we discuss \(\siml\) or \(\simr\) on a group, assume that it is with respect to a particular subgroup \(H\) of \(G\text{.}\)
Now, as equivalence relations on a group \(G\text{,}\) each of \(\siml\) and \(\simr\) gives rise to a partition of \(G\text{.}\) What are the cells of those partitions?
Given \(a\in G\text{,}\) we define
\begin{equation*} aH = \{ah\,:\, h\in H\} \end{equation*}and
\begin{equation*} Ha=\{ha\,:\,h\in H\}. \end{equation*}We call \(aH\) and \(Ha\text{,}\) respectively, the left and right cosets of \(H\) containing \(a\).
If we know that \(G\) is abelian, with operation denoted by \(+\text{,}\) we may denote these left and right cosets by \(a+H\) and \(H+a\text{,}\) respectively.
In the following, we use the notation \(\Leftrightarrow\) to denote the phrase “if and only if.”
Let \(a\in G\text{.}\) Then under \(\siml\text{,}\) \([a]=aH\) while under \(\simr\text{,}\) \([a]=Ha\text{.}\)
We next summarize some facts about the left and right cosets of a subgroup \(H\) of a group \(G\text{:}\)
Let \(G\) be a group with \(H\leq G\) and \(a,b\in G\text{.}\)
The left [right] cosets of \(H\) in \(G\) partition \(G\text{.}\)
and
\begin{equation*} b\in Ha \Leftrightarrow Ha=Hb \Leftrightarrow a\in Hb. \end{equation*}In particular,
\begin{equation*} a\in H \Leftrightarrow aH=H \Leftrightarrow Ha=H. \end{equation*}\(H\) is the only left or right coset of \(H\) that is a subgroup of \(G\text{.}\)
\(|aH|=|H|=|Ha|\text{.}\)
We can use Statements 2 and 3, above, to save some time when computing left and right cosets of a subgroup of a group.
Find the left and right cosets of \(H=\langle (12)\rangle\) in \(S_3\text{.}\)
The left cosets are
\begin{equation*} eH=H=(12)H, \end{equation*} \begin{equation*} (13)H=\{(13),(123)\}=(123)H, \end{equation*} \begin{equation*} \text{ and } (23)H=\{(23),(132)\}=(132)H, \end{equation*}and the right cosets are
\begin{equation*} He=H=H(12), \end{equation*} \begin{equation*} H(13)=\{(13),(132)\}=H(132), \end{equation*} \begin{equation*} \text{ and } H(23)=\{(23),(123)\}=H(123). \end{equation*}Thus, \(\siml\) partitions \(S_3\) into \(\{H,\{(13),(123)\},\{(23), (132)\}\}\) and \(\simr\) partitions \(S_3\) into \(\{H,\{(13),(132)\},\{(23), (123)\}\}\text{.}\)
Find the left and right cosets of \(H=\langle f\rangle\) in \(D_4\text{.}\)
This example is left as an exercise for the reader.
We now draw attention to some very important facts:
For \(a,b\in G\text{:}\)
In general, \(aH \neq Ha\text{!}\)
\(aH=bH\) does not necessarily imply \(a=b\) or that there exists an \(h\in H\) with \(ah=bh\text{;}\) similarly, \(Ha=Hb\) does not necessarily imply \(a=b\) or that there exists an \(h\in H\) with \(ha=hb\text{.}\)
We saw above that in \(S_3\) with \(H=\langle (12)\rangle\text{,}\)
\begin{equation*} (13)H=\{(13),(123)\} \neq \{(13),(132)\}=H(13). \end{equation*}Also, \((13)H=(123)H\) but \((13)e\neq (123)e\) and \((13)(12)\neq (123)(12)\text{.}\)
It turns out that subgroups \(H\) for which \(aH=Ha\) for all \(a\in G\) will be very important to us.
We say that subgroup \(H\) of \(G\) is normal in \(G\) (or is normal subgroup of \(G\)) if \(aH=Ha\) for all \(a\in G\text{.}\) We denote that fact that \(H\) is normal in \(G\) by writing \(H\unlhd G\text{.}\)
If \(H\) is normal in \(G\text{,}\) we may refer to the left and right cosets of \(G\) as simply cosets.
Of course, if \(G\) is abelian, every subgroup of \(G\) is normal in \(G\text{.}\) But there can also be normal subgroups of nonabelian groups: for instance, the trivial and improper subgroups of every group are normal in that group.
Find the cosets of \(5\Z\) in \(\Z\text{.}\)
Notice that in additive notation, the statement “\(a^{-1}b\in H\)” becomes \(-a+b\in H\text{.}\) So for \(a,b\in \Z\text{,}\) \(a\siml b\) if and only if \(-a+b \in 5\Z\text{;}\) that is, if and only if \(5\) divides \(b-a\text{.}\) In other words, \(a\siml b\) if and only if \(a\equiv_5 b\text{.}\) So in this case, \(\siml\) is just congruence modulo \(5\text{.}\) Thus, the cosets of \(5\Z\) in \(\Z\) are
\begin{align*} 5\Z\amp =\{\ldots,-5,0,5,\ldots\}\\ 1+5\Z\amp =\{\ldots,-4, 1, 6,\ldots\},\\ 2+5\Z\amp =\{\ldots,-3,2, 7, \ldots\},\\ 3+5\Z\amp =\{\ldots,-2,3, 8, \ldots\},\\ 4+5\Z\amp =\{\ldots,-1, 4, 9, \ldots\}. \end{align*}Do you see how this example would generalize for \(n\Z\) (\(n \in \Z^+\)) in \(\Z\text{?}\)
Find the cosets of \(H=\langle 12\rangle\) in \(4\Z\text{.}\)
They are
\begin{align*} H\amp =\{\ldots, -12,0,12\ldots\},\\ 4+H \amp = \{\ldots,-8,4,16,\ldots\},\\ 8+H\amp =\{\ldots, -4,8,20,\ldots\}. \end{align*}Find the cosets of \(H=\langle 4\rangle\) in \(\Z_{12}\text{.}\)
They are
\begin{align*} H\amp =\{0,4,8\},\\ 1+H \amp = \{1,5,9\},\\ 2+H\amp =\{2,6,10\}\\ 3+H\amp =\{3,7,11\}. \end{align*}We now consider the set of all left cosets of a subgroup of a group.
We let \(G/H\) be the set of all left cosets of subgroup \(H\) in \(G\text{.}\) We read \(G/H\) as “\(G\) mod \(H\text{.}\)”
(We may denote the set of all right cosets of subgroup \(H\) in \(G\) by \(H\backslash G\text{,}\) but we will not use that notation in this class.)