Section7.2Introduction to cosets and normal subgroups
Definition7.2.1
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*}
Theorem7.2.2
\(\siml\) and \(\simr\) are equivalence relations on \(G\text{.}\)
Proof
First, let \(a\in G\text{.}\) Then \(a^{1}a=e\in H\text{,}\) so \(a\siml a\text{.}\) Thus, \(\siml\) is reflexive.
Next, let \(a,b\in G\) with \(a\siml b\text{.}\) Then \(a^{1}b\in H\text{,}\) so, since \(H\) is a subgroup of \(G\text{,}\) \((a^{1}b)^{1}\in H\text{.}\) But \((a^{1}b)^{1}=b^{1}(a^{1})^{1}=b^{1}a\text{;}\) thus, \(b\siml a\text{,}\) and so \(\siml\) is symmetric.
Finally, let \(a,b,c\in G\) with \(a\siml b\) and \(b\siml c\text{.}\) Then \(a^{1}b\) and \(b^{1}c\) are in \(H\text{.}\) Since \(H\) is a subgroup of \(G\text{,}\) we must then have \((a^{1}b)(b^{1}c)\in H\text{;}\) but \((a^{1}b)(b^{1}c)\) equals \(a^{1}c\text{.}\) Thus, \(a\siml c\text{,}\) and so \(\siml\) is transitive.
Thus, \(\siml\) is an equivalence relation on \(G\text{.}\) The proof that \(\simr\) is an equivalence relation is left as an exercise for the reader.
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?
Definition7.2.4
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.
Theorem7.2.6
Let \(a\in G\text{.}\) Then under \(\siml\text{,}\) \([a]=aH\) while under \(\simr\text{,}\) \([a]=Ha\text{.}\)
Proof
Let \(b\in G\text{.}\) Then \(b\siml a \Leftrightarrow a \siml b
\Leftrightarrow a^{1}b\in H \Leftrightarrow a^{1}b=h\) for some \(h\in H \Leftrightarrow b=ah\) for some \(h\in H \Leftrightarrow b\in
aH\text{.}\) So under \(\siml\) we have \([a]=aH\text{.}\) Similarly, under \(\simr\) we have \([a]=Ha\text{.}\)
We next summarize some facts about the left and right cosets of a subgroup \(H\) of a group \(G\text{:}\)
Theorem7.2.7
Let \(G\) be a group with \(H\leq G\) and \(a,b\in G\text{.}\)
The left [right] cosets of \(H\) in \(G\) partition \(G\text{.}\)

\begin{equation*}
b\in aH \Leftrightarrow aH=bH \Leftrightarrow a\in bH
\end{equation*}
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 \(G\text{.}\)
\(aH=H=Ha\text{.}\)
Proof
Statements 1 and 2 hold because the left and right cosets of \(H\) in \(G\) are equivalence classes. Statement 3 holds because no left or right coset of \(H\) other than \(H\) itself can contain \(e\text{,}\) since the left [right] cosets of \(H\) are mutually disjoint. For Statement 4: Define \(f\,:\,H\to aH\) by \(f(h)=ah\text{.}\) It is straightforward to show that \(f\) is a bijection, so \(H=aH\text{.}\) Similarly, \(Ha=H\text{.}\)
Example7.2.9
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 \(\{H,\{(13),(132)\},\{(23), (123)\}\}\text{.}\)
Example7.2.10
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:
Example7.2.12
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.
Definition7.2.13
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 \(H\unlhd G\text{.}\)
Example7.2.15
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 \(ba\text{.}\) 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 \(\Z\text{?}\)
Example7.2.16
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*}
Example7.2.17
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.
Definition7.2.18
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.)