7.2: Introduction to Cosets and Normal Subgroups

Theorem \(\PageIndex{1}\)

\(\sim_L\) and \(\sim_R\) are equivalence relations on \(G\text{.}\)

Proof

First, let \(a∈G\). Then \(a^{−1}a=e∈H\), so \(a \sim_La\). Thus, \(\sim_L\) is reflexive.

Next, let \(a,b∈G\) with \(a\sim_Lb\). Then \(a^{−1}b∈H\), so, since \(H\) is a subgroup of \(G\), \((a^{−1}b)^{−1}∈H\). But \((a^{−1}b)^{−1}=b^{−1}(a^{−1})^{−1}=b^{−1}a\); thus, \(b \sim_La\), and so \(\sim_L\) is symmetric.

Finally, let \(a,b,c∈G\) with \(a\sim_Lb\) and \(b\sim_Lc\). Then \(a^{−1}b\) and \(b^{−1}c\) are in \(H\). Since \(H\) is a subgroup of \(G\), we must then have \((a^{−1}b)(b^{−1}c)∈H\); but \((a^{−1}b)(b^{−1}c)\) equals \(a^{−1}c\). Thus, \(a\sim_Lc\), and so \(\sim_L\) is transitive.

Thus, \(\sim_L\) is an equivalence relation on \(G\). The proof that \(\sim_R\) is an equivalence relation is left as an exercise for the reader.

Remark

Of course, different subgroups \(H\) and \(K\) in a group \(G\) will give rise to different relations \(\sim_L\) and \(\sim_R\) 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 \(\sim_L\) or \(\sim_R\) 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 \(\sim_L\) and \(\sim_R\) gives rise to a partition of \(G\text{.}\) What are the cells of those partitions?

Definition: Left and Right Cosets

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.

Note

In the following, we use the notation \(\Leftrightarrow\) to denote the phrase “if and only if.”

Theorem \(\PageIndex{2}\)

Let \(a\in G\text{.}\) Then under \(\sim_L\text{,}\) \([a]=aH\) while under \(\sim_R\text{,}\) \([a]=Ha\text{.}\)

Proof: Let \(b∈G\). Then \(b \sim_La⇔a \sim_Lb⇔a^{−1}b∈H⇔a^{−1}b=h\) for some \(h∈H⇔b=ah\) for some \(h∈H⇔b∈aH\). So under \(\sim_L\) we have \([a]=aH\). Similarly, under \(\sim_R\) we have \([a]=Ha\).

We next summarize some facts about the left and right cosets of a subgroup \(H\) of a group \(G\text{:}\)

Theorem \(\PageIndex{3}\)

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 \(e\), since the left [right] cosets of \(H\) are mutually disjoint. For Statement 4: Define \(f:H→aH\) by \(f(h)=ah\). It is straightforward to show that \(f\) is a bijection, so \(|H|=|aH|\). Similarly, \(|Ha|=|H|\).

Remark

We can use Statements 2 and 3, above, to save some time when computing left and right cosets of a subgroup of a group.

Example \(\PageIndex{1}\)

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, \(\sim_L\) partitions \(S_3\) into \(\{H,\{(13),(123)\},\{(23), (132)\}\}\) and \(\sim_R\) partitions \(S_3\) into \(\{H,\{(13),(132)\},\{(23), (123)\}\}\text{.}\)

Example \(\PageIndex{2}\)

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:

Note

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 \(ha=hb\text{.}\)

Example \(\PageIndex{3}\)

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.

Definition: Normal Subgroup

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 \(H\unlhd G\text{.}\)

Remark

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.

Example \(\PageIndex{4}\)

Find the cosets of \(5\mathbb{Z}\) in \(\mathbb{Z}\text{.}\)

Notice that in additive notation, the statement “\(a^{-1}b\in H\)” becomes \(-a+b\in H\text{.}\) So for \(a,b\in \mathbb{Z}\text{,}\) \(a\sim_L b\) if and only if \(-a+b \in 5\mathbb{Z}\text{;}\) that is, if and only if \(5\) divides \(b-a\text{.}\) In other words, \(a\sim_L b\) if and only if \(a\equiv_5 b\text{.}\) So in this case, \(\sim_L\) is just congruence modulo \(5\text{.}\) Thus, the cosets of \(5\mathbb{Z}\) in \(\mathbb{Z}\) are

\begin{align*} 5\mathbb{Z}& =\{\ldots,-5,0,5,\ldots\}\\ 1+5\mathbb{Z}& =\{\ldots,-4, 1, 6,\ldots\},\\ 2+5\mathbb{Z}& =\{\ldots,-3,2, 7, \ldots\},\\ 3+5\mathbb{Z}& =\{\ldots,-2,3, 8, \ldots\},\\ 4+5\mathbb{Z}&=\{\ldots,-1, 4, 9, \ldots\}. \end{align*}

Do you see how this example would generalize for \(n\mathbb{Z}\) (\(n \in \mathbb{Z}^+\)) in \(\mathbb{Z}\text{?}\)

Example \(\PageIndex{5}\)

Find the cosets of \(H=\langle 12\rangle\) in \(4\mathbb{Z}\text{.}\)

They are

\begin{align*} H& =\{\ldots, -12,0,12\ldots\},\\ 4+H & = \{\ldots,-8,4,16,\ldots\},\\ 8+H& =\{\ldots, -4,8,20,\ldots\}. \end{align*}

Example \(\PageIndex{6}\)

Find the cosets of \(H=\langle 4\rangle\) in \(\mathbb{Z}_{12}\text{.}\)

They are

\begin{align*} H&=\{0,4,8\},\\ 1+H & = \{1,5,9\},\\ 2+H& =\{2,6,10\}\\ 3+H& =\{3,7,11\}. \end{align*}

We now consider the set of all left cosets of a subgroup of a group.

Definition: G mod H

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.)