3.7 Quotients by normal subgroups

In this section \(G\) is a group with identity element \(1\) and \(N\) is a normal subgroup of \(G\).

Lemma 3.30

Suppose \(M\) is a magma and \(f:G\to M\) is a surjective magma homomorphism from \(G\) to \(M\). Then \(M\) is a group and \(f\) is a group homomorphism. Moreover,

The identity element of \(M\) is \(f(1)\).
If \(m = f(g)\in M\), then \(m^{-1} = f(g^{-1})\).

Proof ▼

We need to show that \(M\) is associative, with identity \(f(1)\) and that every element of \(M\) is invertible.

To see that \(M\) is associative, pick \(m_1\), \(m_2\) and \(m_3\) in \(M\). Since \(f\) is onto, we can find \(g_1\), \(g_2\) and \(g_3\) in \(G\) such that \(f(g_i) = m_i\) (\(i = 1, 2, 3\)). Then

\begin{align*} m_1(m_2m_3) & = f(g_1)(f(g_2)f(g_3)) = f(g_1)f(g_2g_3)\\ & = f(g_1(g_2g_3)) = f((g_1g_2)g_3) \\ & = f(g_1g_2)f(g_3) = (f(g_1)f(g_2))f(g_3) \\ & = (m_1m_2)m_3. \end{align*}

To see that \(f(1)\) is the identity element of \(M\), pick \(m\in M\) and, using the assumption that \(f\) is onto, pick \(g\in G\) such that \(f(g) = m\). Then \(f(1)m = f(1)f(g) = f(1g) = f(g) = m\), and, similarly, \(mf(1) = m\).

To see that \(M = M^{\times }\), pick \(m\in M\) and write \(m = f(g)\). Then \(f(g^{-1})m = f(g^{-1})f(g) = f(g^{-1}g) = f(1) = f(g g^{-1}) = f(g) f(g^{-1}) = m f^(g^{-1})\). So \(m\) is invertible, with \(m^{-1} = f(g^{-1})\).

Recall from Proposition 2.34, that, if \(M\) is a monoid with identity element \(1\), then the power set \(\mathcal{P}(M)\) is also a monoid with identity element \(\{ 1\} \).

Lemma 3.31

For \(g_1N\), \(g_2N\in G/N\), we have \((g_1N)(g_2N) = g_1g_2 N\). Consequently, the set \(G/N\) is a submagma of \(\mathcal{P}(G)\).

Proof ▼

Suppose \(g_1N\) and \(g_2N\) are elements of \(G/N\). Then, since \(N\unlhd G\), \((g_1N)(g_2N) = g_1 (Ng_2)N = g_1(g_2N) N = (g_1g_2) NN = g_1g_2 N\). So \(G/N\) is closed under the binary operation on \(\mathcal{P}(G)\).

Warning 3.32

Note that Lemma 3.31 does not say that \(G/N\) is a submonoid of \(\mathcal{P}(M)\). In fact, this is true if and only if \(N = \{ 1\} \) is the trivial normal subgroup. The reason is that \(G/N\) contains the identity element \(\{ 1\} \) of \(\mathcal{P}(M)\) if and only if \(N = \{ 1\} \).

Lemma 3.33

The map \(\pi _M:G\to G/N\) given by \(g\mapsto gN\) is a surjective magma homomorhism.

Proof ▼

The map \(\pi _M\) is obviously surjective. To see that it’s a magma homomorphism, compute

\begin{align*} \pi _M(g_1)\pi _M(g_2) & = (g_1N)(g_2N) \\ & = (g_1g_2)N = \pi _M(g_1g_2). \end{align*}

Theorem 3.34

Suppose \(N\unlhd G\). Then \(G/N\) has a group structure with the binary operation given by

\begin{equation} \label{e:quotbinop} (g_1N)(g_2N) = (g_1g_2)N. \end{equation}

3.35

Moreover, the map \(\pi _N:G\to G/N\) given by \(g\mapsto gN\) is a surjective group homomorphism with kernel \(N\).

Proof ▼

Combining Lemma 3.30 with Lemmas 3.33 shows that \(G/N\) is a group and \(\pi _M:G\to G/N\) is a surjective group homomorphism. The formula 3.35 is just taken from Lemma 3.31.

It only remains to show that \(\ker \pi _M = N\). For this, note that, as \(N = 1N\), \(gN = N\) iff \(g = \in N\) (by Lemma 2.85).