3.9 Basic Isomorphism Theorem

Theorem 3.43

Suppose \(\pi :G\to Q\) is a surjective group homomorphism and \(K\) is another group. Then \(\pi ^*(\operatorname{\mathrm{Hom}}(Q,K)) = \{ f:G\to K: \ker \pi \leq \ker f\} \).

Proof ▼

Suppose \(f\in \pi ^*\operatorname{\mathrm{Hom}}(Q,K)\). Then \(f = \pi ^*\psi = \psi \circ \pi \) for some \(\psi \in \operatorname{\mathrm{Hom}}(Q,K)\). So, if \(g\in \ker \pi \), then \(f(g) = \psi (\pi (g)) = \psi (e) = e\). Therefore, \(\ker \pi \leq \ker f\).

On the other hand, suppose \(f\in \operatorname{\mathrm{Hom}}(G,K)\) and \(\ker \pi \leq \ker f\). Set \(\Gamma = \{ (\pi (g), f(g))\in Q\times K: g\in G\} \).

I claim that \(\Gamma \) is the graph of a function \(\psi :Q\to K\). To see this, it suffices to show that, for every \(q\in Q\), there exists exactly one \(k\in K\) such that \((q,k)\in \Gamma \). So suppose \(q\in Q\). Since \(\pi :G\to Q\) is surjective, we can find \(g\in G\) such that \(\pi (g) = q\) and then \((q, f(g))\in K\). On the other hand, suppose \(g'\) is any element of \(G\) such that \(\pi (g') = q\). Then \(\pi (g^{-1}g') = e\). So \(g^{-1}g'\in \ker \pi \). Therefore, \(g^{-1}g'\in \ker f\). So \(f(g) = f(g')\), and it follows that \((q,k)\in \Gamma \) if and only if \(k = f(g)\). So we get a map \(\psi :Q\to K\) with \(\psi (\pi (g)) = f(g)\) for all \(g\in G\).

I claim that \(\psi \in \operatorname{\mathrm{Hom}}(Q,K)\). To see that, suppose \(q_1,q_2\in Q\) and find \(g_1,g_2\in G\) such that \(\pi (g_i) = q_i\). Then \(\psi (q_1q_2) = \psi (\pi (g_1)\pi (g_2)) = \psi (\pi (g_1g_2)) = f(g_1g_2) = f(g_1)f(g_2) = \psi (\pi (g_1))\psi (\pi (g_2)) = \psi (q_1)\psi (q_2)\). So \(\psi :Q\to K\) is a group homomorphism.

Corollary 3.44

Suppose \(\pi :G\to Q\) is a surjective group homomorphism and \(K\) is another group. Let \(f:G\to K\) be a group homomorphism with \(\ker \pi \leq \ker f\). Then there exists a unique group homomorphism \(\psi :Q\to K\) such that \(f = \psi \circ \pi \). In other words there exists a unique group homomorphism \(\psi \in \operatorname{\mathrm{Hom}}(Q,K)\) making the diagram

\begin{equation} \begin{tikzcd} \label{c-isocomm} G \ar[d, two heads, swap, "\pi"]\ar[r, "f"] & K \\ Q \ar[ru, dashed, swap, "\psi"] & \end{tikzcd} \end{equation}

3.45

commute.

Proof ▼

This is just Theorem 3.44 plus the injectivity of \(\pi ^*\), which follows from Proposition 3.39. To say that the “diagram commutes" just means that \(f = \psi \circ \pi \).

Proposition 3.46

Suppose \(\pi :G\to Q\) is a group homomorphism and \(K\) is another group. Let \(f:G\to K\) be a group homomorphism with \(\ker \pi \leq \ker f\), and let \(\psi :Q\to K\) be the homomorphism from Corollary 3.44 such that \(f = \psi \circ \pi \).

\(f(G) = \psi (Q)\). In particular, if \(f:G\to K\) is onto, then so is \(\psi :Q\to K\).
\(\ker \psi = \pi (\ker f)\).

Proof ▼

1 Suppose \(k = f(g)\in f(G)\). Then \(k = f(g) = \psi (\pi (g))\). So \(k\in \psi (Q)\). Therefore, \(f(G) \leq \psi (Q)\). On the other hand, if \(k = \psi (q)\) for some \(q\in Q\), then, since \(\pi \) is onto, we can find \(g\in G\) such that \(\pi (g) = q\). Then \(k = \psi (\pi (g)) = f(g) \in f(G)\). So \(\psi (Q)\leq f(G)\).

2 Suppose \(q\in \ker \psi \). Write \(q = \pi (g)\), which we can do since \(\pi \) is onto. Then \(e = \psi (q) = \psi (\pi (g)) = f(g)\). So \(g\in \ker f\). So \(q = \pi (g) \in \pi (\ker f)\). Therefore, \(\ker \psi \leq \pi (\ker f)\).

On the other hand, suppose \(q\in \pi (\ker f)\). So \(q = \pi (g)\) with \(f(g) = e\). Then \(\psi (q) = \psi (\pi (g)) = f(g) = e\). So \(q\in \ker \psi \).

Corollary 3.47

Suppose \(\phi :G\to Q\) is a group homomorphism with kernel \(N = \ker \phi \). Let \(\pi _N:G\to G/N\) be the projection homomorphisms given by \(g\mapsto gN\). Then there is a unique group homomorphism \(\psi :G/N\to Q\) such that \(\psi \circ \pi = \phi \). Moreover

\(\psi :G/N\to Q\) is one-one.
If \(\phi \) is onto, then \(\psi :G/N\xrightarrow {\sim } Q\) is an isomorphism.

Proof ▼

Since \(\ker \pi _N = \ker \phi \), the existence of \(\psi \) follows from Corollary 3.44. By Proposition 2, we have \(\ker \psi = \pi (\ker \phi ) = \pi (\ker \pi ) = e\). So \(\psi \) is one-one proving 1. Then 2 follows from Proposition 1.

Corollary 3.48

Suppose \(G = \langle g\rangle \) is a cyclic group, and let \(f:\mathbb {Z}\to G\) be the group homomorphism such that \(f(1) = g\) (Theorem 3.16).

If \(|g| = \infty \), then \(f:\mathbb {Z}\to G\) is an isomorphism.
If \(|g| = n {\lt} \infty \), then \(\ker f = n\mathbb {Z}\), and there is a unique group homomorphism \(\psi :\mathbb {Z}/n\mathbb {Z}\to G\) such that \(f = \psi \circ \pi \), where \(\pi :\mathbb {Z}\to \mathbb {Z}/n\mathbb {Z}\) is the quotient homomorphism. Moreover, \(\psi \) is an isomorphism. In particular, \(G\cong \mathbb {Z}/n\mathbb {Z}\).

Proof ▼

1 This follows easily from Theorem 2.52 and is left to the reader.

2 Suppose \(|g| = n\). Then, with the notation of Proposition 2.51, \(n\mathbb {Z} = K(g) = \ker f\). Since \(f:\mathbb {Z}\to G\) is clearly onto, the rest follows from Corollary 3.47 2.