Problem

Let \(\varphi : G \rightarrow H\) be a homomorphism and let \(E\) be a subgroup of \(H\).

1. Prove that \(\varphi^{-1}(E) \le G\) (i.e., the preimage or pullback of a subgroup under a homomorphism is a subgroup).
2. If \(E \trianglelefteq H\), prove that \(\varphi^{-1}(E) \trianglelefteq G\).
3. Deduce that \(\ker \varphi \trianglelefteq G\).

Solution

(1)

\(\varphi\) is a homomorphism, so \(\varphi(1) = 1 \in E \le H\). Thus \(\varphi^{-1}(E) \ne \emptyset\) because \(1 \in \varphi^{-1}(E)\). Also, \(\varphi^{-1}(E) \subseteq G\).

Let \(x, y \in \varphi^{-1}(E)\). Then there exist \(a, b \in E\) such that \(\varphi(x) = a\) and \(\varphi(y) = b\). Given that \(\varphi\) is a homomorphism, that \(a,b \in E\), and that \(E\) is a group,

$$ \varphi(xy^{-1}) = \varphi(x)\varphi(y^{-1}) = \varphi(x)\varphi^{-1}(y) = ab^{-1} \in E. $$

Therefore, \(xy^{-1} \in \varphi^{-1}(E)\). Hence, by the Subgroup Criterion, \(\varphi^{-1}(E) \le G\).

(2)

Suppose \(E \trianglelefteq H\). That is, \(hEh^{-1} = E\) for all \(h \in H\).

Let \(g\) be an arbitrary element of \(G\), and \(x\) be an arbitrary element of \(\varphi^{-1}(E)\). Then there exists \(h_1 \in H\) such that \(\varphi(g) = h_1\), and \(a \in E\) such that \(\varphi(x) = a\). Given that \(\varphi\) is a homomorphism and that \(E \trianglelefteq H\),

$$ \varphi(gxg^{-1}) = \varphi(g)\varphi(x)\varphi(g^{-1}) = \varphi(g)\varphi(x)\varphi^{-1}(g)= h_1ah_1^{-1} \in E. $$

Therefore, for all \(g \in G\) and \(x \in \varphi^{-1}(E)\), \(gxg^{-1} \in \varphi^{-1}(E)\). In other words, \(g\varphi^{-1}(E)g^{-1} \subseteq \varphi^{-1}(E)\) for all \(g \in G\).

Hence, by Theorem 6, \(\varphi^{-1}(E) \trianglelefteq G\).

(3)

Note \(1 \le H\). By definition of kernel and what we showed above, \(\ker\varphi = \varphi^{-1}(1) \trianglelefteq G\).

$$\tag*{$\blacksquare$}$$