Homomorphic mapping of subgroup of domain is subgroup of codomain

Dependencies:

Let $H_1$ and $H_2$ be subgroups of $G_1$ and $G_2$. Let $\phi$ be a homomorphism from $G_1$ to $G_2$.

Then

$\phi(H_1)$ is a subgroup of $G_2$.
$\phi^{-1}(H_2) = \{g_1: \phi(g_1) \in H_2\}$ is a subgroup of $G_1$.

Proof of part 1

$\phi(H_1)$ is a non-empty subset of $G_2$.

Let $\phi(a), \phi(b) \in \phi(H_1)$, where $a, b \in H_1$. Then \[ \phi(a)\phi(b)^{-1} = \phi(a)\phi(b^{-1}) = \phi(ab^{-1}) \in \phi(H_1) \]

Therefore, $H_1$ is a subgroup of $G_2$.

Proof of part 2

Since $H_2$ is non-empty, $\phi^{-1}(H_2)$ is a non-empty subset of $G_1$.

\begin{align} & a, b \in \phi^{-1}(H_2) \\ &\Rightarrow \phi(a), \phi(b) \in H_2 \\ &\Rightarrow \phi(a)\phi(b)^{-1} \in H_2 \\ &\Rightarrow \phi(a)\phi(b^{-1}) \in H_2 \\ &\Rightarrow \phi(ab^{-1}) \in H_2 \\ &\Rightarrow ab^{-1} \in \phi^{-1}(H_2) \end{align}

Therefore, $\phi^{-1}(H_2)$ is a subgroup of $G_1$.

Dependency for:

Homomorphic mapping and inverse mapping of normal subgroup is normal

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