To prove that $f(H)$ is a subgroup of $G_2$, we will check the three subgroup criteria:

1. Non-emptiness:

   • Since $H$ is a subgroup of $G_1$, the identity element $e_{G_1} \in H$.
   • By the property of homomorphism, $f(e_{G_1}) = e_{G_2}$, where $e_{G_2}$ is the identity element in $G_2$.
   • Thus, $e_{G_2} \in f(H)$, so $f(H)$ is non-empty.

2. Closure under group operation:

   • Let $x, y \in f(H)$. Then, there exist $h_1, h_2 \in H$ such that $x = f(h_1)$ and $y = f(h_2)$.
   • Since $H$ is a subgroup, $h_1h_2 \in H$.
   • By the property of homomorphism, $f(h_1h_2) = f(h_1)f(h_2)$.
   • Therefore, $xy = f(h_1)f(h_2) = f(h_1h_2)$, and $f(h_1h_2) \in f(H)$.
   • This shows $f(H)$ is closed under the group operation.

3. Closure under inverses:

   • Let $x \in f(H)$. Then, there exists $h \in H$ such that $x = f(h)$.
   • Since $H$ is a subgroup, $h^{-1} \in H$.
   • By the property of homomorphism, $f(h^{-1}) = (f(h))^{-1}$.
   • Therefore, $x^{-1} = (f(h))^{-1} = f(h^{-1})$, and $f(h^{-1}) \in f(H)$.
   • This shows $f(H)$ is closed under inverses.

Conclusion:

Since $f(H)$ satisfies all three criteria (non-emptiness, closure under group operation, and closure under inverses), it follows that $f(H)$ is a subgroup of $G_2$.

Would you like additional examples or further clarifications?
Here are 5 related questions:

1. What happens if $f$ is not a homomorphism?
2. Can you show that $f(H)$ is normal in $G_2$?
3. Is the kernel of $f$ always a subgroup of $G_1$?
4. Can this result generalize to infinite groups?
5. How does $f(H)$ relate to the image of $f$?

Tip: Always verify the properties of the mapping $f$ before proving subgroup-related properties.