If $f:G\rightarrow G'$ is a homomorphism, then prove $f^{-1}(H')$, where $H' < G'$, is a subgroup of $G$

Question

If $f:G\rightarrow G'$ is a homomorphism, then prove $f^{-1}(H')$, where $H' < G'$, is a subgroup of $G$

Attempt: Let $K = f^{-1}(H')$. We want to prove that $K < G.$ Because $H'$ is a subgroup of $G'$, then $e_{G^{\,'}}\in H'\implies e_G\in K.$ For inverses, we have that becuase $H'<G'$, $g\in G'\implies g^{-1} \in G'\implies f^{-1}(g^{-1})=(f^{-1}(g))^{-1}\in K.$ For associativity, because $K \subseteq G,$ $K$ is associative. But how do you prove closure of $K$?

Answer 1

If $g_1,g_2 \in f^{-1}(H')$, then $f(g_1),f(g_2) \in H'$. Then $f(g_1 g_2) = f(g_1) f(g_2) \in H'$ as well, so $g_1 g_2 \in f^{-1}(H')$.

Answer 2

Let $g_1,g_2 \in f^{-1}(H)$ then $f(g_1),f(g_2) \in H $.

Since $f$ is homomorphism and $H$ is subgroup, $f(g_1)f(g_2)=f(g_1g_2)\in H$.

So $g_1g_2 \in f^{-1}(H)$.