Prove that if $\theta$ is a homomorphism of $G$ onto $H$, $B \triangleleft H$, and $A=\{g \in G: \theta (g) \in B\}$, then $A \triangleleft G$

Question

Prove that if $\theta$ is a homomorphism of $G$ onto $H$, $B \triangleleft H$, and $A=\{g \in G: \theta (g) \in B\}$, then $A \triangleleft G$

So $ \theta :G \to H$ is a homomorphism.
If $B \triangleleft H$, then $A= \{g \in G : \theta(g) \in B\} \triangleleft G$ So $H \triangleleft G,$ for every $a \in G, Ha=aH$ $\iff$ for every $a \in G, h \in H, aha^{-1} \in H$

So this is what I know but I don't know what to do with it next

Answer 1

I think here it's easier to work with the element-wise definition of a normal subgroup: $$ A \lhd G :\Leftrightarrow \forall a\in A, g\in G\colon gag^{-1}\in A. $$

So let $a\in A$ and $g\in G$. To verify that $gag^{-1}\in A$, by definition of $A$, we need to show that $\theta(gag^{-1})\in B$.

• Because $\theta$ is a homomorphism, we have $\theta(gag^{-1})=\theta(g)\theta(a){\theta(g)}^{-1}$
• Because $\theta(a)\in B$ (by definition of $A$), $\theta(g)\in H$ and $B\lhd H$, we have that $\theta(g)\theta(a){\theta(g)}^{-1}\in B$ as well
• Hence, $gag^{-1}\in A$

And we're done.

Answer 2

Let $a \in A$ and $g \in G$ then we have to check $ga g^{-1} \in A$. Using the definition we see $\theta(g a g^{-1}) = \theta(g)\theta(a) \theta(g)^{-1} \in B$ by normality of $B$.