$K \lhd G, \space \space G/K \simeq H_1$, and $K \simeq H_2$ Let $H_1$ and $H_2$ be groups. Define $G = H_1 \times H_2$ and $K = \{(1,h)\space | \space h \in H_2\}$
I'm trying to show that $K \lhd G, \space \space G/K \simeq H_1$, and $K \simeq H_2$
So a normal subgroup is a subgroup that is invariant under conjugation by members of the group.
It is clear that this problem relies on relating $K$ to $G$, but I'm having the hardest time visualizing $K$. If people have thoughts or hints on how to think about G and K (even without relating the two), I can take a crack at solving it and update the problem with my work.
 A: $G = H_1 \times H_2$
Consider a map  $\phi : H_1 \times H_2 \to H_1$ defined by
$$\phi (h_1, h_2) =h_1$$
Then $\phi$ is a onto homomorphism (check!) from
$H_1 \times H_2 \to H_1$.
\begin{align}Ker(\phi) &=\{(h_1, h_2) :\phi(h_1, h_2)=e_{H_1}\}\\
&=\{(e_{H_1},h):h\in H_2\} \\
&=K\end{align}
Since,$\phi : H_1 \times H_2 \to H_1$ is an onto homomorphism with $Ker(\phi) =K$.

*

*Hence, $K$ is a normal subgroup of $G= H_1 \times H_2$


*Then, by first Isomorphism Theorem, $G/K \simeq H_1$


*Define a map $\mu:K \to H_2$ by
$\mu(e_{H_1}, h) =h $ for all $h\in H_2$
This gives an isomorphism from $K$ to $H_2$.
A: So for $K \lhd G$:
Let $a \in H_1$, $b \in H_2$.  Then the conjugate of $(1,h) \in K$ is
$$(a,b)(1,h)(a,b)^{-1} = (a1a^{-1},bhb^{-1}) = (1,bhb^{-1})$$
And you can probably make the final connection yourself.
Now for $G/K \simeq H_1$: You want to find a homomorphism $\phi : G \to H_1$ which is onto, such that $ker \space \phi = K$.  Think about a function that just pulls out the relevant 'coordinate' from $(a,b) \in G$ and sends it to $H_1$, and then just show that that it has the needed kernel and is a homomorphism.  Then the first isomorphism theorem will produce the result.
Finally $K \simeq H_2$: This is almost self-evident.  Considering the elements in the two groups have the form $(1,h)$ on the one hand and $h$ on the other, the 'obvious' mapping between the two will be the isomorphism you need.
