Proving a normal subgroup of a direct product Suppose that $G =H \times K$, where $H$ and $K$ are finite groups. If $N \unlhd H$ then $N \unlhd G$.
I know this result is true and it is easy to prove but I want clarification on how to present the proof
I know that from $G = H \times K$, we get that $H \unlhd G$, $K\unlhd G$, $H \cap K = \{e\}$, and $G = HK$. 

To show $N \unlhd G$, should I state that I am identifying the elements $(h, 1_k)$ as $h$ for $h\in H$ and $(1_H, k)$ as $k$ for $k\in K$?

Then for arbtirary $g = hk \in G$, we have $hkNk^{-1}h^{-1} = hNkk^{-1}h^{-1} =hNh^{-1} = N$

Alternatively, knowing that $N \cong \{(n, 1_K) | n\in N\} = N_0$, should I show that $N_0$ is normal in $H \times K$?

If somebody could please which would be better to use and how I should go about constructing the proof.
 A: The two approaches are equivalent, and a better way to "see" what is going on might be writing the elements of the direct product as $(h,k)$ where the group operation is, obviously
$$(h_1, k_1) (h_2,k_2) = (h_1 h_2, k_1 k_2)$$
As you correctly said, you identify $H$ with $\{(h,1), h \in H\}$ and then show that for all elements in $H \times K$ and for all $n \in N$ you have
$$(h,k)(n,1)(h^{-1},k^{-1} = (hnh^{-1},kk^{-1}) = (hnh^{-1},1) \in N$$
since conjugation by elements of $H$ is an internal operation in $N$.
A: Have you been taught the distinction between internal and external direct products? Using internal direct products, I would say that your first argument is fine: to say that the group $G$ is internal direct product of subgroups $H$ and $K$
means that $H$ and $K$ are normal subgroups of $G$ with $H \cap K = 1_{G}$ and $G = HK.$ It follows from these given facts that $hk = kh$ for all $h \in H,k \in K.$
Then your first argument correctly shows that if $G$ is the internal product of $H$ and $K$ and $N \lhd H,$ then we also have $N \lhd G.$
In general, if $H$ and $K$ are groups ( not necessarily subgroups of some other given group), then the external direct product $H \times K = \{(h,k): h \in H, k \in K \}$ ( with componentwise operations inherited from $H$ and $K$ respectively ) is the internal direct product of its two normal subgroup $\{(h,1_{K}): h \in H \}$
and $\{(1_{H},k): k \in K \},$ the first normal subgroup being obviously isomorphic to $H$ and the second isomorphic to $K$. This is why most texts stop making the distinction between internal and external direct product fairly soon.
