Show that $N \lhd G \times H \not \Rightarrow N = N_1 \times N_2$ with $N_1 \lhd G$ and $N_2 \lhd H$ I'm trying to prove the following assertion: Show that $N \lhd G\times H \not \Rightarrow N = N_1\times N_2$ with $N_1 \lhd G$ and $N_2 \lhd H$
What I tried to do is find $(n_1;n_2) \in N$ such that $(g;h) \circ (n_1;n_2)\circ(g;h)^{-1} \in G\times H$ but $(n_1;n_2) \not \in N_1\times N_2$
 A: A single counterexample will suffice.
Recall that $\mathbb Z_2$ is the additive group of integers modulo $2$, consisting of two elements $\{0,1\}$ where $0$ is the additive identity and $1+1=0$. This group is cyclic with order $2$.
Now let $G = H = \mathbb Z_2$. We have $G \times H = \mathbb Z_2 \times \mathbb Z_2$. This group is abelian (because it is a direct product of abelian groups), so all of its subgroups are normal. Now define $N = \langle (1,1) \rangle = \{(0,0), (1,1)\}$. This subgroup has order $2$.
To see that $N$ cannot be expressed in the form $N_1 \times N_2$, where $N_1 \lhd Z_2$ and $N_2 \lhd Z_2$, note that this would imply $2 = |N| = |N_1 \times N_2| = |N_1||N_2|$. So exactly one of $N_1$ and $N_2$ would need to have order $2$ (and would therefore be $\mathbb Z_2$), and the other would need to have order $1$ (and would therefore be $\{0\}$).
Consequently, $N_1 \times N_2$ would have to be either $\mathbb Z_2 \times \{0\} = \{(0,0), (1,0)\}$ or $\{0\} \times \mathbb Z_2 = \{(0,0), (0,1)\}$, neither of which is the same as our $N$.
