Show that if H is a subgroup of a group G, and N is a normal subgroup in G, then $H \cap N$ is normal in H. Show by an example that $H \cap N$ need not be normal in G.
I can condone the proof hence not asking about it.
(1.) What's the intuition? By means of Fraleigh p. 59 5.54, intersection of subgroups is a subgroup.
Because $\color{magenta}{H, N \leq G}$ and $(H \cap N) \le G \implies (H \cap N) \le \quad \color{magenta}{H \text{ or } N \quad \le G}$,
I believed something similar for normal subgroups: $ (H \cap N) \trianglelefteq \color{magenta}{H} \implies (H \cap N) \trianglelefteq \color{magenta}{G} $.
But this question question proves $\begin{align}H \le G, &\implies (H \cap N) & \trianglelefteq \color{magenta}H \\ N \trianglelefteq G & &\not \trianglelefteq \color{magenta}G \end{align}$. What did I bungle?
Counterexample 1: www.auburn.edu/~huanghu/math5310/alg-hw-ans-13 I think.pdf
$\color{brown}{G = N} = S_3, H = \{ \; id, (1)(2, 3) \; \}$.
The trivial normal subgroup of a group are $\{ id \}$ and itself. Hence $\color{brown}{G = N} \; \trianglelefteq G$.
But $\begin{align} (1,2)(3)H & = \{ \; (1, 2)(3), (1, 2, 3) \; \} \\ & \neq \{ \; (1, 2)(3), (1, 3, 2) \; \} = H(1,2)(3) \end{align}$. Hence $H \not\trianglelefteq G \implies
(H \cap N) \not\trianglelefteq G$?
(2.) How do you envisage and envision this counterexample? Where did it loom from?
Counterexample 2: Let $G = D_4$, let $N = \{ \; id, \text{ 180 deg anticlockwise rotation, horizontal flip, vertical flip } \}$, and let $H = \{id, \text{ horizontal flip } \}$. Then N is normal in G, but $H ∩ N = H$ is not normal in G.
(3.) Where did this magically spring from? Can you envisage all this from Cayley diagrams? Notify me if you want me to post digraphs.
Best Answer
(1.) I don't understand your argument, you need to be more clear
(2., 3.) A general way to produce a counterexample is to consider any group with a non-normal subgroup $H$, then try to obtain a normal subgroup $N$ by "adding" elements to $H$. Thus by construction $H$ is not normal, $N$ is normal, and $H \cap N = H$ is not normal.
In particular, since $G \trianglelefteq G$, we may simply let $N = G$. There is nothing special about $S_3$, it was probably chosen because it is the smallest group with non-normal subgroups.
Counterexample 2 uses a proper normal subgroup (only for educative purposes I guess, again letting $N = D_4$ would have been easier); I think the only thing you can envision here is why $N$ is normal.