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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.

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    $\begingroup$ It is true that if both $H$ and $N$ are normal in $G$, then $H\cap N$ is normal in $G$, but here $H$ is not normal in $G$. $\endgroup$ – fkraiem Dec 5 '14 at 13:05
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(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.

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  • $\begingroup$ Thanks a lot. I rewrote (1.). Understand it now? $\endgroup$ – Group Theory Feb 15 '14 at 15:30
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The most strikingly evident and effortless counterexample would be: Pick any group $G$ and any non-normal subgroup $H$. Then let $N=H$. - The fact that not all subgroups are normal should be known of course ...

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    $\begingroup$ But $N$ is required to be normal. Seems better to just take $N = G$. $\endgroup$ – Ravi Fernando Apr 7 '16 at 18:34

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