If $G$ is a arbitrary group, $H,K,N\leq G$ such that $H\subseteq K$, then $K\cap \langle H\cup N\rangle\subseteq \langle H\cup (K\cap N)\rangle$?. If it is not true, how can I find an counterexample?

$\begingroup$ this is basic set theory. $\endgroup$ – Ehud Meir Nov 10 '15 at 12:36

5$\begingroup$ @EhudMeir No it's not. $\endgroup$ – Todd Trimble♦ Nov 10 '15 at 12:37

$\begingroup$ I am sorry, I misinterpreted the \langle \rangle symbols. My mistake. $\endgroup$ – Ehud Meir Nov 10 '15 at 12:38

9$\begingroup$ In my opinion, any grouptheory question to which $D_8$ is a counterexample is not of research level. I'm voting to close. $\endgroup$ – HJRW Nov 10 '15 at 14:34
I assume $\leq$ means the same thing as the relation $\subseteq$ between subgroups.
This property, which can be formulated for any lattice and is called the modular law, can fail for the lattice of subgroups of a group; for example it fails for the dihedral group of order $8$. It holds though if $G$ is abelian, or if all the subgroups $H, K, N$ are normal (i.e. it holds for the normal subgroup lattice).
The simplest way to find a counterexample, if you didn't already know of one, is by googling with search terms "modular lattice subgroups". One good thing to know about modular lattices is that they have a "forbidden sublattice" characterization: they can't have a pentagon in their Hasse diagram. I'll let you click on the article lattice of subgroups to detect for yourself a pentagon and determine the corresponding subgroups for the $D_8$ example.

$\begingroup$ It was asked (in an answer post, now deleted) why the lattice of ideals of a ring or the lattice of submodules is modular whereas the same is not true for the subgroup lattice. That's a good question. I would say this: a general theorem is that the lattice of congruences of a model $G$ of an algebraic theory $T$ is modular when $T$ is a Mal'cev theory: ncatlab.org/nlab/show/Mal'cev+variety The theories of groups, of rings, and of modules over a fixed ring are all Mal'cev. Second, the lattice of congruences in each of these cases is isomorphic to the lattice of kernels of homomorphisms. $\endgroup$ – Todd Trimble♦ Nov 10 '15 at 16:49
You can obtain a more explicit answer with the following GAP function:
counterexamples := function ( G )
local subs, examples, H, K, N, L, R;
subs := Concatenation(List(ConjugacyClassesSubgroups(G),AsList));
examples := [];
for H in subs do
for K in subs do
if not IsSubset(K,H) then continue; fi;
for N in subs do
L := Intersection(K,ClosureGroup(H,N));
R := ClosureGroup(H,Intersection(K,N));
if not IsSubset(R,L) then Add(examples,[H,K,N]); fi;
od;
od;
od;
return examples;
end;
With this you get for example:
gap> D8 := Group((1,2,3,4),(1,3));;
gap> counterexamples(D8); # Todd Trimble's example made explicit
[ [ Group([ (2,4) ]), Group([ (1,3)(2,4), (2,4) ]), Group([ (1,2)(3,4) ]) ],
[ Group([ (2,4) ]), Group([ (1,3)(2,4), (2,4) ]), Group([ (1,4)(2,3) ]) ],
[ Group([ (1,3) ]), Group([ (1,3)(2,4), (2,4) ]), Group([ (1,2)(3,4) ]) ],
[ Group([ (1,3) ]), Group([ (1,3)(2,4), (2,4) ]), Group([ (1,4)(2,3) ]) ],
[ Group([ (1,2)(3,4) ]), Group([ (1,3)(2,4), (1,2)(3,4) ]), Group([ (2,4) ]) ],
[ Group([ (1,2)(3,4) ]), Group([ (1,3)(2,4), (1,2)(3,4) ]), Group([ (1,3) ]) ],
[ Group([ (1,4)(2,3) ]), Group([ (1,3)(2,4), (1,2)(3,4) ]), Group([ (2,4) ]) ],
[ Group([ (1,4)(2,3) ]), Group([ (1,3)(2,4), (1,2)(3,4) ]), Group([ (1,3) ]) ] ]
gap> counterexamples(AlternatingGroup(4)); # another example
[ [ Group([ (1,2)(3,4) ]), Group([ (1,3)(2,4), (1,2)(3,4) ]), Group([ (2,4,3) ]) ],
[ Group([ (1,2)(3,4) ]), Group([ (1,3)(2,4), (1,2)(3,4) ]), Group([ (1,3,4) ]) ],
[ Group([ (1,2)(3,4) ]), Group([ (1,3)(2,4), (1,2)(3,4) ]), Group([ (1,4,2) ]) ],
[ Group([ (1,2)(3,4) ]), Group([ (1,3)(2,4), (1,2)(3,4) ]), Group([ (1,2,3) ]) ],
[ Group([ (1,3)(2,4) ]), Group([ (1,3)(2,4), (1,2)(3,4) ]), Group([ (2,4,3) ]) ],
[ Group([ (1,3)(2,4) ]), Group([ (1,3)(2,4), (1,2)(3,4) ]), Group([ (1,3,4) ]) ],
[ Group([ (1,3)(2,4) ]), Group([ (1,3)(2,4), (1,2)(3,4) ]), Group([ (1,4,2) ]) ],
[ Group([ (1,3)(2,4) ]), Group([ (1,3)(2,4), (1,2)(3,4) ]), Group([ (1,2,3) ]) ],
[ Group([ (1,4)(2,3) ]), Group([ (1,3)(2,4), (1,2)(3,4) ]), Group([ (2,4,3) ]) ],
[ Group([ (1,4)(2,3) ]), Group([ (1,3)(2,4), (1,2)(3,4) ]), Group([ (1,3,4) ]) ],
[ Group([ (1,4)(2,3) ]), Group([ (1,3)(2,4), (1,2)(3,4) ]), Group([ (1,4,2) ]) ],
[ Group([ (1,4)(2,3) ]), Group([ (1,3)(2,4), (1,2)(3,4) ]), Group([ (1,2,3) ]) ]
]
gap> Length(counterexamples(SymmetricGroup(4))); # S_4 has a lot of counterexamples
348
Thus you see that there are plenty of easytofind counterexamples.