Suppose G is a group, and K is a normal subgroup of G, and L is a normal subgroup of K. Then L is a subgroup of G, and it seems logical to expect that it too will be normal in G. Unfortunately, this is false.
 Finite example

Let G=D_{4}, the dihedral group of the symmetries of a square. Take K to be the copy of a Klein group C_{2}*C_{2} contained in D_{4} (i.e. take K to be the subgroup generated by a reflection of the square and the 180 degree rotation). Since the index of K in G is 2 (it has 4 out of the 8 elements), it's reasonably easy to see that K is a normal subgroup. Now take L to be the subgroup (C_{2}) generated by the reflection. Obviously, L is normal in K (since it has index 2 in K, or alternatively because K is Abelian).
But L is not normal in D_{4}! Denoting reflection by f (so L is {1,f}) and rotation by r (so D_{4}={1, r, r^{2}, r^{1}, f, fr, fr^{2}, fr^{1}}), note that fr != rf:
1 2 f 2 1 r 4 2
> >
3 4 4 3 3 1
1 2 r 3 1 f 1 3
> >
3 4 4 2 2 4
but rL={r,rf}, and Lr={r,fr}, so rL!=Lr.
 Infinite example

Take G to be the group of all orientation preserving isometries of the plane (these are all the isometries which consist of 2 reflections). The group consists of all translations and all rotations of the plane. It turns out that the subgroup K of just the translations is normal (rotating, translating and rotating back is just some other translation!). And since K is Abelian, all its subgroups are normal in it. So take L to be the integer translations, i.e. the translations by (x,y), with both x and y integers. Then rotation by any angle apart from 90, 180 or 270 degrees, followed by integer translation, followed by rotation back, is not an integer translation (for instance, any grid point isn't carried to a grid point by this!), so L isn't normal in G.