Suppose G is a group and K⊲ are normal subgroups of G. Is K a normal subgroup of G, i.e. K\lhd G? If not, what extra conditions on G or H make this possible?

Applying the definitions, we know \{ghg^{-1}\mid h\in H\}=H and \{hkh^{-1}\mid k\in K\}=K, and want \{gkg^{-1}\mid k\in K\}=K. Clearly, the best avenue for a counterexample is if gkg^{-1}\not\in K for some k\in K and g\in G-H.

If no such element exists, \{gkg^{-1}\mid k\in K\}\subseteq K implies \{gkg^{-1}\mid k\in K\}=K because if k’\in K, gk’g^{-1}=k\in K\Rightarrow k’=g^{-1}kg\in\{gkg^{-1}\mid k\in K\}.

**Answer**

Using some suggestions from the other commenters:

The alternating group, A_4, has the set H=\{I,(12)(34),(13)(24),(14)(23)\}\cong V_4 as a subgroup. If f\in S_4\supseteq A_4 is a permutation, then f^{-1}[(12)(34)]f has the effect of swapping f(1) with f(2) and f(3) with f(4). One of these is 1, and depending on which it is paired with, the conjugated element may be any of H-\{I\}, since the other two are also swapped. Thus H\lhd S_4 is normal, so H\lhd A_4 as well. Similarly, H has three nontrivial subgroups, and taking K=\{I,(12)(34)\}\cong C_2, this is normal because V_4 is abelian. But K\not\lhd A_4, since [(123)][(12)(34)][(132)]=(13)(24)\in H-K.

Moreover, this is a minimal counterexample, since |A_4|=12=2\cdot 2\cdot 3 is the next smallest number which factors into three integers, which is required for K\lhd H\lhd G but \{I\}\subset K\subset H\subset G so that [G\,:\,H]>1, [H\,:\,K]>1, |K|>1 and

|G|=[G\,:\,H]\cdot[H\,:\,K]\cdot|K|.

The smallest integer satisfying this requirement is 8, but the only non-abelian groups with |G|=8 are the dihedral group D_4 and the quaternion group Q_8, and neither of these have counterexamples. (Note that if G is abelian, then all subgroups are normal.) Thus A_4 is a minimal counterexample. (Edit: Oops, D_4 has a counterexample, as mentioned in the comments: \langle s\rangle\lhd\langle r^2,s\rangle\lhd \langle r,s\rangle=D_4, but \langle s\rangle\not\lhd D_4.)

However, if H\lhd G and K is a characteristic subgroup of H, then K is normal in G. This is because the group action f defines an automorphism on G, \varphi(g)=f^{-1}gf, and because H is normal, \varphi(H)=H so that \varphi|_H is an automorphism on H. Thus \varphi(K)=K since K is characteristic on H and so \{f^{-1}kf\mid k\in K\}=K\Rightarrow K\lhd G.

**Attribution***Source : Link , Question Author : Mario Carneiro , Answer Author : Michael Hardy*