In this recent question, Iota asked if, given a finite group G and two isomorphic normal subgroups H and K, it would follow that G/H and G/K are isomorphic. This is not true (a simple example given by G=Z2⊕Z4, with H=⟨(1,0)⟩ and K=⟨(0,2)⟩). A sufficient condition to guarantee isomorphic quotients is the existence of an automorphism φ∈Aut(G) such that φ(H)=K.

One can also have G/H≅G/K, but H and K not isomorphic. For example, take again G=Z2⊕Z4, and take H=Z2⊕⟨(0,2)⟩, isomorphic to the Klein 4-group, and K={0}⊕Z4. Then G/H≅G/K≅Z2.

And of course, it is trivial to have H≅K and G/H≅G/K.

Now, here’s the question:

Question.Can we have a finite group G, normal subgroups H and K that are isomorphic as groups, G/H isomorphic to G/K, but no φ∈Aut(G) such that φ(H)=K?It is not hard to come up with examples with infinite G. For example, take

G=Z2⊕(∞⨁i=1Z4),

let H=2G, and let K be the subgroup generated by H and the generator of the cyclic factor Z2. Then both H and K are isomorphic to a direct sum of countably many copies of Z2, as are the quotients G/H and G/K. But H is a verbal subgroup, hence fully invariant, so any automorphism of G maps H to H, so there does not exist any φ∈Aut(G) such that φ(H)=K.Does someone have an example with finite G? Note that I am not requiring that φ be a lift of the given isomorphism of G/H with G/K.

**Answer**

There are lots of examples, see here:

http://groupprops.subwiki.org/wiki/Series-equivalent_not_implies_automorphic

**Attribution***Source : Link , Question Author : Arturo Magidin , Answer Author : Vipul*