Isomorphic quotients by isomorphic normal subgroups

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=Z2Z4, with H=(1,0) and K=(0,2)). A sufficient condition to guarantee isomorphic quotients is the existence of an automorphism φAut(G) such that φ(H)=K.

One can also have G/HG/K, but H and K not isomorphic. For example, take again G=Z2Z4, and take H=Z2(0,2), isomorphic to the Klein 4-group, and K={0}Z4. Then G/HG/KZ2.

And of course, it is trivial to have HK and G/HG/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 φAut(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 φAut(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

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