This is a generalization of my previous question: What are all finite groups such that all isomorphic subgroups are identical?

Specifically, what finite groups $G$ have the following property: For any two subgroups $1 < H, H' < G$ that are isomorphic, then $H \cap H'$ is non-trivial.

Some quick results: This property is equivalent to all prime order cyclic subgroups being unique. Also, each of these subgroups are normal.

If $G$ is an abelian group, it turns out $G$ is cyclic. (Proof: $G=G_1 \times \cdots \times G_n$, each cyclic, with $|G_i|$ dividing $|G_{i+1}|$. Two factors share a common prime $p$, hence two disjoint copies of $C_p$, so no two factors exist.)

However, there are non-abelian groups that satisfies this condition (if I've thought through it right). Take the Quaternion group, which from the subgroup lattice is easily seen to have the desired property.

Can the (non-abelian) groups with this property be easily characterized?


1 Answer 1


The question has an answer here.

[In such a group] every Sylow $p$-subgroup is either cyclic or generalized quaternion (because these are the only $p$-groups that have a unique subgroup of order $p$). Apparently, Zassenhaus (in the 1930s) classified all solvable groups where all the Sylow $p$-subgroups for odd primes are cyclic, and where the Sylow 2-subgroup contains an index $2$ cyclic subgroup (and Michio Suzuki, in the paper ''On finite groups with cyclic Sylow subgroups for all odd primes'' from Amer. J. of Math in 1955 classified the non-solvable groups with the same properties).

This gives a complete classification of the groups that meet your condition.


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