# Sufficient conditions on the isomorphism of two groups

Let $$G_1, G_2$$ be two groups with at least one nontrivial proper subgroup each.

Let $$S_1, S_2$$ be the sets of proper subgroups of, respectively $$G_1, G_2$$.

Suppose there exists a bijective function $$f: S_1 \rightarrow S_2$$ such that $$\forall A\in S_1, f(A)$$ is isomorphic to $$A$$.

When can I conclude that $$G_1, G_2$$ are isomorphic?

I think that, if $$G_1$$ and $$G_2$$ are finite and abelian we can conclude that they are isomorphic, but I can't prove It. Moreover, I haven't found any counterexample for nonabelian finite groups.

• $f(A)$ is a subgroup Jan 28 '19 at 17:53
• If the two groups are not finite we surely can't conclude anything. A counterexample is $\mathbb{Z}_2 \times \mathbb{Z}_4 \times \mathbb{Z}_4...$ and $\mathbb{Z}_4 \times \mathbb{Z}_4...$ Jan 28 '19 at 17:54
• @the_fox Ah, you are right, sets of proper subgroups, this was missing. I am sorry. Still, the question is a bit vague "when can I conclude that $G_1\cong G_2$." Certainly not always, but sometimes. Jan 28 '19 at 17:56

There are two pairs of examples of order $$16$$. These are the smallest examples. One of these two pairs is $$C_4\times C_4$$ and $$C_4\rtimes C_4$$. For both of these, the complete list of proper subgroups is:

• 1 trivial subgroup

• 3 subgroups isomorphic to $$C_2$$

• 6 subgroups isomorphic to $$C_4$$

• 1 subgroup isomorphic to $$C_2\times C_2$$

• 3 subgroups isomorphic to $$C_4\times C_2$$

(See https://groupprops.subwiki.org/wiki/Nontrivial_semidirect_product_of_Z4_and_Z4#Subgroups for the subgroups of $$C_4\rtimes C_4$$.)

Another easy pair of examples is $$C_9\times C_3$$ and $$C_9\rtimes C_3$$.

(It is definitely true for finite abelian groups though, this is an easy consequence of their classification.)

Certainly not always. I'd be surprised if there is a concrete set of conditions which is both necessary and sufficient to conclude isomorphism between the two groups. (My answer refers to finite groups only.)

There are groups which are called $$P$$-groups in Schmidt's book "Subgroup Lattices of Groups" (not be confused with $$p$$-groups) and which are lattice-isomorphic to elementary abelian groups.

Here is the proof in the case of finite abelian groups $$G_1, G_2$$ like above.
Lemma 1 Let $$G^{(n)}$$ be the number of elements in $$G$$ of order $$n$$. $$G^{(n)}$$ is uniquely determined by the number of cyclic subgroups of $$G$$ of order $$n$$.
Proof Every element of order $$n$$ is an element of exactly one cyclic subgroup of $$G$$ of order $$n$$. All the cyclic subgroups of order $$n$$ have $$\phi(n)$$ elements of order $$n$$.
Lemma 2 Let $$p$$ be a prime that divides $$|G|$$ then the numbers $$G^{(p)}, G^{(p^2)},...$$ uniquely determine the p-Sylow of $$G$$.
Proof The p-Sylow, P, of G is of the form $$\mathbb{Z}_{p^{a_1}} \times ... \times \mathbb{Z}_{p^{a_n}}$$. Moreover, let $$P^{(\leq p^k)}$$ be the number of elements of P that have an order less or equal to $$p^k$$. $$P^{(\leq p^k)}=\Pi_{i\leq n}{\min (p^{a_i}, p^k)}$$ Then we can determine $$a_1,...,a_n$$.