4
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ $f(A)$ is a subgroup $\endgroup$
    – the_fox
    Jan 28 '19 at 17:53
  • 1
    $\begingroup$ 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 $\mathbb{Z}_4 \times \mathbb{Z}_4...$ $\endgroup$ Jan 28 '19 at 17:54
  • 1
    $\begingroup$ @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. $\endgroup$ Jan 28 '19 at 17:56
2
$\begingroup$

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.)

$\endgroup$
1
$\begingroup$

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.

enter image description here enter image description here


Added for clarity:

enter image description here

$\endgroup$
0
0
$\begingroup$

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 $a_1,...,a_n$.

Then the thesis follows easily.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.