Assume I have two finite groups $G$ and $H$ of equal order. Further assume I have found minimal generating sets $A$ and $B$ for a the two groups respectively (of equal size) and additionally (see comments) I know at least one decomposition into generating elements of all $g \in G$.

I now want to find out if the two groups are isomorphic give this extra information of the minimal generating sets.

Is there an approach along these lines:

Define a bijective function $f: A\to B$. Extend the function to all of $G$ in the following manner:
For $g \in G$ find a decomposition of $g = a_1 + \cdots + a_m$ where $a_i \in A$ and define $f(g) = \sum_{i=1}^m f(a_i)$. If this extension fulfills the homomorphism property $f(g_1 + g_2) = f(g_1) + f(g_2)$ for all $g_1, g_2 \in G$ then it is an isomorphism.

Question 1. Is the above statement correct? Do I need to check $f(G) = H$ or is this already implicitly true?

Question 2. Here I need to expand the function for all $g \in G$ and check the homomorphism property for all pairs of elements from $G$.
Given the information of two minimal generating sets, can I reduce the amount of checking I have to do?
(Checking only the pairs of generators is obviously not enough since the extension fulfills the homomorphism property for elements from $A$ by construction)

  • $\begingroup$ Are you assuming your groups are abelian? How do you know that that the extension of $f$ that you define is independent of the choice of decomposition of $g$? If, say, $a_1$ has order $3$ and $f(a_1)=b_1$ has order $2$ then $f(a_1^2)=b_1^2=e_H$ so your function isn't even a set theoretic bijection. $\endgroup$ – lulu Jan 29 at 11:38
  • $\begingroup$ No I don't assume commutativity. Good point, I had not thought about that. Is there a way to rescue this? $\endgroup$ – elfeck Jan 29 at 11:41
  • $\begingroup$ Well, I don't know. It's hard to say much about a group just knowing a generating set. $\endgroup$ – lulu Jan 29 at 11:43
  • $\begingroup$ If the extension of $f$ were an isomorphism, then it must also be independent of the decomposition of any $g$. So perhaps it is not necessary to require (a-priori) that the extension of $f$ is bijective $\endgroup$ – elfeck Jan 29 at 11:45
  • $\begingroup$ You haven't even defined an extension of $f$, your definition depends on that independence. $\endgroup$ – lulu Jan 29 at 11:46

(I should state at the start that the ideas in the question are close to pretty standard ideas. All I am doing here is trying to explain the link to these standard ideas.)

There are two things you seem to be missing.

  1. If $f$ is a homomorphism then its extension to the whole group is not necessarily a bijection. You need to verify that it is injective or surjective (or both if your group is infinite).
  2. The image group $H$ may have lots of minimal generating sets. So you need to find all minimal generating sets $B_i$ (for speed you can take them up to conjugacy, or up to automorphism). For example, consider $\mathbb{Z}_3\times\mathbb{Z}_{18}$. This is isomorphic to $\mathbb{Z}_6\times\mathbb{Z}_9$. One group is generated by an element of order $3$ and an element of order $18$, while the other is generated by an element of order $6$ and an element of order $9$.

The following is then true:

Theorem. Let $A$ be a minimal generating set for $G$, and let $\{B_i\}$ be the set of all minimal generating sets of $H$ which have cardinality $|A|$. Then $G\cong H$ if and only if there exists a bijection $f_{i, j}: A\rightarrow B_i$ which extends to an isomorphism $G\rightarrow H$ in the way you describe.

Note that you have to check that the extension is bijective and is a homomorphism. These are non-trivial tasks*. Also, the subscript $j$ is because there are lots of bijections between $A$ and $B_i$, but not all of these will extend to be a homomorphism.

*If $G$ is given by a presentation $\langle \mathbf{x}\mid\mathbf{r}\rangle$ (alternatively this presentation can be computed, but not necessarily quickly!, as your group is finite) then determining if $f$ is a homomorphism is relatively easy, as you just need to verify that $f(R)=_H1$ for all $R\in \mathbf{r}$.

  • $\begingroup$ Thank you for your answer. So what you are implying that is not even sufficient to fix two generating sets $A$ and $B$ of G and H and check all extensions of all bijections $f: A \to B$ for bijectivity and homomorphism? Edit: So there might not even exists a bijective homomorphisms $G \to H$ that maps $A$ to $B$. $\endgroup$ – elfeck Jan 29 at 12:05
  • 1
    $\begingroup$ Nope! For example, consider $\mathbb{Z}_3\times\mathbb{Z}_{18}$. This is isomorphic to $\mathbb{Z}_6\times\mathbb{Z}_9$. One group is generated by an element of order $3$ and an element of order $18$, while the other is generated by an element of order $6$ and an element of order $9$. $\endgroup$ – user1729 Jan 29 at 12:08
  • $\begingroup$ (I've added the above comment into the answer.) $\endgroup$ – user1729 Jan 29 at 12:10
  • $\begingroup$ Aiaiai, okay, I see. I assume restricting the generating sets to contain elements of the same order is not sufficient either, although coming up with a counter example might be a bit tricky? $\endgroup$ – elfeck Jan 29 at 12:10

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