# Existence and construction of isomorphism between finite groups

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)

• 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. – lulu Jan 29 at 11:38
• No I don't assume commutativity. Good point, I had not thought about that. Is there a way to rescue this? – elfeck Jan 29 at 11:41
• Well, I don't know. It's hard to say much about a group just knowing a generating set. – lulu Jan 29 at 11:43
• 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 – elfeck Jan 29 at 11:45
• You haven't even defined an extension of $f$, your definition depends on that independence. – lulu Jan 29 at 11:46

## 1 Answer

(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 $$R\in \mathbf{r}$$.

• 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 $B$. – elfeck Jan 29 at 12:05
• 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 $9$. – user1729 Jan 29 at 12:08
• (I've added the above comment into the answer.) – user1729 Jan 29 at 12:10
• 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? – elfeck Jan 29 at 12:10