# Group action defined on generators.

In a question for an assignment (specific group but I've abstracted the question for academic integrity). I was given a group $G=\langle\ a,b \mid R\ \rangle$ where $R$ is some set of relations. Then, the question defines a group action on $X$ for each generating element. The first question is then "Verify that this gives a group action on $X$".

My problem is that it seems like by defining a group action on generators inherently assumes that $e(x)=x$ and $a(b(x))=ab(x)$ from the beginning, so it seems like there is nothing left to show?

Since the question only defined the functions $a(x)=?$ and $b(c)=?$, I can't verify that $ab(x)=a(b(x))$, since I wasn't given the action defined by $ab$.

Is this a misunderstanding on my part, or do you think it is likely that the question itself is ambiguous?

## 2 Answers

What you need to check is that every relator will act (by concatenating the action for the generators as given in the relator) as the identity.

An example to show that this is needed would be a cyclic group $\langle g \rangle$ of order $3$, which acts on $4$ points via $g\mapsto (1,2,3,4)$.

Then $g^3$, the identity, would have to act as $(1,4,3,2)$, which is forbidden.

• Okay, yes - that makes sense! I did check that the equations given satisfied the $a^n=e$ and $b^m=e$ relations out of habit, so I'm glad my instincts weren't entirely wrong. Thank you very much! – Floyd Everest Apr 30 '18 at 15:11

The answer of ahulpke tells you how to show your map is a homomorphism in practice. I just wanted to tell you that you have seen this before.

In particular, one way to think about your question is that a group action of $G$ on a set $X$ corresponds to a homomorphism $G\rightarrow \operatorname{Sym}(X)$. You are, I am sure, well used to group homomorphisms being given only by the generators*. So the question you have is exactly the same game as showing a map is a homomorphism, just disguised slightly.

*e.g. "show that the map $\phi: \mathbb{Z}\rightarrow\mathbb{Z}, 1\mapsto -1$ is an isomorphism".