Showing that one axiom of the concept group action implies another (similar to group homomorphisms)

[Beware that my question is at the bottom of all this text.]

We defined in a group theory course a group action to be a mapping $$\nu:G\times X \rightarrow X,$$ $(G,\cdot)$ group and $X$ some set, such that the axioms $$\textrm{(GA1)} \ \ \ \nu(1,x)=x$$ and $$\textrm{(GA2)} \ \ \ \nu(g,\nu(h,x))=\nu(g h, x)$$ hold for every $g,h\in G$ and $x\in X$.

We defined a group homomorphism to be a map $$f:G\rightarrow H,$$ such that the axioms $$\textrm{(HOM1)} \ \ \ f(1_G)=1_H$$ and $$\textrm{(HOM2)} \ \ \ f(gh)=f(g)f(h)$$ hold for all $g,h \in G$.

Now, obviously, these axioms for the homomorphism aren't minimal, meaning (HOM2) $\Rightarrow$ (HOM1). So, naturally, I asked myself, if the same doesn't happen with the group action axioms, meaning (GA2) $\Rightarrow$ (GA1).

Now the thing is, I can show that this is the case by showing that a mapping $\nu$ satisfies (GA2) iff the mapping $$\tau: G\rightarrow X_{\leftrightarrow} ,$$ where $X_{\leftrightarrow}$ is the group of all bijections on $X$ together with the composition, defined by $$\tau(g)(x)=\nu(g,x),$$ is a homomorphism:

$(\tau(g)\circ \tau(h))(x)=\tau(g)(\nu(h,x))=\nu(g,\nu(h,x)),$ which by (GA2) equals $\nu(gh,x)=\tau(gh)(x)$, and $\nu(g,\nu(h,x))=\tau(g)(\tau(h)(x))$ which by (HOM2) equals $\tau(gh)(x)=\nu(gh,x)$ for all $g,h\in G$ and $x\in X$.

From that I can now show (GA1) by the following implications: $$\textrm{(GA2)} \Leftrightarrow \textrm{(HOM2)} \Rightarrow \textrm{(HOM1)} \Rightarrow \textrm{(GA1)}.$$

Now (finally) my question: Can't I show that (GA1) follows from (GA2) directly ? I tried doing that, but couldn't succeed, because unlike showing $\textrm{(HOM2)} \Rightarrow \textrm{(HOM1)}$ I don't have an thing like an "inverse to $\nu(1,x)$" that I can apply upon $\nu(1,\nu(1,x))=\nu(1,x)$ to get $\nu(,x)=1$.

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You want $\nu:G\times X\to X$. –  Brian M. Scott Nov 1 '11 at 12:16
Fixed.${}{}{}{}$ –  joriki Nov 1 '11 at 12:36
Refixed, after somebody changed it again. –  el le Nov 1 '11 at 12:53

Let $G=\{1_G\}$ be the trivial group, and let $X=\mathbb{Z}$. Define $\nu:G\times X\to X$ by $$\nu(1_G,n)=\begin{cases} n,&n\text{ is even}\\ n-1,&n\text{ is odd}\;. \end{cases}$$
Clearly (GA$1$) isn’t satisfied, but for any $n\in\mathbb{Z}$ we have $$\nu(1_G,\nu(1_G,n)) = \nu(1_G,n) = \begin{cases} n,&n\text{ is even}\\ n-1,&n\text{ is odd}\;, \end{cases}$$ so (GA$2$) is satisfied.
Since this implies there's an error in the argument in the question, it might be worth pointing out where it is: You said that $\tau$ maps $G$ to the group of bijections, but you didn't prove that the maps you defined were actually bijections. Brian's example shows that they needn't be. But composition of non-bijective maps isn't a group operation, so the part where you use the inverse to go from (HOM2) to (HOM1) doesn't work out. By the way, the action of $1_G$ could be defined to be any non-trivial idempotent operation, e.g. a projection in a vector space. –  joriki Nov 1 '11 at 12:43