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[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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    Refixed, after somebody changed it again.2011-11-01

1 Answers 1

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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.

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    Thanks, great example - and good to know where the error in my argument was hidden.2011-11-01