I would appreciate guidance on the following problem.
Show that the additive group $\mathbb{Z}$ acts on itself by $xy = x+y$ and find all $x\in\mathbb{Z}$ such that $xy = y$ for all $y\in\mathbb{Z}$.
I'm having difficulty actually understanding exactly what $xy = x+y$ means in terms of a group action. I understand that there are two axioms that a group action need comply with, namely:
- $e\cdot a = a$
- that $g_1\cdot (g_2.a) = (g_1\circ g_2).a$, where $g_1, g_2 \in$ group $G$, $a\in$ set $A$, and "$\cdot$" is the group action with "$\circ"$ being the composition in G. Also I'm aware of the fact that the action is additive (so "$\cdot$" is "$+$" ) and that $G= A$ in this instance.
Any assistance/guidance would be welcomed.
throughout my solution, $\cdot$ will denote a group action
We want to find all $x \in \mathbb{Z}$ such that $x \cdot y := x + y = y$. Since $\mathbb{Z}$ is a gruop we know that the only element which satisfy the relation must be the identity since $$ 0 + y = y \quad \forall y \in \mathbb{Z} $$ $$ x + y = y \Rightarrow x = 0 $$ Actually, you can prove a more general fact: Given a group $G$, we can define a $G$ action over $G$ such that $$ g \cdot h := gh $$ It is trivial to show that this is an action. More intresting is that $$ g \cdot h = h \iff g = e $$ where $e$ is the identity element. Again we can easily prove it remembering that each element of a group has an inverse, so $$ gh = h \iff ghh^{-1} = h ^{-1} \iff g = e $$