# Group Action of $\mathbb{Z}$ on Itself

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: 1) $e.a = a$ and 2) that $g_1.(g_2.a) = (g_1\circ g_2).a$, where $g_1, g_2 \in$ group $G$, $a\in$ set $A$, and "." is the group action with "$\circ"$ being the composition in G. Also I'm aware of the fact that the action is additive (so "." is "+" ) and that G = A in this instance. Any assistance/guidance would be welcomed.

• Let's add a symbol to make things clearer. When you write $xy$, you are thinking of $x$ as in the group $\mathbb{Z}$ whereas $y$ is in the set $\mathbb{Z}$. Therefore, $xy$ could be rewritten as $x(y)$. Therefore, the function determined by $x$ is applied to the set $y$. – Michael Burr Jan 18 '18 at 12:49
• $xy$ is being used for the result of $x$ acting on $y$, i.e., what you are writing as $x.y$ in your two axioms. The equation means that "$x$ acts on $y$ by adding $x$ to it". i.e. $x.y$ is equal to $x + y$. (it is also possible to read $xy$ as the result of $y$ acting on $x$ as Michael Burr did - so the statement isn't very clear out of context.) – Rob Arthan Jan 18 '18 at 12:49
• MB & RA: Thank you so much for your concise explanation(s) of "xy". I spent hours trying to figure out what this meant – Gismho Jan 18 '18 at 14:20

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 $$gh = h \iff ghh^{-1} = h ^{-1} \iff g = e$$