Does this form a group ? From Nathan Carter's Visual Group Theory A penny and nickel are placed side by side on a table along with Dime (placed to the right of the two coins). The only allowed action is to perform is the swapping of penny and nickel. Does this forms a group ?
Now, we have two elements in group basically in the set. Are viz {e,a}. Now we have predefined action which is swapping, and swapping is reversible also,deterministic, also any sequence of swaps is valid. Is this correct ?
 A: Interpreting $a$ as swapping the coins and $e$ as not swapping the coins, we get the set $G = \{e, a\}$ equipped with operation laws
$$ee = e, \\
ea = a, \\
ae = a, \\
aa = e.$$
The axioms you provided are very much appealing to intuition, making it impossible to more formally than you that they are satisfied. What we can check formally are the formal group axioms:


*

*If $x, y \in G$, then $xy \in G$

*There exists an identity element $e$ satisfying that for any $x \in G$ we have $ex = xe = x$

*Any element $x \in G$ has an inverse $x^{-1}$ satisfying $xx^{-1} = x^{-1}x = e$

*The group operation is associative, i.e., for $x,y,z \in G$ we have $(xy)z = x(yz)$.


(Note that these do not exactly correspond to your axioms, but altogether they are equivalent).
The first axiom is satisfied, as can be seen by looking at the operation laws above. Moreover, applying $e$ on either side does not change anything, so $e$ is the identity element in the second axiom. For the third axiom, note that $e^{-1} = e$ and $a^{-1} = a$. Finally, for the fourth axiom, we need to check eight things, e.g.,
$$(ea)a = (a)a = aa = e = e(e) = e(aa).$$
