Answer to Question #158911 in Abstract Algebra for kv sing

Question

Determine the operation tables for group G with orders 1, 2, 3 and 4 using the elements a, b, c, and

e as the identity element in an appropriate way.

Answer 1

1. There is only one group of order 1, as it consists only of identity element $e$. There is no need to do a multiplication table in this case.
2. For a group of order 2 we have two elements - $e, a$. As in a group any element should have an inverse, we will forcefully have $a\cdot a = e$ (as $a\cdot e = e\cdot a = a$, so the identity element can not be inverse of $a$)
3. For a group of order 3 we also have only one possible group. We have elements $e,a,b$. We will study the possible values of $a \cdot b$ : it can not be $a$ or $b$ as in this case we would have $a\cdot b = a \Rightarrow b=e$ (or $a\cdot b = b \Rightarrow a=e$), therefore we have $a\cdot b = e$. This also means that $a\cdot a \neq e$ as the inverse is unique and thus we have $a\cdot a = b$. The same reasoning gives us $b\cdot b = a$.
4. For a group of order 4 we, actually, have two possibilities. Let's start by studying the value of $a\cdot a$. If $a \cdot a = b$, then we forcefully have $a\cdot b = c$. Indeed, we can not have $a\cdot b = a$ or $=b$ and if $a\cdot b=e$, then $a\cdot c$ can not be neither $e$ (as inverse of $a$ is $b$), not $a,c$ and it can not be $b$, as $b = a\cdot a$. Therefore we have a group $e, a, a^2,a^3$ with $a^4 =e$ and the multiplication table becomes trivial. The cases $a\cdot a =c$, $b\cdot b =a$, $b\cdot b =c$, $c\cdot c =a$, $c\cdot c =b$ are completely analogous as it is just enough to rename the letters a,b and c. Therefore we now need to treat the case when $a^2 = b^2 =c^2=e$. But in this case the multiplication table trivially becomes $b\cdot a = a\cdot b =c$, $c\cdot a = a\cdot c =b$, $b\cdot c = c\cdot b =a$.