This is a smaller part of a larger question to prove that the group of order 6 looks like $\{ e_G, a, a^2, b, ab, a^2b\}$. I must prove that $ab \neq e_G$, $ab \neq a$, $ab \neq a^2$, and $ab \neq b$. Given that $\langle a \rangle = \{e_G, a, a^2\}$ and $\langle b \rangle = \{e_G, b\}$ the only element these two sets share in common is $e_G$ thus $a \neq b$, $a^2 \neq b$. This is what I attempted:
For $ab \neq e_G$: Assume $ab = e_G$. Then $$a = a e_G = ab^2 = ab(b) = e_G (b) = b$$ Which is a contradiction since $a \neq b$. Thus $ab \neq e_G$.
For $ab \neq a$: Assume $ab = a$. Then $b = e_G \in \langle a \rangle$ which is a contradiction since $b \notin \langle a \rangle$. Thus, $ab \neq a$.
For $ab \neq a^2$: Assume $ab = a^2$. Then $$a^2 = a^2b(b) = (a)ab = ab = a^2 = (a)a^2 = e_G(b) = b$$ Which is a contradiction since $a^2 \neq b$. Thus, $ab \neq a^2$.
For $ab \neq b$: Assume $ab = b$. Then $a = e_G \in \langle b \rangle$ which is a contradiction since $a \notin \langle b \rangle$. Thus $ab \neq b$.
Does this make sense/is this sufficient to convince you that these elements are unique? Thanks!
Edit: Thanks everyone! I see now that it is a bit redundant to use contradiction in some cases.
It's fine, if a little verbose.
Here's an alternative approach . . .
If $ab=e_G$, then $a=b^{-1}=b$ by multiplying on the right by $b^{-1}$. But $a$ and $b$ have different orders. Hence $ab\neq e_G$.
If $ab=a,$ then, by multiplying on the left by $a^{-1}$, $b=e_G$. But $b$ has order two, so cannot be $e_G$. Hence $ab\neq a$. Similarly, $ab\neq b$.
If $ab=a^2$, then, by multiplying on the left by $a^{-1}$, $b=a$. Hence $ab\neq a^2$.