Determining the elements in a group Say $G$ is a group with exactly six elements $\{a,b,c,d,e,f\}$.
Further, the following is defined: $a^2=b^2=d^2=d$, $e^2=f$, $ab=e$.
I have to determine the following elements in $G$.


*

*$bc$

*$ac$

*$f^{-1}$

*$df$


I assume that $d$ is the neutral element in $G$, since $d^2=d$. So far I have the following.


*

*$bc=bc$

*$ac=ac$


Is this correct? And how can I determine the elements of $f^{-1}$ and $df=$?
For $f^{-1}$, do I have to find the inverse of $f$ such that $ff^{-1}=d$? Some hints about how to proceed would be enough. Thank you in advance!
 A: 
Any group of order $6$ is either isomorphic to $\Bbb Z_6$ or $S_3$.

The group here is $S_3$ ( clear from the operations between the elements given here)where the elements are 
$a=(12),b=(23),c=(13),e=(123),f=(132)$ and $d$ is the identity
Now just compute the expressions 
A: Here's a way to solve this without making an explicit isomorphism to $\mathcal{S}_3$:
First consider the order of each element. You deduced correctly that $d$ is the neutral element, hence $\text{ord}(d) = 1$. This also means that $\text{ord}(a) = \text{ord}(b) =2$.
Now, we know that $e^2 = f \neq d$. Because the group has order $6$, we know that $d = e^6 = f^3$, so $\text{ord}(f) = 3$. And we also know that $e$ does not have order $1$ or $2$, so it must have order $3$ or $6$.
If $e$ were to have order $6$, then our group would be cyclic. However, in that case, $e$ would have order $6$, $e^2$ order $3$, $e^3$ order $2$, $e^4$ order $3$ and $e^5$ order $6$. However, we have two distinct elements of order $2$, so this cannot be. Hence $\text{ord}(e) = 3$.
Note that we also know now that $f^{-1} = f^2 = e^4 = e$, meaning that $e$ and $f$ are inverses!
Now consider $c$, which must have order $2$ or $3$. If $c$ were to have order $3$, then its inverse $c^{-1}$ also has order $3$. So $c^{-1}$ must be either $e$ or $f$. But if $c^{-1} = e$, then $c = f$; and if $c^{-1} = f$, then $c = e$. Both lead to a contradiction, so $\text{ord}(c) = 2$.
Now, let's go through your cases.


*

*$bc$. This is the only hard part, really. Suppose first that $bc = a$, then $abc = a^2$, but since $a$ has order $2$ and $ab = e$ we then find $ec = d$. But $e$ and $c$ are not inverses, so $bc \neq a$. Next, we try $bc = f$. Then $bc = e^{-1} = (ab)^{-1} = ba$, and then $c = a$, contradiction. So $bc = e$.

*$ac$. Use that $ac = a(b^2)c = (ab)(bc)$.

*$f^{-1}$. We already determined this in the previous steps somewhere.

*$df$. Use that $d$ is the neutral element.

A: Since $G$ has order six, we know that $G$ is isomorphic to either $\mathbb Z_6$, $\mathbb Z_3 \times \mathbb Z _2$, or $S_3$. As noted in the comments above, $d$ is the trivial element, so $a$ and $b$ have order $2$. Thus, $G$ must be isomorphic to $S_3$, so we are good to go, if we can find that isomorphism, $\phi$. Assume that it maps $a$ to $(1 2)$ and $b$ to $(1 3)$. It follows that $\phi(e) = \phi(ab) = (1 2 3)$, and $\phi(f) = \phi(e^2) = (1 3 2)$. Finally, as we have excluded all other possibilities, this implies that $\phi(c) = (2 3)$, the only remaining non-trivial element. Note now that $\phi$ is actually a well-defined bijective homomorphism and thus one of the possible isomorphisms to $S_3$.
Now, for instance, $\phi(bc) = (1 2 3) = \phi(e)$, so $bc = e$.
