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The definition in my book gives it as $\lambda_x$ such that $\lambda_xg=xg$.I understand this much. But my question asks to find the left regular representation of $\mathbb{Z}_3$ in $S_3$.

I don't undertand how to do this at all. If we are working under addition, it doesn't matter whether or not we add from the left or right, correct? I am missing how to connect this to permutations. Any tips/advice would help.

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Given any finite group $G$ you can view the left regular representation as a homomorphism $G \rightarrow S_{|G|} \cong Sym(G)$, since left multiplication just permutes group elements.

Therefore, taking $1 \in \mathbb{Z}_3$ ($0$ of course acts as the identity and is rather uninteresting), we have:

\begin{align*}\lambda_1 0 = 1+0=1\\ \lambda_1 1 = 1+1=2\\ \lambda_1 2 = 1+2 =0\end{align*}

So, the action of $1$ on $Sym(G)$ is given by $(0,1,2)$. Can you see how this would generalise for $\lambda_2$/what subgroup of $S_3$ this generates? Hint: it's quite similar to what you began with...

As for the right/left distinction, in this case it doesn't matter since the group is abelian. The left regular representation corresponds to rows (or columns, I can never remember) of the Cayley table, and the right corresponds to columns (or rows). They are of course isomorphic in general, but the distinction is important, since we can get different permutations in $Sym(G)$ by taking the left/right action respectively.

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    $\begingroup$ Yes, when the group is abelian they are exactly the same, but in general they are equivalent/isomorphic (perhaps a nice exercise to show). Indeed for $2$. What subgroup of $S_3$ have you found? $\endgroup$ – ah11950 Mar 28 '14 at 0:29
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    $\begingroup$ Yep! Hence the left regular representation has image isomorphic to...? $\endgroup$ – ah11950 Mar 28 '14 at 0:42
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    $\begingroup$ No, you are correct! Any $3$-cycle, $\sigma$ in $S_n$ generates a cyclic subgroup $\langle \sigma \rangle \cong C_3 (\cong \mathbb{Z}_3)$. In more generality, for $k\leq n$, a $k$-cycle in $S_n$ generates a cyclic subgroup isomorphic to $C_k$. $\endgroup$ – ah11950 Mar 28 '14 at 0:48
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    $\begingroup$ It is isomorphic to the subgroup generated by a $3$-cycle yes. That doesn't mean the elements of $\mathbb{Z}_3$ lie inside $S_n$ though. Are you happy with the fact that we can identify the left regular representation with the permutations $\{id, (0,1,2) , (0,2,1)\}$ in $S_3$? $\endgroup$ – ah11950 Mar 28 '14 at 0:52
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    $\begingroup$ Okay, are you happy with how each $\lambda_i$ can be viewed as an element of $Sym(\mathbb{Z}_3)$ and why we can identify $Sym(\mathbb{Z}_3)$ with $S_3$? If you think about how the left regular representation is defined, it gives for each group element a bijection $G \rightarrow G$, and thus can be thought of as a permutation on the set $G$. This is where this identification arises from. It might help to read up a bit more on group actions, as this is essentially all we're dealing with here. $\endgroup$ – ah11950 Mar 28 '14 at 1:00

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