# Calculate the characters of the left and right regular representationsof an arbitrary finite group.

Calculate the characters of the left and right regular representations of an arbitrary finite group.

The answer of the question is given below:

But I do not know why the character of the left and right regular representations takes this form ..... could anyone explain for me why?

Thanks!

Edit:

A left regular representation of a finite group $$G$$ is given by sending each $$g \in G$$ to the permutation matrix of $$\sigma$$, where $$\sigma$$ is the permutation induced on the elements of $$G$$ by left multiplication by $$g$$. In other words, we label the elements of $$G$$ as $$g_1,\ldots,g_n$$, where $$n = \vert G \vert$$. Then given $$g \in G$$, we let $$\sigma_g \in S_n$$ such that $$gg_i = g_{\sigma_g(i)}$$ for each $$i = 1,\ldots,n$$. Then a left regular representation for $$G$$ is given by sending $$g$$ to the permutation matrix $$M_{\sigma_g}$$ of $$\sigma_g$$, i.e., the matrix whose $$i$$th column has $$1$$ in position $$\sigma_g(i)$$ and $$0$$ elsewhere. If $$g = e$$, then of course $$M_{\sigma_g}$$ is the identity matrix, which has trace equal to $$n$$. On the other hand, if $$g \neq e$$, then $$gg_i \neq g_i$$ for all $$i = 1\ldots,n$$. Therefore the diagonal entries of $$M_{\sigma_g}$$ are all $$0$$, hence $$M_{\sigma_g}$$ has trace $$0$$.
The right regular representation is the same except you multiply by $$g$$ on the right instead of on the left.