Show that $\tau \sum_{\sigma \in S_k} f = \sum_{\sigma \in S_k} \sigma f$

Question

Let $f$ be a $k$-linear form $f(v_1,\cdots,v_k)$. Let $S_k$ be the permutation group of permutations on set $\{1,\cdots,k\}$.

Define a new $k$-linear form as

$$S_k f := \sum_{\sigma \in S_k} \sigma f\left(v_1,\cdots,v_k\right) = \sum_{\sigma \in \S_k} f\left(v_{\sigma(1)},\cdots,v_{\sigma(k)}\right)$$

I am wondering why is it true that for any $\tau \in S_k$, we have

$$\tau \sum_{\sigma \in S_k} \sigma f\left(v_1,\cdots,v_k\right) = \sum_{\sigma\in\Sigma} \sigma f\left(v_1,\cdots,v_k\right)\,?$$

I think it all boils down to showing that for any $j \in \{1,\cdots,k\}$, we have

$$\tau\left(\sigma_1 (j)\right) + \cdots + \tau\left(\sigma_{k!} (j)\right) = \sigma_1(j) + \cdots + \sigma_{k!}(j)$$

But I am not sure how to show this.

Answer 1

Left multiplication by $\tau$ is a bijection $S_k \to S_k$, so multiplying your sum on the left by $\tau$ just reorders the summands and reordering a finite sum doesn't change its value. There's not really anything you need to show notationally, just note that $\{\tau\sigma \ | \ \sigma \in S_k\} = S_k$.