Let $1\leq n\in \mathbb{N}$ and $\pi\in \text{Sym}(n)$. For $1\leq k\in \mathbb{N}$ we define $\pi^{-n}:=\left (\pi^n\right )^{-1}$.
Show for all $k,\ell\in \mathbb{Z}$ the equation $\pi^k\circ \pi^{\ell}=\pi^{k+\ell}$.
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Since $\pi^{-n}$ is defined I suppose that we have to show first the above equality for natural exponents, right?
But how exactly can we show that? Could you give me a hint?
Consider the statement $$\tag{$\Phi_\ell$} \forall k\in\Bbb Z\colon \pi^k\circ \pi ^\ell=\pi^{k+\ell}.$$ As $\pi^0=\operatorname{id}$, we know that $\Phi_0$ is true.
By the recursive definition of power, $$\tag1\pi^{k+1}=\pi^k\circ \pi=\pi\circ \pi^k$$ holds at least for integers $k\ge 0$. From this $\pi^{k}=\pi\circ \pi^{k-1}$ for $k\ge 1$; by taking inverses, $\pi^{-k}=\pi^{1-k}\circ \pi^{-1}$; by rearranging $\pi^{-k}\circ \pi=\pi^{-k+1}$. Together, this shows that $\Phi_1$ holds.
If $\Phi_{\ell_1}$ and $\Phi_{\ell_2}$ are true, then so is $\Phi_{\ell_1+\ell_2}$: $$ \pi^k\circ \pi^{\ell_1+\ell_2}\stackrel{\Phi_{\ell_2}}= \pi^k\circ (\pi^{\ell_1}\circ \pi^{\ell_2})\stackrel{ass.}=(\pi^k\circ \pi^{\ell_1})\circ \pi^{\ell_2}\stackrel{\Phi_{\ell_1}}=\pi^{k+\ell_1}\circ\pi^{\ell_2}\stackrel{\Phi_{\ell_2}}= \pi^{k+\ell_1+\ell_2}.$$
If $\Phi_\ell$ is true, then so is $\Phi_{-\ell}$: $$ \pi^k\circ \pi^{-\ell}=\pi^k\circ(\pi^\ell)^{-1}=\pi^{(k-\ell)+\ell}\circ(\pi^\ell)^{-1}=\pi^{k-\ell}\circ\pi^{\ell}\circ(\pi^\ell)^{-1}=\pi^{k-\ell}$$
From the above, we conclude that $\{\,\ell\in\Bbb Z\mid \Phi_\ell\,\}$ is closed under addition and additive inverses and contains $1$. Then it must be all of $\Bbb Z$, as desired