Show that the decryption transformation for the El Gamal cryptosystem works. Want to show that, if $P$ is the original plaintext block and $(\gamma^a)'$ is the inverse of $\gamma^a$ modulo $p$, then
$$(\gamma^a)'\delta \equiv P \pmod p$$
So, we have:


*

*$\gamma = \alpha^k \pmod p$

*$\delta = P(\alpha^a)^k mod p$

*$C = (\gamma, \delta)$


Plugging the values above to $(\gamma^a)'\delta$
$$(\gamma^a)'\delta = ((\alpha^k)^a)' P(\alpha^a)^k \equiv P \pmod p$$
$$((\alpha^k)^a)' P(\alpha^a)^k \equiv ((\alpha^k)^a)' (\alpha^a)^k  \equiv 1 \pmod p$$
Not sure if I'm doing this correctly...

Hint: We want to show that if $(\gamma^a)'$ is the inverse of $\gamma^a$ modulo $p$, then
$(\gamma^a)'\delta \equiv P \pmod p$.
We have: 


*

*$\gamma = \alpha^k \pmod p$

*$\delta = P(\alpha^a)^k \pmod p$

*$C = (\gamma, \delta)$
So, we need to find: $(\gamma^a)'\delta$
Plug in the above values and what do end up with? 
$$(\gamma^a)'\delta = ((\alpha^k)^a)' P(\alpha^a)^k \equiv \alpha^{-ak} \cdot \alpha^{ak} P \equiv P \pmod p $$
What allows me to swap and cancel those values?
