# Pohlig-Hellman Algorithm: Adding up the solution via CRT

I have a question about the Pohlig-Hellman Algorithm for the discrete log problem. I understand the concept, but doing the exact calculations I get confused at one point; to illustrate, let's look at the equation $2^x=3$ in the group $(\mathbb{Z}/29\mathbb{Z})^*$.

If my calculations are correct, we get that $x \equiv 1$ in a subgroup of order 4 and $x \equiv 5$ in a subgroup of order 7. Using the isomorphism $((\mathbb{Z}/29\mathbb{Z})^*, \cdot) \cong (\mathbb{Z}/28\mathbb{Z}, +)$, we get the equations $$x \equiv 1 \mod 4 \\ x \equiv 5 \mod 7$$ which is then pasted together using the Chinese Remainder Theorem to get $x \equiv 5 \mod 28$.

Now here is where I am confused: The results x=1 and x=5 are derived using calculations in $((\mathbb{Z}/29\mathbb{Z})^*, \cdot)$. Why can we just 'transfer' the very same numbers to the respective subgroups of $(\mathbb{Z}/28\mathbb{Z}, +)$, when we do not know the image of these elements under the isomorphism? Is it just because we are technically already in the group $(\mathbb{Z}/28\mathbb{Z}, +)$ when we are dealing with the exponents? Or am I missing something more crucial here?

Because $x$ is defined modulo $28$ ($2^{28} = 2^0$ in $(\mathbf{Z}/29\mathbf{Z})^*$), you can view $x$ as an element of $\mathbf{Z}/28\mathbf{Z}$, while $2$ and $3$ are elements of $(\mathbf{Z}/29\mathbf{Z})^*$.