Cheating in a commitment scheme based on discrete log Question:

Consider the following commitment scheme:
Public parameters: large primes $q$ and $p$ such that $p = 2\cdot q + 1$, and two generators $g, g'$ of a $q$-order subgroup of $\mathbb Z_p^*$.  


Alice commits to $t$ in $\mathbb Z_q$ by uniformly picking $r$ from $\mathbb Z_q$, and sending $g^t \cdot g'^r$.  
She opens the commitment by sending $s$ and $r$.  
Suppose both parties are poly-time bounded.  
If Bob gets to pick the public parameters, can he somehow cheat? (Alice verifies the parameters in poly time)  




Attempted solution:

If he can cheat, I think Bob needs to choose $q$ and $p$ in a smart way, because even if he picks $g$ and $g'$ to be equal, he is stuck with computing discrete log, which isn't possible in poly-time.  

I thought maybe Bob can pick $q$ to be a Carmichael number, then Alice will think it is actually prime (by checking with Miller–Rabin algorithm).  

Two problem I ran into with this:  


I'm not sure how much it helps him to solve the problem — he can solve it modulo the prime factors of $q$ and use CRT, but if the factors aren't small enough its still exponential.  
Maybe there is no prime $p$ such that $p=2q+1$ for a Carmichael number $q$.




I think my solution fails because of the problems above... would appreciate a pointer in the right direction.
Assuming that Alice verifies that:


$p = 2q + 1$ with $p, q$ prime
$g, g'$ are generators of order $q$


then Bob cannot cheat. 

First note that $g$ and $g'$ generate the same subgroup, such that $g' = g^\alpha$ for some integer $\alpha$. Given a signature value $s = g^t g'^r = g^{t + \alpha r}$, there are $q$ possible messages $t' \bmod q$, and for every $t'$ there exists a unique value $r' \bmod q$ such that

$$r' \equiv \frac{t - t' + \alpha r}{\alpha} \pmod q$$

and:

$$g^tg'^r \equiv g^{t'}g'^{r'} \pmod p$$

Without further knowledge of $t$ or $r$, every pair $(t', r')$ that satisfies the signature is equally likely. So if Bob only knows $(p, q, g, g', s)$, he cannot deduce any information from the signature.

Note that if Alice knows $\alpha$, she can cheat by calculating $r'$ for any message $t'$ and revealing $(t', r')$ instead of $(t, r)$.